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yhonGffUZ1g1RreXDYZq2 | maths | mathematical-reasoning | logical-statement | The contrapositive of the following statement,
<br/><br/>“If the side of a square doubles, then its area increases four times”, is : | [{"identifier": "A", "content": "If the side of a square is not doubled, then its area does not increase four times."}, {"identifier": "B", "content": "If the area of a square increases four times, then its side is doubled."}, {"identifier": "C", "content": "If the area of a square increases four times, then its side is not doubled."}, {"identifier": "D", "content": "If the area of a square does not increase four times, then its side is not doubled.\n"}] | ["D"] | null | Contrapositive of p $$ \to $$ q is $$ \sim $$q $$ \to $$ $$ \sim $$p.
<br><br>Here,
<br><br>Let
<br><br> p = Side of a square is doubles.
<br><br> q = Area of square increases four times.
<br><br>$$ \therefore $$ $$ \sim $$q $$ \to $$ $$ \sim $$p = If the area of a square does not increase four times, then its side is not doubled. | mcq | jee-main-2016-online-10th-april-morning-slot |
5pVUlnZkfL3JgxJQQJldg | maths | mathematical-reasoning | logical-statement | Contrapositive of the statement
<br/><br/>‘If two numbers are not equal, then their squares are not equal’, is : | [{"identifier": "A", "content": "If the squares of two numbers are equal, then the numbers are equal."}, {"identifier": "B", "content": "If the squares of two numbers are equal, then the numbers are not equal."}, {"identifier": "C", "content": "If the squares of two numbers are not equal, then the numbers are not equal.\n"}, {"identifier": "D", "content": "If the squares of two numbers are not equal, then the numbers are equal."}] | ["A"] | null | Let,
<br><br>p : two numbers are not equal
<br><br>q : squares of two numbers are not equal
<br><br>Contrapositive of p $$ \to $$ q is $$ \sim $$q $$ \to $$ $$ \sim $$p.
<br><br>$$ \therefore $$ $$ \sim $$q $$ \to $$ $$ \sim $$p means "If the squares of two numbers are equal, then the numbers are equal". | mcq | jee-main-2017-online-9th-april-morning-slot |
v9SVtI3nIde0ejoVgeEB0 | maths | mathematical-reasoning | logical-statement | Consider the following two statements :
<br/><br/><b>Statement p :</b>
<br/>The value of sin 120<sup>o</sup> can be derived by taking $$\theta = {240^o}$$ in the equation
<br/>2sin$${\theta \over 2} = \sqrt {1 + \sin \theta } - \sqrt {1 - \sin \theta } $$
<br/><br/><b>Statement q :</b>
<br/>The angles A, B, C and D of any quadrilateral ABCD satisfy the equation
<br/>cos$$\left( {{1 \over 2}\left( {A + C} \right)} \right) + \cos \left( {{1 \over 2}\left( {B + D} \right)} \right) = 0$$
<br/><br/>Then the truth values of p and q are respectively : | [{"identifier": "A", "content": "F, T"}, {"identifier": "B", "content": "T, F"}, {"identifier": "C", "content": "T, T"}, {"identifier": "D", "content": "F, F"}] | ["A"] | null | <b>Statement p :</b>
<br>sin 120<sup>o</sup> = cos 30<sup>o</sup> = $${{\sqrt 3 } \over 2}$$ $$ \Rightarrow $$ 2 sin 120<sup>o</sup> = $$\sqrt 3 $$
<br><br>So, $$\sqrt {1 + \sin {{240}^o}} - \sqrt {1 - \sin {{240}^o}} $$
<br><br> $$ = \sqrt {{{1 - \sqrt 3 } \over 2}} - \sqrt {{{1 + \sqrt 3 } \over 2}} \ne \sqrt 3 $$
<br><br><b>Statement q :</b>
<br>So, A + B + C + D = 2$$\pi $$
<br><br>$$ \Rightarrow $$ $${{A + C} \over 2} + {{B + D} \over 2} = \pi $$
<br><br>$$ \Rightarrow $$ cos$$\left( {{{A + C} \over 2}} \right) + \cos \left( {{{B + D} \over 2}} \right)$$
<br><br>= cos $$\left( {{{A + C} \over 2}} \right)$$ $$-$$ cos$$\left( {{{A + C} \over 2}} \right) = 0$$
<br><br>Therefore, statement p is false and statement q is true. | mcq | jee-main-2018-online-15th-april-evening-slot |
WGPsr0o7H2TKzUFjIMsK9 | maths | mathematical-reasoning | logical-statement | Consider the statement : "P(n) : n<sup>2</sup> – n + 41 is prime". Then which one of the following is true ? | [{"identifier": "A", "content": "P(5) is false but P(3) is true"}, {"identifier": "B", "content": "Both P(3) and P(5) are true"}, {"identifier": "C", "content": "P(3) is false but P(5) is true"}, {"identifier": "D", "content": "Both P(3) and P(5) are false"}] | ["B"] | null | P(n) : n<sup>2</sup> $$-$$ n + 41 is prime
<br><br>P(5) = 61 which is prime
<br><br>P(3) = 47 which is also prime | mcq | jee-main-2019-online-10th-january-morning-slot |
XGWuG9V4dktxwFEkSEcmf | maths | mathematical-reasoning | logical-statement | Consider the following three statements :
<br/><br/>P : 5 is a prime number
<br/><br/>Q : 7 is a factor of 192
<br/><br/>R : L.C.M. of 5 and 7 is 35
<br/><br/>Then the truth value of which one of the following statements is true ? | [{"identifier": "A", "content": "(P $$ \\wedge $$ Q) $$ \\vee $$ ($$ \\sim $$ R)"}, {"identifier": "B", "content": "P $$ \\vee $$ ($$ \\sim $$ Q $$ \\wedge $$ R)"}, {"identifier": "C", "content": "(~ P) $$ \\wedge $$ ($$ \\sim $$ Q $$ \\wedge $$ R)"}, {"identifier": "D", "content": "($$ \\sim $$ P) $$ \\vee $$ (Q $$ \\wedge $$ R)"}] | ["B"] | null | It is obvious | mcq | jee-main-2019-online-10th-january-evening-slot |
oBwMmAnQrawvLlbPhGr5h | maths | mathematical-reasoning | logical-statement | Contrapositive of the statement " If two numbers are not equal, then their squares are not equal." is : | [{"identifier": "A", "content": "If the squares of two numbers are equal, then the numbers are not equal"}, {"identifier": "B", "content": "If the squares of two numbers are equal, then the numbers are equal "}, {"identifier": "C", "content": "If the squares of two numbers are not equal, then the numbers are equal"}, {"identifier": "D", "content": "If the squares of two numbers are not equal, then the numbers are not equal"}] | ["B"] | null | Let,
<br><br>p : two numbers are not equal
<br><br>q : squares of two numbers are not equal
<br><br>Contrapositive of p $$ \to $$ q is $$ \sim $$q $$ \to $$ $$ \sim $$p.
<br><br>$$ \therefore $$ $$ \sim $$q $$ \to $$ $$ \sim $$p means "If the squares of two numbers are equal, then the numbers are equal". | mcq | jee-main-2019-online-11th-january-evening-slot |
lCnssXZcN2ubzefec6xFx | maths | mathematical-reasoning | logical-statement | The contrapositive of the statement "If you are
born in India, then you are a citizen of India", is :
| [{"identifier": "A", "content": "If you are not a citizen of India, then you are\nnot born in India."}, {"identifier": "B", "content": "If you are born in India, then you are not a\ncitizen of India."}, {"identifier": "C", "content": "If you are a citizen of India, then you are born\nin India."}, {"identifier": "D", "content": "If you are not born in India, then you are not\na citizen of India"}] | ["A"] | null | Let p = you are born in India.
<br><br>q = you are a citizen of India.
<br><br>$$ \therefore $$ $$ \sim $$ p = you are not born in India.
<br><br>$$ \sim $$ q = you are not a citizen of India.
<br><br>We know Contrapositive of p $$ \to $$ q is ~q $$ \to $$ ~p
<br><br>So contrapositive of statement will be :
<br><br>“If you are not a citizen of India, then you are
not born in India.”
| mcq | jee-main-2019-online-8th-april-morning-slot |
dA7B1752ytqOXgSd827k9k2k5irgtwi | maths | mathematical-reasoning | logical-statement | Negation of the statement :
<br/><br/>$$\sqrt 5 $$ is an integer or 5 is an irrational is : | [{"identifier": "A", "content": "$$\\sqrt 5 $$ is not an integer and 5 is not irrational."}, {"identifier": "B", "content": "$$\\sqrt 5 $$ is irrational or 5 is an integer."}, {"identifier": "C", "content": "$$\\sqrt 5 $$ is an integer and 5 is irrational."}, {"identifier": "D", "content": "$$\\sqrt 5 $$ is not an integer or 5 is not irrational."}] | ["A"] | null | p = $$\sqrt 5 $$ is an integer.
<br><br>q : 5 is irrational
<br><br>$$ \sim $$$$\left( {p \vee q} \right)$$ $$ \equiv $$ $$ \sim $$p $$ \wedge $$ $$ \sim $$q
<br><br>= $$\sqrt 5 $$ is not an integer and 5 is not irrational. | mcq | jee-main-2020-online-9th-january-morning-slot |
hLzGTd37OiUHlAyvkcjgy2xukg38l1kq | maths | mathematical-reasoning | logical-statement | Consider the statement : <br/>‘‘For an integer n, if n<sup>3</sup> – 1 is even, then n is odd.’’<br/> The contrapositive statement of this statement is :
| [{"identifier": "A", "content": "For an integer n, if n is even, then n<sup>3</sup> \u2013 1 is even."}, {"identifier": "B", "content": "For an integer n, if n<sup>3</sup> \u2013 1 is not even, then n is not odd."}, {"identifier": "C", "content": "For an integer n, if n is odd, then n<sup>3</sup> \u2013 1 is even."}, {"identifier": "D", "content": "For an integer n, if n is even, then n<sup>3</sup> \u2013 1 is odd."}] | ["D"] | null | Let,
p : n<sup>3</sup>–1 is even,
<br>q : n is odd
<br><br>Contrapositive of p $$ \to $$ q = $$ \sim $$q $$ \to $$ $$ \sim $$p
<br><br>$$ \Rightarrow $$ If n is not odd then n<sup>3</sup> – 1 is not even.
<br><br>$$ \Rightarrow $$ For an integer n, if n is even, then n<sup>3</sup> – 1 is odd. | mcq | jee-main-2020-online-6th-september-evening-slot |
mQhzgCsAFvfKM3rNxDjgy2xukfah5akq | maths | mathematical-reasoning | logical-statement | Contrapositive of the statement :<br/>
‘If a function f is differentiable at a, then it is also continuous at a’, is: | [{"identifier": "A", "content": "If a function f is continuous at a, then it is not differentiable at a."}, {"identifier": "B", "content": "If a function f is not continuous at a, then it is differentiable at a."}, {"identifier": "C", "content": "If a function f is not continuous at a, then it is not differentiable at a.\n"}, {"identifier": "D", "content": "If a function f is continuous at a, then it is differentiable at a."}] | ["C"] | null | p = function is differentiable at a
<br><br>q = function is continuous at a
<br><br>Contrapositive of statements p $$ \to $$ q is
<br><br>$$ \sim $$q $$ \to $$ $$ \sim $$p
<br><br>$$ \therefore $$ Contrapositive statement is :
<br><br> If a function f is not continuous at a, then it is not differentiable at a.
| mcq | jee-main-2020-online-4th-september-evening-slot |
DnY3UxFkGprb04C8H1jgy2xukewqg1g8 | maths | mathematical-reasoning | logical-statement | The contrapositive of the statement <br/>"If I reach the
station in time, then I will catch the train" is : | [{"identifier": "A", "content": "If I will catch the train, then I reach the station\nin time."}, {"identifier": "B", "content": "If I do not reach the station in time, then I will\nnot catch the train."}, {"identifier": "C", "content": "If I will not catch the train, then I do not reach\nthe station in time."}, {"identifier": "D", "content": "If I do not reach the station in time, then I will\ncatch the train."}] | ["C"] | null | Let p denotes statement
<br><br>p : I reach the station in time.
<br><br>q : I will catch the train.
<br><br>Contrapositive of p $$ \to $$ q
is $$ \sim $$q $$ \to $$ $$ \sim $$p
<br><br>$$ \sim $$q $$ \to $$ $$ \sim $$p : If I will not catch the train, then I do not reach the station in time. | mcq | jee-main-2020-online-2nd-september-morning-slot |
LWR517jHskS2az6BJG1klt77hyt | maths | mathematical-reasoning | logical-statement | The contrapositive of the statement "If you will work, you will earn money" is : | [{"identifier": "A", "content": "If you will not earn money, you will not work"}, {"identifier": "B", "content": "If you will earn money, you will work"}, {"identifier": "C", "content": "You will earn money, if you will not work"}, {"identifier": "D", "content": "To earn money, you need to work"}] | ["A"] | null | Contrapositive of p $$ \to $$ q is ~q $$ \to $$ ~p
<br><br>p : you will work
<br><br>q : you will earn money
<br><br>~q : you will not earn money
<br><br>~p : you will not work
<br><br>$$ \therefore $$ ~q $$ \to $$ ~p : If you will not earn money, you will not work | mcq | jee-main-2021-online-25th-february-evening-slot |
1krrp2k1g | maths | mathematical-reasoning | logical-statement | Consider the following three statements :<br/><br/>(A) If 3 + 3 = 7 then 4 + 3 = 8<br/><br/>(B) If 5 + 3 = 8 then earth is flat.<br/><br/>(C) If both (A) and (B) are true then 5 + 6 = 17.<br/><br/>Then, which of the following statements is correct? | [{"identifier": "A", "content": "(A) is false, but (B) and (C) re true"}, {"identifier": "B", "content": "(A) and (C) are true while (B) is false"}, {"identifier": "C", "content": "(A) is true while (B) and (C) are false"}, {"identifier": "D", "content": "(A) and (B) are false while (C) is true"}] | ["B"] | null | Truth Table <br><br><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-baqh">P</th>
<th class="tg-baqh">q</th>
<th class="tg-baqh">P$$ \to $$q</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">T</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
<td class="tg-baqh">T</td>
</tr>
<tr>
<td class="tg-baqh">F</td>
<td class="tg-baqh">F</td>
<td class="tg-baqh">T</td>
</tr>
</tbody>
</table> | mcq | jee-main-2021-online-20th-july-evening-shift |
1kryepvv0 | maths | mathematical-reasoning | logical-statement | Which of the following is the negation of the statement "for all M > 0, there exists x$$\in$$S such that x $$\ge$$ M" ? | [{"identifier": "A", "content": "there exists M > 0, such that x < M for all x$$\\in$$S"}, {"identifier": "B", "content": "there exists M > 0, there exists x$$\\in$$S such that x $$\\ge$$ M"}, {"identifier": "C", "content": "there exists M > 0, there exists x$$\\in$$S such that x < M"}, {"identifier": "D", "content": "there exists M > 0, such that x $$\\ge$$ M for all x$$\\in$$S"}] | ["A"] | null | P : for all M > 0, there exists x$$\in$$S such that x $$\ge$$ M.<br><br>$$ \sim $$ P : there exists M > 0, for all x$$\in$$S<br><br>Such that x < M<br><br>Negation of 'there exists' is 'for all'. | mcq | jee-main-2021-online-27th-july-evening-shift |
1krzlvswd | maths | mathematical-reasoning | logical-statement | Consider the statement "The match will be played only if the weather is good and ground is not wet". Select the correct negation from the following : | [{"identifier": "A", "content": "The match will not be played and weather is not good and ground is wet."}, {"identifier": "B", "content": "If the match will not be played, then either weather is not good or ground is wet."}, {"identifier": "C", "content": "The match will be played and weather is not good or ground is wet."}, {"identifier": "D", "content": "The match will not be played or weather is good and ground is not wet."}] | ["C"] | null | p : weather is good<br><br>q : ground is not wet<br><br>$$\sim$$ (p $$ \wedge $$ q) $$ \equiv $$ $$\sim$$ p $$ \vee $$ $$\sim$$ q<br><br>$$\equiv$$ weather is not good or ground is wet | mcq | jee-main-2021-online-25th-july-evening-shift |
1l5baqdkw | maths | mathematical-reasoning | logical-statement | <p>Consider the following statements:</p>
<p>A : Rishi is a judge.</p>
<p>B : Rishi is honest.</p>
<p>C : Rishi is not arrogant.</p>
<p>The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is</p> | [{"identifier": "A", "content": "B $$\\to$$ (A $$\\vee$$ C)"}, {"identifier": "B", "content": "($$\\sim$$B) $$\\wedge$$ (A $$\\wedge$$ C)"}, {"identifier": "C", "content": "B $$\\to$$ (($$\\sim$$A) $$\\vee$$ ($$\\sim$$C))"}, {"identifier": "D", "content": "B $$\\to$$ (A $$\\wedge$$ C)"}] | ["B"] | null | <p>$$\because$$ Given statement is</p>
<p>(A $$\wedge$$ C) $$\to$$ B</p>
<p>Then its negation is</p>
<p>$$\sim$$ {(A $$\wedge$$ C) $$\to$$ B}</p>
<p>or $$\sim$$ {$$\sim$$ (A $$\wedge$$ C) $$\vee$$ B}</p>
<p>$$\therefore$$ (A $$\wedge$$ C) $$\wedge$$ ($$\sim$$ B)</p>
<p>or ($$\sim$$ B) $$\wedge$$ (A $$\wedge$$ C)</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l6f32pyv | maths | mathematical-reasoning | logical-statement | <p>Consider the following statements:</p>
<p>P : Ramu is intelligent.</p>
<p>Q : Ramu is rich.</p>
<p>R : Ramu is not honest.</p>
<p>The negation of the statement "Ramu is intelligent and honest if and only if Ramu is not rich" can be expressed as:</p> | [{"identifier": "A", "content": "$$((P \\wedge(\\sim R)) \\wedge Q) \\wedge((\\sim Q) \\wedge((\\sim P) \\vee R))$$"}, {"identifier": "B", "content": "$$((P \\wedge R) \\wedge Q) \\vee((\\sim Q) \\wedge((\\sim P) \\vee(\\sim R)))$$"}, {"identifier": "C", "content": "$$((P \\wedge R) \\wedge Q) \\wedge((\\sim Q) \\wedge((\\sim P) \\vee(\\sim R)))$$"}, {"identifier": "D", "content": "$$((P \\wedge(\\sim R)) \\wedge Q) \\vee((\\sim Q) \\wedge((\\sim P) \\vee R))$$"}] | ["D"] | null | <p>P : Ramu is intelligent</p>
<p>Q : Ramu is rich</p>
<p>R : Ramu is not honest</p>
<p>Given statement, "Ramu is intelligent and honest if and only if Ramu is not rich"</p>
<p>$$ = (P \wedge \sim R) \Leftrightarrow \, \sim Q$$</p>
<p>So, negation of the statement is</p>
<p>$$ \sim [(P \wedge \sim R) \Leftrightarrow \, \sim Q]$$</p>
<p>$$ = \sim [\{ \sim (P \wedge \sim R) \vee \, \sim Q\} \wedge \{ Q \vee (P \wedge \sim R)\} ]$$</p>
<p>$$ = ((P \wedge \sim R) \wedge Q) \vee ( \sim Q \wedge ( \sim P \vee R))$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6notty1 | maths | mathematical-reasoning | logical-statement | <p>Let</p>
<p>$$\mathrm{p}$$ : Ramesh listens to music.</p>
<p>$$\mathrm{q}$$ : Ramesh is out of his village.</p>
<p>$$\mathrm{r}$$ : It is Sunday.</p>
<p>$$\mathrm{s}$$ : It is Saturday.</p>
<p>Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as</p> | [{"identifier": "A", "content": "$$((\\sim q) \\wedge(r \\vee s)) \\Rightarrow p$$"}, {"identifier": "B", "content": "$$(\\mathrm{q} \\wedge(\\mathrm{r} \\vee \\mathrm{s})) \\Rightarrow \\mathrm{p}$$"}, {"identifier": "C", "content": "$$p \\Rightarrow(q \\wedge(r \\vee s))$$"}, {"identifier": "D", "content": "$$\\mathrm{p} \\Rightarrow((\\sim \\mathrm{q}) \\wedge(\\mathrm{r} \\vee \\mathrm{s}))$$"}] | ["D"] | null | <p>p : Ramesh listens to music</p>
<p>q : Ramesh is out of his village</p>
<p>r : It is Sunday</p>
<p>s : It is Saturday</p>
<p>p $$\to$$ q conveys the same p only if q</p>
<p>Statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday"</p>
<p>$$p \Rightarrow (( \sim \,q)\, \wedge \,(r\, \vee \,s))$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
ldqvxf4d | maths | mathematical-reasoning | logical-statement | <p>Consider the following statements:</p>
<p>P : I have fever</p>
<p>Q: I will not take medicine</p>
<p>$\mathrm{R}$ : I will take rest.</p>
<p>The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to :</p> | [{"identifier": "A", "content": "$((\\sim P) \\vee \\sim Q) \\wedge((\\sim P) \\vee \\sim R)$"}, {"identifier": "B", "content": "$(P \\vee \\sim Q) \\wedge(P \\vee \\sim R)$"}, {"identifier": "C", "content": "$((\\sim P) \\vee \\sim Q) \\wedge((\\sim P) \\vee R)$"}, {"identifier": "D", "content": "$(P \\vee Q) \\wedge((\\sim P) \\vee R)$"}] | ["C"] | null | <p>The given expression is</p>
<p>$$P \to \sim Q \wedge R$$</p>
<p>$$ \equiv ( \sim P) \vee ( \sim Q \wedge R)$$</p>
<p>$$ \equiv ( \sim P \vee \sim Q) \wedge ( \sim P \vee R)$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1lguwsge5 | maths | mathematical-reasoning | logical-statement | <p>The number of ordered triplets of the truth values of $$p, q$$ and $$r$$ such that the truth value of the statement $$(p \vee q) \wedge(p \vee r) \Rightarrow(q \vee r)$$ is True, is equal to ___________.</p> | [] | null | 7 | $$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{r} & \boldsymbol{p} \vee \boldsymbol{q} & \boldsymbol{p} \vee \boldsymbol{r} & \begin{array}{c}
(\boldsymbol{p} \vee \boldsymbol{q}) \wedge \\
(\boldsymbol{p} \vee \boldsymbol{r})
\end{array} & \boldsymbol{q} \vee \boldsymbol{r} & \begin{array}{c}
(\boldsymbol{p} \vee \boldsymbol{q}) \wedge(\boldsymbol{p} \vee \boldsymbol{r}) \\
\Rightarrow(\boldsymbol{q} \vee \boldsymbol{r})
\end{array} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}
$$
<br/><br/>Hence, the total number of ordered triplets are 7. | integer | jee-main-2023-online-11th-april-morning-shift |
QFyqBI1sBBGMrWsu | maths | matrices-and-determinants | adjoint-of-a-matrix | If $$A = \left[ {\matrix{
{5a} & { - b} \cr
3 & 2 \cr
} } \right]$$ and $$A$$ adj $$A=A$$ $${A^T},$$ then $$5a+b$$ is equal to : | [{"identifier": "A", "content": "$$4$$ "}, {"identifier": "B", "content": "$$13$$"}, {"identifier": "C", "content": "$$-1$$ "}, {"identifier": "D", "content": "$$5$$"}] | ["D"] | null | $$A\left( {Adj\,\,A} \right) = A\,{A^T}$$
<br><br>$$ \Rightarrow {A^{ - 1}}A\left( {adj\,\,A} \right) = {A^{ - 1}}A\,{A^T}$$
<br><br>$$Adj\,\,A = {A^T}$$
<br><br>$$ \Rightarrow \left[ {\matrix{
2 & b \cr
{ - 3} & {5a} \cr
} } \right] = \left[ {\matrix{
{5a} & 3 \cr
{ - b} & 2 \cr
} } \right]$$
<br><br>$$ \Rightarrow a = {2 \over 5}\,\,$$ and $$\,\,b = 3$$
<br><br>$$ \Rightarrow 5a + b = 5$$ | mcq | jee-main-2016-offline |
2bwNHWvvDdm8rsiP | maths | matrices-and-determinants | adjoint-of-a-matrix | If $$A = \left[ {\matrix{
2 & { - 3} \cr
{ - 4} & 1 \cr
} } \right]$$,
<br/><br/>then adj(3A<sup>2</sup> + 12A) is equal to | [{"identifier": "A", "content": "$$\\left[ {\\matrix{\n {51} & {63} \\cr \n {84} & {72} \\cr \n\n } } \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {\\matrix{\n {51} & {84} \\cr \n {63} & {72} \\cr \n\n } } \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {\\matrix{\n {72} & {-63} \\cr \n {-84} & {51} \\cr \n\n } } \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {\\matrix{\n {72} & {-84} \\cr \n {-63} & {51} \\cr \n\n } } \\right]$$"}] | ["A"] | null | We have, $$A = \left[ {\matrix{
2 & { - 3} \cr
{ - 4} & 1 \cr
} } \right]$$
<br><br>$$ \therefore $$ A<sup>2</sup> = A.A = $$\left[ {\matrix{
2 & { - 3} \cr
{ - 4} & 1 \cr
} } \right]\left[ {\matrix{
2 & { - 3} \cr
{ - 4} & 1 \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
{4 + 12} & { - 6 - 3} \cr
{ - 8 - 4} & {12 + 1} \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
{16} & { - 9} \cr
{ - 12} & {13} \cr
} } \right]$$
<br><br>Now, 3A<sup>2</sup> + 12A
<br><br>= $$3\left[ {\matrix{
{16} & { - 9} \cr
{ - 12} & {13} \cr
} } \right] + 12\left[ {\matrix{
2 & { - 3} \cr
{ - 4} & 1 \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
{48} & { - 27} \cr
{ - 36} & {39} \cr
} } \right] + \left[ {\matrix{
{24} & { - 36} \cr
{ - 48} & {12} \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
{72} & { - 63} \cr
{ - 84} & {51} \cr
} } \right]$$
<br><br>$$ \therefore $$ adj(3A<sup>2</sup> + 12A) = $$\left[ {\matrix{
{51} & {63} \cr
{84} & {72} \cr
} } \right]$$ | mcq | jee-main-2017-offline |
8yFgTFWWqZrfjWfpecKpP | maths | matrices-and-determinants | adjoint-of-a-matrix | Let A be any 3 $$ \times $$ 3 invertible matrix. Then which one of the following is <b>not</b> always true ? | [{"identifier": "A", "content": "adj (A) = $$\\left| \\right.$$A$$\\left| \\right.$$.A<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "adj (adj(A)) = $$\\left| \\right.$$A$$\\left| \\right.$$.A"}, {"identifier": "C", "content": "adj (adj(A)) = $$\\left| \\right.$$A$$\\left| \\right.$$<sup>2</sup>.(adj(A))<sup>$$-$$1</sup> "}, {"identifier": "D", "content": "adj (adj(A)) = $$\\left| \\, \\right.$$A $$\\left| \\, \\right.$$.(adj(A))<sup>$$-$$1</sup>"}] | ["D"] | null | We know, the formula
<br><br>A<sup>-1</sup> = $${{adj\left( A \right)} \over {\left| A \right|}}$$
<br><br>$$ \therefore $$ adj (A) = $$\left| \right.$$A$$\left| \right.$$.A<sup>$$-$$1</sup>
<br><br><b>So, Option (A) is true.</b>
<br><br>We know, the formula
<br><br>adj (adj (A)) = $${\left| A \right|^{n - 2}}.A$$
<br><br>Now if we put n = 3 as given that A is a 3 $$ \times $$ 3 matrix, we get
<br><br>adj (adj (A)) = $${\left| A \right|^{3 - 2}}.A$$ = $$\left| A \right|.A$$
<br><br><b>So, Option (B) is also true.</b>
<br><br>We know, the formula
<br><br>adj (adj (A)) = $${\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$
<br><br>Now if we put n = 3 as given that A is a 3 $$ \times $$ 3 matrix, we get
<br><br>adj (adj (A)) = $${\left| A \right|^{3 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$ = $${\left| A \right|^{2}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$
<br><br><b>So, Option (C) is also true.</b>
<br><br>Now in this formula
<br><br>adj (adj (A)) = $${\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$
<br><br>if we put n = 2, we get
<br><br>adj (adj (A)) = $${\left| A \right|^{2 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}$$ = $${\left| A \right|}{\left( {adj\left( A \right)} \right)^{ - 1}}$$
<br><br>But as A is a 3 $$ \times $$ 3 matrix so we can not take n = 2, so we can say for a 3 $$ \times $$ 3 matrix option (D) is not true.
<br><br><b>So, Option (D) is false.</b>
| mcq | jee-main-2017-online-8th-april-morning-slot |
1lgpy4kyx | maths | matrices-and-determinants | adjoint-of-a-matrix | <p>Let $$B=\left[\begin{array}{lll}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right], \alpha > 2$$ be the adjoint of a matrix $$A$$ and $$|A|=2$$. Then
$$\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c}\alpha \\ -2 \alpha \\ \alpha\end{array}\right]$$ is equal to :</p> | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "$$-$$16"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "16"}] | ["B"] | null | $$
B=\left[\begin{array}{lll}
1 & 3 & \alpha \\
1 & 2 & 3 \\
\alpha & \alpha & 4
\end{array}\right], \alpha>2
$$
<br/><br/>And $\operatorname{adj}(A)=B,|A|=2$
<br/><br/>$$
\begin{aligned}
& \Rightarrow|\operatorname{adj}(A)|=|B| \\\\
& \Rightarrow 2^2=(8-3 \alpha)-3(4-3 \alpha)+\alpha(-\alpha) \\\\
& \Rightarrow \alpha^2-6 \alpha+8=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\Rightarrow & (\alpha-4)(\alpha-2)=0 \\\\
& \alpha=4,2 \text { but } \alpha>2 \text { so } \alpha=4
\end{aligned}
$$
<br/><br/>Now
<br/><br/>$$
\begin{aligned}
& {\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c}
\alpha \\
-2 \alpha \\
\alpha
\end{array}\right]=\left[\begin{array}{lll}
4-8 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 3 & 4 \\
1 & 2 & 3 \\
4 & 4 & 4
\end{array}\right]\left[\begin{array}{c}
4 \\
-8 \\
4
\end{array}\right]} \\\\
& =\left[\begin{array}{lll}
12 & 12 & 8
\end{array}\right]\left[\begin{array}{c}
4 \\
-8 \\
4
\end{array}\right] \\\\
& = {48-96+32=-16}
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgzxiiqh | maths | matrices-and-determinants | adjoint-of-a-matrix | <p>Let $$A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]$$. If $$|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))|=(16)^{n}$$, then $$n$$ is equal to :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}] | ["C"] | null | We have,
<br/><br/>$$
\begin{aligned}
& |\mathrm{A}|=\left|\begin{array}{ccc}
2 & 1 & 0 \\
1 & 2 & -1 \\
0 & -1 & 2
\end{array}\right|=2(4-1)-1(2-0)+0 \\\\
& =6-2=4 \\\\
& \text { So, }|2 \mathrm{~A}|=2^3|\mathrm{~A}|=8 \times 4=32 \\\\
& \text { Now, }|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A}))|=|2 \mathrm{~A}|^{(n-1)^3} \\\\
& =(32)^{2^3}=32^8 \\\\
& \Rightarrow 16^n=(32)^8=2^8 \times 16^8 \\\\
& \Rightarrow 16^n=16^{2+8} \Rightarrow n=10
\end{aligned}
$$
<br/><br/><b>Concepts :</b>
<br/><br/>(a) $|k \mathrm{~A}|=k^n|\mathrm{~A}|$
<br/><br/>(b) $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1}$ | mcq | jee-main-2023-online-8th-april-morning-shift |
jaoe38c1lsd58dpi | maths | matrices-and-determinants | adjoint-of-a-matrix | <p>Let A be a $$3 \times 3$$ matrix and $$\operatorname{det}(A)=2$$. If $$n=\operatorname{det}(\underbrace{\operatorname{adj}(\operatorname{adj}(\ldots . .(\operatorname{adj} A))}_{2024-\text { times }}))$$, then the remainder when $$n$$ is divided by 9 is equal to __________.</p> | [] | null | 7 | <p>$$\begin{aligned}
& |\mathrm{A}|=2 \\
& \underbrace{\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \ldots . .(\mathrm{a})))}_{2024 \text { times }}=|\mathrm{A}|^{(\mathrm{n}-1)^{2024}} \\
& =|\mathrm{A}|^{2024} \\
& =2^{2^{2024}}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& 2^{2024}=\left(2^2\right) 2^{2022}=4(8)^{674}=4(9-1)^{674} \\
& \Rightarrow 2^{2024} \equiv 4(\bmod 9) \\
& \Rightarrow 2^{2024} \equiv 9 \mathrm{~m}+4, \mathrm{~m} \leftarrow \text { even } \\
& 2^{9 \mathrm{~m}+4} \equiv 16 \cdot\left(2^3\right)^{3 \mathrm{~m}} \equiv 16(\bmod 9) \\
& \quad \equiv 7
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
luy9cloc | maths | matrices-and-determinants | adjoint-of-a-matrix | <p>Let $$A$$ be a non-singular matrix of order 3. If $$\operatorname{det}(3 \operatorname{adj}(2 \operatorname{adj}((\operatorname{det} A) A)))=3^{-13} \cdot 2^{-10}$$ and $$\operatorname{det}(3\operatorname{adj}(2 \mathrm{A}))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}$$, then $$|3 \mathrm{~m}+2 \mathrm{n}|$$ is equal to _________.</p> | [] | null | 14 | <p>$$|\operatorname{adj}(2 \operatorname{adj}(|A| A))|=3^{-13} \cdot 2^{-10}$$</p>
<p>Let $$|A| A=B \Rightarrow|B|=\| A|A|=|A|^3|A|=|A|^4$$</p>
<p>$$\begin{aligned}
\Rightarrow \quad & \operatorname{adj}(|A| A)=(\operatorname{adj} B) \\
\Rightarrow \quad & 2 \operatorname{adj}(|A| A)=(2 \operatorname{adj} B)=C \text { (say) } \\
& |\operatorname{3adj}(C)|=3^3 \cdot|C|^2
\end{aligned}$$</p>
<p>$$\begin{aligned}
& |C|=|(2 \operatorname{adj} B)|=2^3|B|^2=2^3 \cdot\left|A^4\right|^2=2^3 \cdot|A|^8 \\
& \Rightarrow|\operatorname{3adj} C|=3^3 \cdot\left(2^3|A|^8\right)^2=3^{-13} \cdot 2^{-10} \\
& \quad=2^6|A|^{16}=3^{-16} \cdot 2^{-10} \\
& \Rightarrow|A|^{16}=(3 \cdot 2)^{-16}=\left(\frac{1}{6}\right)^{16} \\
& \Rightarrow|A|= \pm \frac{1}{6}
\end{aligned}$$</p>
<p>$$\begin{array}{r}
\mid \text { 3adj }\left.2 A\left|=3^3\right| 2 A\right|^2=3^3 \cdot\left(2^3|A|\right)^2=3^3 \cdot 2^6|A|^2 \\
=3^3 \cdot 2^6 \cdot \frac{1}{36}=\frac{27 \times 64}{36}=48
\end{array}$$</p>
<p>$$
\begin{aligned}
& \Rightarrow 2^m \cdot 3^n=2^4 \cdot 3^1 \Rightarrow m=4 \\
& \qquad n=1 \\
& \Rightarrow|3 \times 4+2 \times 1|=14 \\
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-morning-shift |
lv0vxbzn | maths | matrices-and-determinants | adjoint-of-a-matrix | <p>Let $$\alpha \in(0, \infty)$$ and $$A=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]$$. If $$\operatorname{det}\left(\operatorname{adj}\left(2 A-A^T\right) \cdot \operatorname{adj}\left(A-2 A^T\right)\right)=2^8$$, then $$(\operatorname{det}(A))^2$$ is equal to:</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "49"}, {"identifier": "D", "content": "1"}] | ["A"] | null | <p>$$\begin{aligned}
& \left|\operatorname{adj}\left(A-2 A^T\right) \cdot \operatorname{adj}\left(2 A-A^T\right)\right|=2^8 \\
& P=A-2 A^{\top} \\
& Q=2 A^T-A \Rightarrow Q^T=2 A^T-A=-P \\
& |\operatorname{adj}(P) \operatorname{adj}(Q)| \Rightarrow|P Q|=-2^4 \\
& \Rightarrow|P|(-|P|)=-2^4 \Rightarrow|P|=4 \text { and }|Q|=-4 \\
& \left|A-2 A^T\right|=4 \\
& A-2 A^T=\left[\begin{array}{lll}
1 & 2 & \alpha \\
1 & 0 & 1 \\
0 & 1 & 2
\end{array}\right]-2\left[\begin{array}{lll}
1 & 1 & 0 \\
2 & 0 & 1 \\
\alpha & 1 & 2
\end{array}\right]=\left[\begin{array}{ccc}
-1 & 0 & \alpha \\
-3 & 0 & -1 \\
-2 \alpha & -1 & -2
\end{array}\right] \\
& \Rightarrow\left|A-2 A^T\right|=1+3 \alpha=4 \Rightarrow \alpha=1 \Rightarrow|A|=-4 \\
& \Rightarrow|A|^2=16
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv2eryn4 | maths | matrices-and-determinants | adjoint-of-a-matrix | <p>Let $$A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$$ and $$B=I+\operatorname{adj}(A)+(\operatorname{adj} A)^2+\ldots+(\operatorname{adj} A)^{10}$$.
Then, the sum of all the elements of the matrix $$B$$ is:</p> | [{"identifier": "A", "content": "$$-$$110"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "$$-$$124"}, {"identifier": "D", "content": "$$-$$88"}] | ["D"] | null | <p>$$\begin{aligned}
& \operatorname{adj}(A)=\left[\begin{array}{ll}
1 & -2 \\
0 & 1
\end{array}\right] \\
& (\operatorname{adj} A)^2=\left[\begin{array}{ll}
1 & -4 \\
0 & 1
\end{array}\right] \\
& (\operatorname{adj} A)^3=\left[\begin{array}{cc}
1 & -6 \\
0 & 1
\end{array}\right] \\
& (\operatorname{adj} A)^4=\left[\begin{array}{cc}
1 & -8 \\
0 & 1
\end{array}\right] \\
& (\operatorname{adj} A)^r=\left[\begin{array}{cc}
1 & (-2 r) \\
0 & 1
\end{array}\right]
\end{aligned}$$</p>
<p>$$\begin{aligned}
& B=\sum_{r=0}^{10}(\operatorname{adj} A)^r=\left[\begin{array}{ll}
\sum_\limits{r=0}^{10} 1 & \sum_\limits{r=0}^{10}(-2 r) \\
\sum_\limits{r=0}^{10}(0) & \sum_\limits{r=0}^{10}(1)
\end{array}\right] \\
& B=\left[\begin{array}{cc}
11 & -110 \\
0 & 11
\end{array}\right]
\end{aligned}$$</p>
<p>$$\text { Sum of elements }=-110+11+11=-88$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv9s20az | maths | matrices-and-determinants | adjoint-of-a-matrix | <p>Let $$\alpha \beta \neq 0$$ and $$A=\left[\begin{array}{rrr}\beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha\end{array}\right]$$. If $$B=\left[\begin{array}{rrr}3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta\end{array}\right]$$ is the matrix of cofactors of the elements of $$A$$, then $$\operatorname{det}(A B)$$ is equal to :</p> | [{"identifier": "A", "content": "64"}, {"identifier": "B", "content": "343"}, {"identifier": "C", "content": "125"}, {"identifier": "D", "content": "216"}] | ["D"] | null | <p>$$A=\left[\begin{array}{ccc}
\beta & \alpha & 3 \\
\alpha & \alpha & \beta \\
-\beta & \alpha & 2 \alpha
\end{array}\right], B=\left[\begin{array}{ccc}
3 \alpha & -9 & 3 \alpha \\
-\alpha & 7 & -2 \alpha \\
-2 \alpha & 5 & -2 \beta
\end{array}\right]$$</p>
<p>Cofactor of $$A$$-matrix is</p>
<p>$$=\left[\begin{array}{ccc}
2 \alpha^2-\alpha \beta & -\left(2 \alpha^2+\beta^2\right) & \alpha^2+\alpha \beta \\
-\left(2 \alpha^2-3 \alpha\right) & (2 \alpha \beta+3 \beta) & -(2 \alpha \beta) \\
\alpha \beta-3 \alpha & -\left(\beta^2-3 \alpha\right) & \beta \alpha-\alpha^2
\end{array}\right]$$</p>
<p>which is equal to matrix $$B$$</p>
<p>So, by comparing elements of two matrix</p>
<p>$$\begin{aligned}
& \Rightarrow \alpha \beta-3 \alpha=-2 \alpha \\
& \Rightarrow \alpha \beta-\alpha=0 \\
& \Rightarrow \alpha(\beta-1)=0 \\
& \Rightarrow \alpha=0 \text { or } \beta=1[\because \alpha \text { cannot be } 0] \\
& \Rightarrow \beta=1 \\
& \text { and }-\beta^2+3 \alpha=5 \\
& \Rightarrow 3 \alpha=6 \\
& \Rightarrow \alpha=2 \\
& A=\left[\begin{array}{ccc}
1 & 2 & 3 \\
2 & 2 & 1 \\
-1 & 2 & 4
\end{array}\right] \\
& \operatorname{Det}(A B)=|A||B|=|A|\left|(\operatorname{adj} A)^{\top}\right| \\
& =|A| \cdot|A|^2 \\
& =|A|^3 \\
& =(6-18+18)^3 \\
& =6^3 \\
& =216 \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
lvb294f8 | maths | matrices-and-determinants | adjoint-of-a-matrix | <p>If $$A$$ is a square matrix of order 3 such that $$\operatorname{det}(A)=3$$ and $$\operatorname{det}\left(\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 \mathrm{~A})^{-1}\right)\right)\right)\right)\right)=2^{\mathrm{m}} 3^{\mathrm{n}}$$, then $$\mathrm{m}+2 \mathrm{n}$$ is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "6"}] | ["B"] | null | <p>$$\begin{aligned}
& |A|=3 \\
& \left|\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}(2 A)^{-1}\right)\right)\right)\right| \\
& =\left|-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 A)^{-1}\right)\right)\right)\right|^2
\end{aligned}$$</p>
<p>$$\begin{aligned}
& =4^6\left|-3 \operatorname{adj}\left(\operatorname{aadj}\left((2 A)^{-1}\right)\right)\right|^4 \\
& =4^6 \cdot 3^{12}\left|\operatorname{aadj}\left((2 A)^{-1}\right)\right|^8 \\
& =4^6 \cdot 3^{12} \cdot 3^{24}\left|(2 A)^{-1}\right|^{16} \\
& =4^6 \cdot 3^{36} 2^{-48}\left|A^{-1}\right|^{16} \\
& =\frac{2^{-36} 3^{36}}{3^{16}}=2^{-36} 3^{20} \\
& m=-36 \\
& n=20 \\
& m+2 n=4
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
vLYN6IboT26J7IDy | maths | matrices-and-determinants | basic-of-matrix | The number of $$3 \times 3$$ non-singular matrices, with four entries as $$1$$ and all other entries as $$0$$, is : | [{"identifier": "A", "content": "$$5$$ "}, {"identifier": "B", "content": "$$6$$ "}, {"identifier": "C", "content": "at least $$7$$ "}, {"identifier": "D", "content": "less than $$4$$ "}] | ["C"] | null | $$\left[ {\matrix{
1 & {...} & {...} \cr
{...} & 1 & {...} \cr
{...} & {...} & 1 \cr
} } \right]\,\,$$ are $$6$$ non-singular matrices because $$6$$
<br><br>blanks will be filled by $$5$$ zeros and $$1$$ one.
<br><br>Similarly, $$\left[ {\matrix{
{...} & {...} & 1 \cr
{...} & 1 & {...} \cr
1 & {...} & {...} \cr
} } \right]\,\,$$ are $$6$$ non-singular matrices.
<br><br>So, required cases are more than $$7,$$ non-singular $$3 \times 3$$ matrices. | mcq | aieee-2010 |
1l6f3ahmk | maths | matrices-and-determinants | basic-of-matrix | <p>Let $$A=\left[\begin{array}{lll}
1 & a & a \\
0 & 1 & b \\
0 & 0 & 1
\end{array}\right], a, b \in \mathbb{R}$$. If for some <br/><br/>$$n \in \mathbb{N}, A^{n}=\left[\begin{array}{ccc}
1 & 48 & 2160 \\
0 & 1 & 96 \\
0 & 0 & 1
\end{array}\right]
$$ then $$n+a+b$$ is equal to ____________.</p> | [] | null | 24 | <p>$$A = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right] + \left[ {\matrix{
0 & a & a \cr
0 & 0 & b \cr
0 & 0 & 0 \cr
} } \right] = I + B$$</p>
<p>$${B^2} = \left[ {\matrix{
0 & a & a \cr
0 & 0 & b \cr
0 & 0 & 0 \cr
} } \right] + \left[ {\matrix{
0 & a & a \cr
0 & 0 & b \cr
0 & 0 & 0 \cr
} } \right] = \left[ {\matrix{
0 & 0 & {ab} \cr
0 & 0 & 0 \cr
0 & 0 & 0 \cr
} } \right]$$</p>
<p>$${B^3} = 0$$</p>
<p>$$\therefore$$ $${A^n} = {(1 + B)^n} = {}^n{C_0}I + {}^n{C_1}B + {}^n{C_2}{B^2} + {}^n{C_3}{B^3} + \,\,....$$</p>
<p>$$ = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right] + \left[ {\matrix{
0 & {na} & {na} \cr
0 & 0 & {nb} \cr
0 & 0 & 0 \cr
} } \right] + \left[ {\matrix{
0 & 0 & {{{n(n - 1)ab} \over 2}} \cr
0 & 0 & 0 \cr
0 & 0 & 0 \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
1 & {na} & {na + {{n(n - 1)} \over 2}ab} \cr
0 & 1 & {nb} \cr
0 & 0 & 1 \cr
} } \right] = \left[ {\matrix{
1 & {48} & {2160} \cr
0 & 1 & {48} \cr
0 & 0 & 1 \cr
} } \right]$$</p>
<p>On comparing we get $$na = 48$$, $$nb = 96$$ and</p>
<p>$$na + {{n(n - 1)} \over 2}ab = 2160$$</p>
<p>$$ \Rightarrow a = 4,n = 12$$ and $$b = 8$$</p>
<p>$$n + a + b = 24$$</p> | integer | jee-main-2022-online-25th-july-evening-shift |
Ame1VqiXGsSrDpFU | maths | matrices-and-determinants | expansion-of-determinant | If $$a>0$$ and discriminant of $$\,a{x^2} + 2bx + c$$ is $$-ve$$, then
<br/>$$\left| {\matrix{
a & b & {ax + b} \cr
b & c & {bx + c} \cr
{ax + b} & {bx + c} & 0 \cr
} } \right|$$ is equal to | [{"identifier": "A", "content": "$$+ve$$ "}, {"identifier": "B", "content": "$$\\left( {ac - {b^2}} \\right)\\left( {a{x^2} + 2bx + c} \\right)$$ "}, {"identifier": "C", "content": "$$-ve$$"}, {"identifier": "D", "content": "$$0$$ "}] | ["C"] | null | We have $$\left| {\matrix{
a & b & {ax + b} \cr
b & c & {bx + c} \cr
{ax + b} & {bx + c} & 0 \cr
} } \right|$$
<br><br>By $$\,\,\,{R_3} \to {R_3} - \left( {x{R_1} + {R_2}} \right);$$
<br><br>$$ = \left| {\matrix{
a & b & {ax + b} \cr
b & c & {bx + c} \cr
0 & 0 & { - \left( {a{x^2} + 2bx + C} \right)} \cr
} } \right|$$
<br><br>$$ = \left( {a{x^2} + 2bx + c} \right)\left( {{b^2} - ac} \right)$$
<br><br>$$ = \left( + \right)\left( - \right) = - ve.$$ | mcq | aieee-2002 |
tULOO1HXFB32hWVY | maths | matrices-and-determinants | expansion-of-determinant | If $$1,$$ $$\omega ,{\omega ^2}$$ are the cube roots of unity, then
<p>$$\Delta = \left| {\matrix{
1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr
{{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr
{{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr
} } \right|$$ is equal to</p> | [{"identifier": "A", "content": "$${\\omega ^2}$$ "}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$\\omega $$ "}] | ["B"] | null | $$\Delta = \left| {\matrix{
1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr
{{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr
{{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr
} } \right|$$
<br><br>$$ = 1\left( {{\omega ^{3n}} - 1} \right) - {\omega ^n}\left( {{\omega ^{2n}} - {\omega ^{2n}}} \right) + $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^{2n}}\left( {{\omega ^n} - {\omega ^{4n}}} \right)$$
<br><br>$$ = {\omega ^{3n}} - 1 - 0 + {\omega ^{3n}} - {\omega ^{6n}}$$
<br><br>$$ = 1 - 1 + 1 - 1 = 0$$ $$\left[ {} \right.$$ as $$\,\,\,\,\,$$ $${\omega ^{3n}} = 1$$ $$\left. {} \right]$$ | mcq | aieee-2003 |
cR2lVel1EGtJOygR | maths | matrices-and-determinants | expansion-of-determinant | If $${a_1},{a_2},{a_3},.........,{a_n},......$$ are in G.P., then the value of the determinant
<p>$$\left| {\matrix{
{\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr
{\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr
{\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr
} } \right|,$$ is </p> | [{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$0$$"}] | ["D"] | null | $$\left| {\matrix{
{\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr
{\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr
{\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr
} } \right|$$
<br><br>$$ = \left| {\matrix{
{\log {a_1}r{}^{n - 1}} & {\log {a_1}r{}^n} & {\log {a_1}r{}^{n + 1}} \cr
{\log {a_1}r{}^{n + 2}} & {\log {a_1}r{}^{n + 3}} & {\log {a_1}r{}^{n + 4}} \cr
{\log {a_1}r{}^{n + 5}} & {\log {a_1}r{}^{n + 6}} & {\log {a_1}r{}^{n + 7}} \cr
} } \right|$$
<br><br>$$ = \left| {\matrix{
{\log {a_1} + \left( {n - 1} \right)\log r} & {\log {a_1} + n\log r} & {\log {a_1} + \left( {n + 1} \right)\log r} \cr
{\log {a_1} + \left( {n + 2} \right)\log r} & {\log {a_1} + \left( {n + 3} \right)\log r} & {\log {a_1} + \left( {n + 4} \right)\log r} \cr
{\log {a_1} + \left( {n + 5} \right)\log r} & {\log {a_1} + \left( {n + 6} \right)\log r} & {\log {a_1} + \left( {n + 7} \right)\log r} \cr
} } \right|$$
<br><br>$$ = 0\left[ \, \right.$$ Apply $$\,\,\,\,{c_2} \to {c_2} - {1 \over 2}{c_1} - {1 \over 2}{c_3}\,\left. \, \right]$$ | mcq | aieee-2004 |
GpuKipzHONqFXg6t | maths | matrices-and-determinants | expansion-of-determinant | If $${a_1},{a_2},{a_3},........,{a_n},.....$$ are in G.P., then the determinant
$$$\Delta = \left| {\matrix{
{\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr
{\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr
{\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr
} } \right|$$$
<br/>is equal to : | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$4$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["B"] | null | As $$\,\,\,\,{a_1},{a_2},{a_3},.........$$ are in $$G.P.$$
<br><br>$$\therefore$$ Using $${a_n} = a{r^{n - 1}},\,\,\,$$ we get the given determinant,
<br><br>as $$\,\,\,\,\,\,\,\left| {\matrix{
{\log a{r^{n - 1}}} & {\log a{r^n}} & {\log a{r^{n + 1}}} \cr
{\log a{r^{n + 2}}} & {\log a{r^{n + 3}}} & {\log a{r^{n + 4}}} \cr
{\log a{r^{n + 5}}} & {\log a{r^{n + 6}}} & {\log a{r^{n + 7}}} \cr
} } \right|$$
<br><br>Operating $${C_3} - {C_2}$$ and $${C_2} - {C_1}$$ and using
<br><br>$$\log m - \log n = \log {m \over n}\,\,\,\,$$ we get
<br><br>$$ = \left| {\matrix{
{\log a{r^{n - 1}}} & {\log r} & {\log r} \cr
{\log a{r^{n + 2}}} & {\log r} & {\log r} \cr
{\log a{r^{n + 5}}} & {\log r} & {\log r} \cr
} } \right| $$
<br><br>$$=0$$ (two columns being identical) | mcq | aieee-2005 |
xFgehdfYl6MahZIw | maths | matrices-and-determinants | expansion-of-determinant | If $${a^2} + {b^2} + {c^2} = - 2$$ and
<p>f$$\left( x \right) = \left| {\matrix{
{1 + {a^2}x} & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr
{\left( {1 + {a^2}} \right)x} & {1 + {b^2}x} & {\left( {1 + {c^2}} \right)x} \cr
{\left( {1 + {a^2}} \right)x} & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr
} } \right|,$$</p>
<p>then f$$(x)$$ is a polynomial of degree :</p> | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$2$$"}] | ["D"] | null | Applying, $${C_1} \to {C_1} + {C_2} + {C_3}\,\,\,$$ we get
<br><br>$$f\left( x \right) = \left| {\matrix{
{1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr
{1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \cr
{1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr
} } \right|$$
<br><br>$$ = \left| {\matrix{
1 & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr
1 & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \cr
1 & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr
} } \right|$$
<br><br>$$\left[ \, \right.$$ As given that $${a^2} + {b^2} + {c^2} = - 2$$ $$\left. {} \right]$$
<br><br>$$\therefore$$ $${a^2} + {b^2} + {c^2} + 2 = 0$$
<br><br>Applying $${R_1} \to {R_1} - {R_2},\,\,\,{R_2} \to {R_2} - {R_3}$$
<br><br>$$\therefore$$ $$f\left( x \right) = \left| {\matrix{
0 & {x - 1} & 0 \cr
0 & {1 - x} & {x - 1} \cr
1 & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr
} } \right|$$
<br><br>$$f\left( x \right) = {\left( {x - 1} \right)^2}$$
<br><br>Hence degree $$=2.$$ | mcq | aieee-2005 |
qp2BEp4dd9iOFnFq | maths | matrices-and-determinants | expansion-of-determinant | If $$D = \left| {\matrix{
1 & 1 & 1 \cr
1 & {1 + x} & 1 \cr
1 & 1 & {1 + y} \cr
} } \right|$$ for $$x \ne 0,y \ne 0,$$ then $$D$$ is : | [{"identifier": "A", "content": "divisible by $$x$$ but not $$y$$"}, {"identifier": "B", "content": "divisible by $$y$$ but not $$x$$"}, {"identifier": "C", "content": "divisible by neither $$x$$ nor $$y$$"}, {"identifier": "D", "content": "divisible by both $$x$$ and $$y$$"}] | ["D"] | null | Given, $$D = \left| {\matrix{
1 & 1 & 1 \cr
1 & {1 + x} & 1 \cr
1 & 1 & {1 + y} \cr
} } \right|$$
<br><br>Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$
<br><br>and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$
<br><br>$$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{
1 & 1 & 1 \cr
0 & x & 0 \cr
0 & 0 & y \cr
} } \right| = xy$$
<br><br>Hence, $$D$$ is divisible by both $$x$$ and $$y$$ | mcq | aieee-2007 |
2Zr5gIOJoHIbdmBP | maths | matrices-and-determinants | expansion-of-determinant | Let $$a, b, c$$ be such that $$b\left( {a + c} \right) \ne 0$$ if
<p>$$\left| {\matrix{
a & {a + 1} & {a - 1} \cr
{ - b} & {b + 1} & {b - 1} \cr
c & {c - 1} & {c + 1} \cr
} } \right| + \left| {\matrix{
{a + 1} & {b + 1} & {c - 1} \cr
{a - 1} & {b - 1} & {c + 1} \cr
{{{\left( { - 1} \right)}^{n + 2}}a} & {{{\left( { - 1} \right)}^{n + 1}}b} & {{{\left( { - 1} \right)}^n}c} \cr
} } \right| = 0$$
<br/><br>then the value of $$n$$ :</br></p> | [{"identifier": "A", "content": "any even integer "}, {"identifier": "B", "content": "any odd integer "}, {"identifier": "C", "content": "any integer "}, {"identifier": "D", "content": "zero"}] | ["B"] | null | $$\left| {\matrix{
a & {a + 1} & {a - 1} \cr
{ - b} & {b + 1} & {b - 1} \cr
c & {c - 1} & {c + 1} \cr
} } \right| + \left| {\matrix{
{a + 1} & {b + 1} & {c - 1} \cr
{a - 1} & {b - 1} & {c + 1} \cr
{{{\left( { - 1} \right)}^{n + 2}}a} & {{{\left( { - 1} \right)}^{n + 1}}b} & {{{\left( { - 1} \right)}^n}c} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \left| {\matrix{
a & {a + 1} & {a - 1} \cr
{ - b} & {b + 1} & {b - 1} \cr
c & {c - 1} & {c + 1} \cr
} } \right| + \left| {\matrix{
{a + 1} & {a - 1} & {{{\left( { - 1} \right)}^{n + 2}}a} \cr
{b + 1} & {b - 1} & {{{\left( { - 1} \right)}^{n + 1}}b} \cr
{c - 1} & {c + 1} & {{{\left( { - 1} \right)}^n}c} \cr
} } \right| = 0$$
<br><br>(Taking transpose of second determinant)
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_1} \Leftrightarrow {C_3}$$
<br><br>$$ \Rightarrow \left| {\matrix{
a & {a + 1} & {a - 1} \cr
{ - b} & {b + 1} & {b - 1} \cr
c & {c - 1} & {c + 1} \cr
} } \right| - \left| {\matrix{
{{{\left( { - 1} \right)}^{n + 2}}a} & {a - 1} & {a + 1} \cr
{{{\left( { - 1} \right)}^{n + 2}}\left( { - b} \right)} & {b - 1} & {b + 1} \cr
{{{\left( { - 1} \right)}^{n + 2}}c} & {c + 1} & {c - 1} \cr
} } \right| = 0$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} \Leftrightarrow {C_3}$$
<br><br>$$ \Rightarrow \left| {\matrix{
a & {a + 1} & {a - 1} \cr
{ - b} & {b + 1} & {b - 1} \cr
c & {c - 1} & {c + 1} \cr
} } \right| + {\left( 1 \right)^{n + 2}}\left| {\matrix{
a & {a + 1} & {a - 1} \cr
{ - b} & {b + 1} & {b - 1} \cr
c & {c - 1} & {c + 1} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{
a & {a + 1} & {a - 1} \cr
{ - b} & {b + 1} & {b - 1} \cr
c & {c - 1} & {c + 1} \cr
} } \right| = 0$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} - {C_1},{C_3} - {C_1}$$
<br><br>$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{
a & 1 & { - 1} \cr
{ - b} & {2b + 1} & {2b - 1} \cr
c & { - 1} & 1 \cr
} } \right| = 0$$
<br><br>$${R_1} + {R_3}$$
<br><br>$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{
{a + c} & 0 & 0 \cr
{ - b} & {2b + 1} & {2b - 1} \cr
c & { - 1} & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left( {a + c} \right)\left( {2b + 1 + 2b - 1} \right) = 0$$
<br><br>$$ \Rightarrow 4b\left( {a + c} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right] = 0$$
<br><br>$$ \Rightarrow 1 + {\left( { - 1} \right)^{n + 2}} = 0$$ $$\,\,\,\,\,$$ as $$\,\,\,\,\,b\left( {a + c} \right) \ne 0$$
<br><br>$$ \Rightarrow n$$ should be an odd integer. | mcq | aieee-2009 |
r4Fv71k1mBq9dYh2 | maths | matrices-and-determinants | expansion-of-determinant | If $$\alpha ,\beta \ne 0,$$ and $$f\left( n \right) = {\alpha ^n} + {\beta ^n}$$ and
$$$\left| {\matrix{
3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr
{1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr
{1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr
} } \right|$$$
<br/> $$ = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2},$$ then $$K$$ is equal to : | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$-1$$"}, {"identifier": "C", "content": "$$\\alpha \\beta $$ "}, {"identifier": "D", "content": "$${1 \\over {\\alpha \\beta }}$$ "}] | ["A"] | null | Consider
<br><br>$$\left| {\matrix{
3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr
{1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr
{1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr
} } \right|$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{
{1 + 1 + 1} & {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} \cr
{1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} \cr
{1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} & {1 + {\alpha ^4} + {\beta ^4}} \cr
} } \right|$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{
1 & 1 & 1 \cr
1 & \alpha & \beta \cr
1 & {{\alpha ^2}} & {{\beta ^2}} \cr
} } \right| \times \left| {\matrix{
1 & 1 & 1 \cr
1 & \alpha & \beta \cr
1 & {{\alpha ^2}} & {{\beta ^2}} \cr
} } \right|$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = {\left| {\matrix{
1 & 1 & 1 \cr
1 & \alpha & \beta \cr
1 & {{\alpha ^2}} & {{\beta ^2}} \cr
} } \right|^2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = {\left[ {\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {\alpha - \beta } \right)} \right]^2}$$
<br><br>So, $$k=1$$ | mcq | jee-main-2014-offline |
3HijCKA6R6fQjXo8tvuQN | maths | matrices-and-determinants | expansion-of-determinant | If A = $$\left[ {\matrix{
{ - 4} & { - 1} \cr
3 & 1 \cr
} } \right]$$,
<br/><br/>then the determinant of the matrix (A<sup>2016</sup> − 2A<sup>2015</sup> − A<sup>2014</sup>) is :
| [{"identifier": "A", "content": "2014"}, {"identifier": "B", "content": "$$-$$ 175"}, {"identifier": "C", "content": "2016"}, {"identifier": "D", "content": "$$-$$ 25"}] | ["D"] | null | Given,
<br><br>$$A = \left[ {\matrix{
{ - 4} & { - 1} \cr
3 & 1 \cr
} } \right]$$
<br><br>$${A^2} = \left[ {\matrix{
{ - 4} & { - 1} \cr
3 & 1 \cr
} } \right]\left[ {\matrix{
{ - 4} & { - 1} \cr
3 & 1 \cr
} } \right]$$
<br><br>$$ = \left[ {\matrix{
{13} & 3 \cr
{ - 9} & { - 2} \cr
} } \right]$$
<br><br>A<sup>2</sup> $$-$$ 2A $$-$$ I
<br><br>$$ = \left[ {\matrix{
{13} & 3 \cr
{ - 9} & { - 2} \cr
} } \right] - \left[ {\matrix{
{ - 8} & { - 2} \cr
6 & 2 \cr
} } \right] - \left[ {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right]$$
<br><br>$$ = \left[ {\matrix{
{20} & 5 \cr
{ - 15} & { - 5} \cr
} } \right]$$
<br><br>$$\left| A \right| = \left| {\matrix{
{ - 4} & { - 1} \cr
3 & 1 \cr
} } \right|$$ $$=$$ $$-$$ 4 + 3 $$=$$ $$-$$ 1
<br><br>Now,
<br><br>$$\left| {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right|$$
<br><br>$$=$$ $${\left| A \right|^{2014}}\left| {{A^2} - 2A - {\rm I}} \right|$$
<br><br>$$ = {\left( { - 1} \right)^{2014}}\left| {\matrix{
{20} & 5 \cr
{ - 15} & { - 5} \cr
} } \right|$$
<br><br>$$=$$ 1 $$ \times $$ ($$-$$ 100 + 75)
<br><br>$$=$$ $$-$$ 25 | mcq | jee-main-2016-online-10th-april-morning-slot |
RS4yHNVUSJJseZvzGSGiK | maths | matrices-and-determinants | expansion-of-determinant | The number of distinct real roots of the equation,
<br/><br/>$$\left| {\matrix{
{\cos x} & {\sin x} & {\sin x} \cr
{\sin x} & {\cos x} & {\sin x} \cr
{\sin x} & {\sin x} & {\cos x} \cr
} } \right| = 0$$ in the interval $$\left[ { - {\pi \over 4},{\pi \over 4}} \right]$$ is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["C"] | null | Given,
<br><br>$$\left| {\matrix{
{\cos x} & {\sin x} & {\sin x} \cr
{\sin x} & {\cos x} & {\sin x} \cr
{\sin x} & {\sin x} & {\cos x} \cr
} } \right| = 0$$
<br><br>R<sub>1</sub> $$ \to $$ R<sub>1</sub> $$-$$ R<sub>3</sub>
<br><br>R<sub>1</sub> $$ \to $$ R<sub>2</sub> $$-$$ R<sub>3</sub>
<br><br>$$\left| {\matrix{
{\cos x - \sin x} & 0 & {\sin x - \cos x} \cr
0 & {\cos x - \sin x} & {\sin x - \cos x} \cr
{\sin x} & {\sin x} & { \cos x} \cr
} } \right| = 0$$
<br><br>C<sub>3</sub> $$ \to $$ C<sub>3</sub> + C<sub>2</sub>
<br><br>$$\left| {\matrix{
{\cos x - \sin x} & 0 & {\sin x - \cos x} \cr
0 & {\cos x - \sin x} & 0 \cr
{\sin x} & {\sin x} & {\sin x + \cos x} \cr
} } \right| = 0$$
<br><br>Expanding using first column,
<br><br>(cosx $$-$$ sinx)(cos $$-$$ sinx) (sinx + cos x)
<br><br> + sinx (cosx $$-$$ sinx) (sinx $$-$$ cosx) = 0
<br><br>$$ \Rightarrow $$ (cosx $$-$$ sinx)<sup>2</sup> (sinx + cosx)
<br><br> $$+$$ sinx (cosx $$-$$ sinx)<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ (cosx $$-$$ sinx)<sup>2</sup> (sinx + cosx $$+$$ sinx) = 0
<br><br>$$ \Rightarrow $$ (2sinx + cosx )(cosx $$-$$ sinx)<sup>2</sup> = 0
<br><br>$$ \therefore $$ cosx = -2sinx or cosx = sinx
<br><br>$$ \Rightarrow $$ tanx = $$ - {1 \over 2}$$ or tanx = 1
<br><br>$$ \therefore $$ x = $$ - {\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$, $${\pi \over 4}$$
<br><br>$$ \therefore $$ Number of solutions = 2
| mcq | jee-main-2016-online-9th-april-morning-slot |
eun9UQWukCJ25rjmhngyg | maths | matrices-and-determinants | expansion-of-determinant | If
<br/><br/>$$S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| {\matrix{
0 & {\cos x} & { - \sin x} \cr
{\sin x} & 0 & {\cos x} \cr
{\cos x} & {\sin x} & 0 \cr
} } \right| = 0} \right\},$$
<br/><br/>then $$\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)} $$ is equal to : | [{"identifier": "A", "content": "$$4 + 2\\sqrt 3 $$"}, {"identifier": "B", "content": "$$ - 2 + \\sqrt 3 $$"}, {"identifier": "C", "content": "$$ - 2 - \\sqrt 3 $$"}, {"identifier": "D", "content": "$$-\\,\\,4 - 2\\sqrt 3 $$"}] | ["C"] | null | Given,
<br><br> $$\left| {\matrix{
0 & {\cos x} & { - \sin x} \cr
{\sin x} & 0 & {\cos x} \cr
{\cos x} & {\sin x} & 0 \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 0 (0 $$-$$ cosx sinx) $$-$$ cosx (0 $$-$$ cos<sup>2</sup>x) $$-$$ sinx(sin<sup>2</sup>x) = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ cos<sup>3</sup>x $$-$$ sin<sup>3</sup> x = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ tan<sup>3</sup>x = 1
<br><br>$$ \Rightarrow $$$$\,\,\,$$ tanx = 1
<br><br>$$ \therefore $$ $$\,\,\,$$ $$\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)} $$
<br><br>= $${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$$
<br><br>= $${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$$
<br><br>= $${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$$
<br><br>= $${{1 + 3 + 2\sqrt 3 } \over { - 2}}$$
<br><br>= $$-$$ 2 $$-$$ $$\sqrt 3 $$ | mcq | jee-main-2017-online-8th-april-morning-slot |
eHnuQRvur8S3cbRr4ps0G | maths | matrices-and-determinants | expansion-of-determinant | Let $$A$$ be a matrix such that $$A.\left[ {\matrix{
1 & 2 \cr
0 & 3 \cr
} } \right]$$ is a scalar matrix and |3A| = 108.
<br/>Then A<sup>2</sup> equals : | [{"identifier": "A", "content": "$$\\left[ {\\matrix{\n 4 & { - 32} \\cr \n 0 & {36} \\cr \n\n } } \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {\\matrix{\n {36} & 0 \\cr \n { - 32} & 4 \\cr \n\n } } \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {\\matrix{\n 4 & 0 \\cr \n { - 32} & {36} \\cr \n\n } } \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {\\matrix{\n {36} & { - 32} \\cr \n 0 & 4 \\cr \n\n } } \\right]$$"}] | ["D"] | null | According to questions, <br/><br>
A. $$\left[ {\matrix{
1 & 2 \cr
0 & 3 \cr
} } \right]$$ = $$\left[ {\matrix{
\lambda & 0 \cr
0 & \lambda \cr
} } \right]$$<br><br>
$$ \Rightarrow $$ A = $$\left[ {\matrix{
\lambda & 0 \cr
0 & \lambda \cr
} } \right]$$ $$\left[ {\matrix{
1 & 2 \cr
0 & 3 \cr
} } \right]^{-1}$$<br><br>
$$ \Rightarrow $$ A = $$1 \over 3$$$$\left[ {\matrix{
\lambda & 0 \cr
0 & \lambda \cr
} } \right]$$ $$\left[ {\matrix{
3 & {-2} \cr
0 & 1 \cr
} } \right]$$<br><br>
$$ \Rightarrow $$ A = $$\left[ {\matrix{
\lambda & 0 \cr
0 & \lambda \cr
} } \right]$$ $$\left[ {\matrix{
1 & { - {2 \over 3}} \cr
0 & {{1 \over 3}} \cr
} } \right]$$<br><br>
$$ \Rightarrow $$ A = $$\left[ {\matrix{
\lambda & { - {2 \over 3}\lambda } \cr
0 & {{\lambda \over 3}} \cr
} } \right]$$<br><br>
As $$\left| {3A} \right|$$ = 108<br><br>
$$ \Rightarrow $$ 108 = $$\left| {\matrix{
{3\lambda } & { - 2\lambda } \cr
0 & \lambda \cr
} } \right|$$
<br><br>
$$ \Rightarrow $$ 3$$\lambda $$<sup>2</sup> = 108<br><br>
$$ \Rightarrow $$ $$\lambda $$<sup>2</sup> = 36<br><br>
$$ \Rightarrow $$ $$\lambda $$ = $$ \pm $$6<br><br>
When $$\lambda $$ = +6<br><br>
then A = $$\left[ {\matrix{
6 & { - 4} \cr
0 & 2 \cr
} } \right]$$<br><br>
$$ \Rightarrow $$ A<sup>2</sup> = $$\left[ {\matrix{
{36} & { - 32} \cr
0 & 4 \cr
} } \right]$$<br><br>
For $$\lambda $$ = -6<br><br>
A = $$\left[ {\matrix{
{ - 6} & 4 \cr
0 & { - 2} \cr
} } \right]$$<br><br>
$$ \Rightarrow $$ A<sup>2</sup> = $$\left[ {\matrix{
{36} & { - 32} \cr
0 & 4 \cr
} } \right]$$ | mcq | jee-main-2018-online-15th-april-morning-slot |
jgExl7kan7jrpclX | maths | matrices-and-determinants | expansion-of-determinant | If $$\left| {\matrix{
{x - 4} & {2x} & {2x} \cr
{2x} & {x - 4} & {2x} \cr
{2x} & {2x} & {x - 4} \cr
} } \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}$$
<br/><br/>then the ordered pair (A, B) is equal to : | [{"identifier": "A", "content": "(4, 5)"}, {"identifier": "B", "content": "(-4, -5)"}, {"identifier": "C", "content": "(-4, 3)"}, {"identifier": "D", "content": "(-4, 5)"}] | ["D"] | null | $$\left| {\matrix{
{x - 4} & {2x} & {2x} \cr
{2x} & {x - 4} & {2x} \cr
{2x} & {2x} & {x - 4} \cr
} } \right|$$
<br><br>Applying c<sub>1</sub> $$ \to $$ c<sub>1</sub> + c<sub>2</sub> + c<sub>3</sub>
<br><br>$$ = \,\,\,\,\left| {\matrix{
{5x - 4} & {2x} & {2x} \cr
{5x - 4} & {x - 4} & {2x} \cr
{5x - 4} & {2x} & {x - 4} \cr
} } \right|$$
<br><br>Taking common (5x $$-$$ 4) from c<sub>1</sub>
<br><br>$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{
1 & {2x} & {2x} \cr
1 & {x - 4} & {2x} \cr
1 & {2x} & {x - 4} \cr
} } \right|$$
<br><br>Apply R<sub>2</sub> $$ \to $$R<sub>2</sub> $$-$$ R<sub>1</sub> and R<sub>3</sub> $$ \to $$R<sub>3</sub> $$-$$ R<sub>1</sub>
<br><br>$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{
1 & {2x} & {2x} \cr
0 & { - \left( {x + 4} \right)} & 0 \cr
0 & 0 & { - \left( {x + 4} \right)} \cr
} } \right|$$
<br><br>$$ = \,\,\,\,\left( {5x - 4} \right){\left( {x + 4} \right)^2}$$
<br><br>So, (A + Bx) (x $$-$$ A)<sup>2</sup> = (5x $$-$$ 4) (x + 4)<sup>2</sup>
<br><br>By comparing both sides we get, A = $$-$$ 4 and B = 5 | mcq | jee-main-2018-offline |
fP5s0hXTeIEiyzNQRTlYJ | maths | matrices-and-determinants | expansion-of-determinant | If $$A = \left[ {\matrix{
{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr
{{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr
{{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr
} } \right]$$
<br/><br/>then A is : | [{"identifier": "A", "content": "invertible for all t$$ \\in $$<b>R</b>."}, {"identifier": "B", "content": "invertible only if t $$=$$ $$\\pi $$"}, {"identifier": "C", "content": "not invertible for any t$$ \\in $$<b>R</b>"}, {"identifier": "D", "content": "invertible only if t $$=$$ $${\\pi \\over 2}$$."}] | ["A"] | null | $$A = \left[ {\matrix{
{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr
{{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr
{{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr
} } \right]$$
<br><br>$$\left| A \right| = {e^t}.\,{e^{ - t}}.{e^{ - t}}\left| {\matrix{
1 & {\cos t} & {\sin t} \cr
1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr
1 & {2\sin t} & { - 2\cos t} \cr
} } \right|$$
<br><br>Apply operations R<sub>2</sub> < R<sub>2</sub> $$-$$R<sub>1</sub>, R<sub>3</sub> < R<sub>3</sub> $$-$$ R<sub>1</sub>, R<sub>1</sub> < R<sub>1</sub>
<br><br>$$\left| A \right| = {e^{ - t}}\left| {\matrix{
1 & {\cos t} & {\sin t} \cr
0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr
0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr
} } \right|$$
<br><br>Open the determinant by R<sub>1</sub>
<br><br>$$\left| A \right| = 5{e^{ - t}}$$
<br><br>Invertible for all t $$ \in $$ R | mcq | jee-main-2019-online-9th-january-evening-slot |
8OQG5ZHmm0j8UPp8rP3rsa0w2w9jxb4ij7c | maths | matrices-and-determinants | expansion-of-determinant | A value of $$\theta \in \left( {0,{\pi \over 3}} \right)$$, for which
<br/>$$\left| {\matrix{
{1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr
{{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr
{{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr
} } \right| = 0$$, is : | [{"identifier": "A", "content": "$${\\pi \\over {18}}$$"}, {"identifier": "B", "content": "$${\\pi \\over {9}}$$"}, {"identifier": "C", "content": "$${{7\\pi } \\over {24}}$$"}, {"identifier": "D", "content": "$${{7\\pi } \\over {36}}$$"}] | ["B"] | null | $$\left| {\matrix{
{1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr
{{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr
{{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr
} } \right| = 0$$<br><br>
R<sub>1</sub> $$ \to $$ R<sub>1</sub> - R<sub>2</sub>, R<sub>2</sub> $$ \to $$ R<sub>2</sub> - <sub>R3</sub><br><br>
$$ \Rightarrow \left| {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
{{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr
} } \right| = 0$$<br><br>
C<sub>2</sub> $$ \to $$ C<sub>2</sub> + C<sub>1</sub><br><br>
$$ \Rightarrow \left| {\matrix{
1 & 0 & 0 \cr
0 & 1 & { - 1} \cr
{{{\cos }^2}\theta } & 1 & {1 + 4\cos 6\theta } \cr
} } \right| = 0$$<br><br>
$$ \Rightarrow 1 + 4\cos 6\theta + 1 = 0$$<br><br>
$$ \Rightarrow 2\cos 6\theta = - 1 \Rightarrow \cos 6\theta = - {1 \over 2}$$ = $$\cos {{2\pi } \over 3}$$<br><br>
$$ \Rightarrow 6\theta = 2n\pi \pm {{2\pi } \over 3}$$<br><br>
$$ \Rightarrow \theta = {{n\pi } \over 3} \pm {\pi \over 9}\,\,\,n \in 1$$<br><br>
$$ \Rightarrow \theta = {\pi \over 9},{{2\pi } \over 9},{{4\pi } \over 9}$$
| mcq | jee-main-2019-online-12th-april-evening-slot |
NvW5EmQOQ4vTpvOt9L3rsa0w2w9jx65olji | maths | matrices-and-determinants | expansion-of-determinant | If $$B = \left[ {\matrix{
5 & {2\alpha } & 1 \cr
0 & 2 & 1 \cr
\alpha & 3 & { - 1} \cr
} } \right]$$ is the inverse of a 3 × 3 matrix A, then the sum of all values of $$\alpha $$ for which
det(A) + 1 = 0, is :
| [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "- 1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["D"] | null | Given |A| + 1 = 0
<br><br>$$ \Rightarrow $$ |A| = -1
<br><br>$$\left| B \right| = \left| {{A^{ - 1}}} \right| = {1 \over {\left| A \right|}} = - 1$$<br><br>
$$\left| {\matrix{
5 & {2\alpha } & 1 \cr
0 & 2 & 1 \cr
\alpha & 3 & { - 1} \cr
} } \right| $$ = -1
<br><br>$$ \Rightarrow $$ $$ 5( - 2 - 3) + 2\alpha (\alpha ) + 1( - 2\alpha ) = - 1$$<br><br>
$$ \Rightarrow $$ $$2{\alpha ^2} - 2\alpha - 24 = 0$$<br><br>
$$ \therefore $$ sum of value of $$\alpha $$ = $${{ - ( - 2)} \over 2} = 1$$
| mcq | jee-main-2019-online-12th-april-morning-slot |
72ZcOZKf5Yr1mCFPHU3rsa0w2w9jx23i3gx | maths | matrices-and-determinants | expansion-of-determinant | The sum of the real roots of the equation
<br/>$$\left| {\matrix{
x & { - 6} & { - 1} \cr
2 & { - 3x} & {x - 3} \cr
{ - 3} & {2x} & {x + 2} \cr
} } \right| = 0$$, is equal to : | [{"identifier": "A", "content": "- 4"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "6"}] | ["B"] | null | x(-3x $$ \times $$ (x + 2) - 2x(x - 3)) + (– 6) (2(x + 2) + 3 (x – 3)) + (–1) (4x + 3 (–3x))<br><br>
$$ \Rightarrow $$ – 5x<sup>3</sup> + 30x –30 + 5x = 0<br><br>
$$ \Rightarrow $$ x<sup>3</sup> – 7x + 6 = 0<br><br>
$$ \therefore $$ sum of roots = 0 | mcq | jee-main-2019-online-10th-april-evening-slot |
2QtHVXivwg1Ceho7jC3rsa0w2w9jwxkqlfg | maths | matrices-and-determinants | expansion-of-determinant | If $${\Delta _1} = \left| {\matrix{
x & {\sin \theta } & {\cos \theta } \cr
{ - \sin \theta } & { - x} & 1 \cr
{\cos \theta } & 1 & x \cr
} } \right|$$ and
<br/>$${\Delta _2} = \left| {\matrix{
x & {\sin 2\theta } & {\cos 2\theta } \cr
{ - \sin 2\theta } & { - x} & 1 \cr
{\cos 2\theta } & 1 & x \cr
} } \right|$$, $$x \ne 0$$ ;
<br/><br/> then for all $$\theta \in \left( {0,{\pi \over 2}} \right)$$ : | [{"identifier": "A", "content": "$${\\Delta _1} - {\\Delta _2}$$ = x (cos 2$$\\theta $$ \u2013 cos 4$$\\theta $$)"}, {"identifier": "B", "content": "$${\\Delta _1} + {\\Delta _2}$$ = - 2x<sup>3</sup>"}, {"identifier": "C", "content": "$${\\Delta _1} + {\\Delta _2}$$ = \u2013 2(x<sup>3</sup> + x \u20131)"}, {"identifier": "D", "content": "$${\\Delta _1} - {\\Delta _2}$$ = - 2x<sup>3</sup>"}] | ["B"] | null | $${\Delta _1} = \left| {\matrix{
x & {\sin \theta } & {\cos \theta } \cr
{ - \sin \theta } & { - x} & 1 \cr
{\cos \theta } & 1 & x \cr
} } \right|$$<br><br>
= x(–x<sup>2</sup> –1) – sin$$\theta $$(–xsin$$\theta $$ – cos$$\theta $$) + cos$$\theta $$(–sin$$\theta $$+ xcos$$\theta $$)<br><br>
= –x<sup>3</sup> – x + xsin<sup>2</sup>$$\theta $$ + sin$$\theta $$cos$$\theta $$ – cos$$\theta $$sin$$\theta $$
+ xcos<sup>2</sup>$$\theta $$<br><br>
= –x<sup>3</sup> – x + x = –x<sup>3</sup><br><br>
Similarly $${\Delta _2} = - {x^3}$$<br><br>
$${\Delta _1} + {\Delta _2} = - 2{x^3}$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
MGCkKbAWU9TkkCJ1F318hoxe66ijvwp7qik | maths | matrices-and-determinants | expansion-of-determinant | Let $$\alpha $$ and $$\beta $$ be the roots of the equation
x<sup>2</sup> + x + 1 = 0. Then for y $$ \ne $$ 0 in R,<br/>
$$$\left| {\matrix{
{y + 1} & \alpha & \beta \cr
\alpha & {y + \beta } & 1 \cr
\beta & 1 & {y + \alpha } \cr
} } \right|$$$
is equal to | [{"identifier": "A", "content": "y(y<sup>2</sup> \u2013 1)"}, {"identifier": "B", "content": "y(y<sup>2</sup> \u2013 3)"}, {"identifier": "C", "content": "y<sup>3</sup>"}, {"identifier": "D", "content": "y<sup>3</sup> \u2013 1"}] | ["C"] | null | $$\alpha $$ and $$\beta $$ are the roots of the equation
x<sup>2</sup> + x + 1 = 0.
<br><br>$$ \therefore $$ $$\alpha $$ = $$\omega $$ and $$\beta $$ = $${\omega ^2}$$
<br><br>$$\left| {\matrix{
{y + 1} & \alpha & \beta \cr
\alpha & {y + \beta } & 1 \cr
\beta & 1 & {y + \alpha } \cr
} } \right|$$
<br><br>= $$\left| {\matrix{
{y + 1} & \omega & {{\omega ^2}} \cr
\omega & {y + {\omega ^2}} & 1 \cr
{{\omega ^2}} & 1 & {y + \omega } \cr
} } \right|$$
<br><br>C<sub>1</sub> $$ \to $$ C<sub>1</sub> + C<sub>2</sub> + C<sub>3</sub>
<br><br>= $$\left| {\matrix{
{y + 1 + \omega + {\omega ^2}} & \omega & {{\omega ^2}} \cr
{y + 1 + \omega + {\omega ^2}} & {y + {\omega ^2}} & 1 \cr
{y + 1 + \omega + {\omega ^2}} & 1 & {y + \omega } \cr
} } \right|$$
<br><br>= $$\left| {\matrix{
y & \omega & {{\omega ^2}} \cr
y & {y + {\omega ^2}} & 1 \cr
y & 1 & {y + \omega } \cr
} } \right|$$
<br><br>As $$1 + \omega + {\omega ^2}$$ = 0
<br><br>= $$y\left| {\matrix{
1 & \omega & {{\omega ^2}} \cr
1 & {y + {\omega ^2}} & 1 \cr
1 & 1 & {y + \omega } \cr
} } \right|$$
<br><br>R<sub>2</sub> $$ \to $$ R<sub>2</sub> - R<sub>1</sub>
<br>R<sub>3</sub> $$ \to $$ R<sub>3</sub> - R<sub>1</sub>
<br><br>= $$y\left| {\matrix{
1 & \omega & {{\omega ^2}} \cr
0 & {y + {\omega ^2} - \omega } & {1 - {\omega ^2}} \cr
0 & {1 - \omega } & {y + \omega - {\omega ^2}} \cr
} } \right|$$
<br><br>= y$$\left[ {\left( {y + {\omega ^2} - \omega } \right)\left( {y + \omega - {\omega ^2}} \right) - \left( {1 - {\omega ^2}} \right)\left( {1 - \omega } \right)} \right]$$
<br><br>= y(y<sup>2</sup>) = y<sup>3</sup> | mcq | jee-main-2019-online-9th-april-morning-slot |
JRTEaKMvpR1dXBveR2RUC | maths | matrices-and-determinants | expansion-of-determinant | Let the number 2,b,c be in an A.P. and<br/>
A = $$\left[ {\matrix{
1 & 1 & 1 \cr
2 & b & c \cr
4 & {{b^2}} & {{c^2}} \cr
} } \right]$$. If det(A) $$ \in $$ [2, 16], then c
lies in the interval : | [{"identifier": "A", "content": "[2, 3)"}, {"identifier": "B", "content": "[4, 6]"}, {"identifier": "C", "content": "(2 + 2<sup>3/4</sup>, 4)"}, {"identifier": "D", "content": "[3, 2 + 2<sup>3/4</sup>]"}] | ["B"] | null | 2, b, c are in AP.
<br><br>Let common difference = d
<br><br>$$ \therefore $$ b = 2 + d and c = 2 + 2d
<br><br>|A| = $$\left[ {\matrix{
1 & 1 & 1 \cr
2 & b & c \cr
4 & {{b^2}} & {{c^2}} \cr
} } \right]$$
<br><br>C<sub>2</sub> = C<sub>2</sub> - C<sub>1</sub>
<br><br>C<sub>3</sub> = C<sub>3</sub> - C<sub>1</sub>
<br><br>= $$\left| {\matrix{
1 & 0 & 0 \cr
2 & {b - 2} & {c - 2} \cr
4 & {{b^2} - 4} & {{c^2} - 4} \cr
} } \right|$$
<br><br>= $$\left( {b - 2} \right)\left( {c - 2} \right)\left| {\matrix{
1 & 0 & 0 \cr
2 & 1 & 1 \cr
4 & {b + 2} & {c + 2} \cr
} } \right|$$
<br><br>= $$\left( {b - 2} \right)\left( {c - 2} \right)\left[ {c + 2 - b - 2} \right]$$
<br><br>= $$\left( {b - 2} \right)\left( {c - 2} \right)\left( {c - b} \right)$$
<br><br>[ As b = 2 + d and c = 2 + 2d, then b - 2 = 4, c - 2 = 2d and c - b = d]
<br><br>= (d) (2d) (d)
<br><br>= 2d<sup>3</sup>
<br><br>Given |A| $$ \in $$ [2, 16]
<br><br>$$ \therefore $$ 2d<sup>3</sup> $$ \in $$ [2, 16]
<br><br>$$ \Rightarrow $$ d<sup>3</sup> $$ \in $$ [1, 8]
<br><br>$$ \Rightarrow $$ d $$ \in $$ [1, 2]
<br><br>As c = 2 + 2d
<br><br>then c $$ \in $$ [4, 6] | mcq | jee-main-2019-online-8th-april-evening-slot |
FIXxORXt1C1lR1l1ZhxXD | maths | matrices-and-determinants | expansion-of-determinant | If A = $$\left[ {\matrix{
1 & {\sin \theta } & 1 \cr
{ - \sin \theta } & 1 & {\sin \theta } \cr
{ - 1} & { - \sin \theta } & 1 \cr
} } \right]$$;
<br/><br/>then for all $$\theta $$ $$ \in $$ $$\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right)$$, det (A) lies in the interval : | [{"identifier": "A", "content": "$$\\left( {{3 \\over 2},3} \\right]$$"}, {"identifier": "B", "content": "$$\\left( {0,{3 \\over 2}} \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {{5 \\over 2},4} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {1,{5 \\over 2}} \\right]$$"}] | ["A"] | null | $$\left| A \right| = \left| {\matrix{
1 & {\sin \theta } & 1 \cr
{ - \sin \theta } & 1 & {\sin \theta } \cr
{ - 1} & { - \sin \theta } & 1 \cr
} } \right|$$
<br><br>= 2(1 + sin<sup>2</sup>$$\theta $$)
<br><br>$$\theta $$ $$ \in $$ $$\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right) \Rightarrow {1 \over {\sqrt 2 }} < \sin \theta < {1 \over {\sqrt 2 }}$$
<br><br> $$ \Rightarrow $$ 0 $$ \le $$ sin<sup>2</sup>$$\theta $$ < $${1 \over 2}$$
<br><br>$$ \therefore $$ $$\left| A \right| \in \left[ {2,3} \right)$$ | mcq | jee-main-2019-online-12th-january-evening-slot |
o5MdSV4OLnSSysFVE1ztM | maths | matrices-and-determinants | expansion-of-determinant | If $$\left| {\matrix{
{a - b - c} & {2a} & {2a} \cr
{2b} & {b - c - a} & {2b} \cr
{2c} & {2c} & {c - a - b} \cr
} } \right|$$
<br/><br/> = (a + b + c) (x + a + b + c)<sup>2</sup>, x $$ \ne $$ 0,
<br/><br/>then x is equal to : | [{"identifier": "A", "content": "\u20132(a + b + c)"}, {"identifier": "B", "content": "2(a + b + c)"}, {"identifier": "C", "content": "abc"}, {"identifier": "D", "content": "\u2013(a + b + c)"}] | ["A"] | null | $$\left| {\matrix{
{a - b - c} & {2a} & {2a} \cr
{2b} & {b - c - a} & {2b} \cr
{2c} & {2c} & {c - a - b} \cr
} } \right|$$
<br><br>R<sub>1</sub> $$ \to $$ R<sub>1</sub> + R<sub>2</sub> + R<sub>3</sub>
<br><br>$$ = \left| {\matrix{
{a + b + c} & {a + b + c} & {a + b + c} \cr
{2b} & {b - c - a} & {2b} \cr
{2c} & {2c} & {c - a - b} \cr
} } \right|$$
<br><br>$$ = \left( {a + b + c} \right)\left| {\matrix{
1 & 0 & 0 \cr
{2b} & { - \left( {a + b + c} \right)} & 0 \cr
{2c} & {2c} & {c - a - b} \cr
} } \right|$$
<br><br>$$=$$ (a + b + c) (a + b + c)<sup>2</sup>
<br><br>$$ \Rightarrow $$ x $$=$$ $$-$$ 2(a + b + c) | mcq | jee-main-2019-online-11th-january-evening-slot |
jeywfxj9iOKHSYmfc5TKS | maths | matrices-and-determinants | expansion-of-determinant | Let A = $$\left[ {\matrix{
2 & b & 1 \cr
b & {{b^2} + 1} & b \cr
1 & b & 2 \cr
} } \right]$$ where b > 0.
<br/><br/>Then the minimum value of $${{\det \left( A \right)} \over b}$$ is - | [{"identifier": "A", "content": "$$\\sqrt 3 $$"}, {"identifier": "B", "content": "$$-$$ $$2\\sqrt 3 $$"}, {"identifier": "C", "content": "$$ - \\sqrt 3 $$"}, {"identifier": "D", "content": "$$2\\sqrt 3 $$"}] | ["D"] | null | A = $$\left[ {\matrix{
2 & b & 1 \cr
b & {{b^2} + 1} & b \cr
1 & b & 2 \cr
} } \right]$$ (b > 0)
<br><br>$$\left| A \right|$$ = 2(2b<sup>2</sup> + 2 $$-$$ b<sup>2</sup>) $$-$$ b(2b $$-$$ b) + 1(b<sub>2</sub> $$-$$ b<sub>2</sub> $$-$$ 1)
<br><br>$$\left| A \right|$$ = 2(b<sup>2</sup> + 2) $$-$$ b<sup>2</sup> $$-$$ 1
<br><br>$$\left| A \right|$$ = b<sup>2</sup> + 3
<br><br>$${{\left| A \right|} \over b} = b + {3 \over b} \Rightarrow {{b + {3 \over b}} \over 2} \ge \sqrt 3 $$
<br><br>$$b + {3 \over b} \ge 2\sqrt 3 $$ | mcq | jee-main-2019-online-10th-january-evening-slot |
gZOLcTTnTzFV5f4sTVG6J | maths | matrices-and-determinants | expansion-of-determinant | Let d $$ \in $$ R, and
<br/><br/>$$A = \left[ {\matrix{
{ - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \cr
1 & {\left( {\sin \theta } \right) + 2} & d \cr
5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \cr
} } \right],$$
<br/><br/>$$\theta \in \left[ {0,2\pi } \right]$$ If the minimum value of det(A) is 8,
then a value of d is - | [{"identifier": "A", "content": "$$-$$ 7"}, {"identifier": "B", "content": "$$2\\left( {\\sqrt 2 + 2} \\right)$$ "}, {"identifier": "C", "content": "$$-$$ 5"}, {"identifier": "D", "content": "$$2\\left( {\\sqrt 2 + 1} \\right)$$"}] | ["C"] | null | $$\det A = \left| {\matrix{
{ - 2} & {4 + d} & {\sin \theta - 2} \cr
1 & {\sin \theta + 2} & d \cr
5 & {2\sin \theta - d} & { - \sin \theta + 2 + 2d} \cr
} } \right|$$
<br><br>(R<sub>1</sub> $$ \to $$ R<sub>1</sub> + R<sub>3</sub> $$-$$ 2R<sub>2</sub>)
<br><br>$$ = \left| {\matrix{
1 & 0 & 0 \cr
1 & {\sin \theta + 2} & d \cr
5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \cr
} } \right|$$
<br><br>= (2 + sin $$\theta $$) ( 2 + 2d $$-$$ sin$$\theta $$) $$-$$ d(2sin$$\theta $$ $$-$$ d)
<br><br>= 4 + 4d $$-$$ 2sin$$\theta $$ + 2sin$$\theta $$ + 2dsin$$\theta $$ $$-$$ sin<sup>2</sup>$$\theta $$ $$-$$ 2dsin$$\theta $$ + d<sup>2</sup>
<br><br>d<sup>2</sup> + 4d + 4 $$-$$ sin<sup>2</sup>$$\theta $$
<br><br>= (d + 2)<sup>2</sup> $$-$$ sin<sup>2</sup>$$\theta $$
<br><br>For a given d, minimum value of
<br><br>det(A) = (d + 2)<sup>2</sup> $$-$$ 1 = 8
<br><br>$$ \Rightarrow $$ d = 1 or $$-$$ 5 | mcq | jee-main-2019-online-10th-january-morning-slot |
HLdljjorhu7gjuwSpEjgy2xukf0wdn0c | maths | matrices-and-determinants | expansion-of-determinant | If $$\Delta $$ = $$\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
{2x - 3} & {3x - 4} & {4x - 5} \cr
{3x - 5} & {5x - 8} & {10x - 17} \cr
} } \right|$$ =
<br/><br/>Ax<sup>3</sup> + Bx<sup>2</sup> + Cx + D, then B + C is equal to : | [{"identifier": "A", "content": "-1"}, {"identifier": "B", "content": "-3"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "1"}] | ["B"] | null | $$\Delta $$ = $$\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
{2x - 3} & {3x - 4} & {4x - 5} \cr
{3x - 5} & {5x - 8} & {10x - 17} \cr
} } \right|$$
<br><br>R<sub>2</sub> $$ \to $$ R<sub>2</sub> – R<sub>1</sub>
<br>R<sub>3</sub> $$ \to $$ R<sub>3</sub> – R<sub>2</sub>
<br><br>= $$\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
{x - 1} & {x - 1} & {x - 1} \cr
{x - 2} & {2\left( {x - 2} \right)} & {6\left( {x - 2} \right)} \cr
} } \right|$$
<br><br>= $$\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{
{x - 2} & {2x - 3} & {3x - 4} \cr
1 & 1 & 1 \cr
1 & 2 & 6 \cr
} } \right|$$
<br><br>C<sub>1</sub> $$ \to $$ C<sub>1</sub> - C<sub>2</sub>
<br>C<sub>2</sub> $$ \to $$ C<sub>2</sub> - C<sub>3</sub>
<br><br>= $$\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{
{ - x + 1} & { - x + 1} & {3x - 4} \cr
0 & 0 & 1 \cr
{ - 1} & { - 4} & 6 \cr
} } \right|$$
<br><br>= -(x - 1)(x - 2)[-4(1 - x) + 1(1 - x)]
<br><br>= -(x<sup>2</sup> - 3x + 2)[3x - 3]
<br><br>= -3x<sup>3</sup> + 9x<sup>2</sup> - 6x + 3x<sup>2</sup> - 9x + 6
<br><br>= -3x<sup>3</sup> + 12x<sup>2</sup> - 15x + 6 = Ax<sup>3</sup> + Bx<sup>2</sup> + Cx + D
<br><br>$$ \therefore $$ A = -3, B = 12, C = -15
<br><br>$$ \therefore $$ B + C = 12 – 15 = – 3 | mcq | jee-main-2020-online-3rd-september-morning-slot |
Mq79PZTU58QxMIUvbvjgy2xukg0cuvx2 | maths | matrices-and-determinants | expansion-of-determinant | Let $$\theta = {\pi \over 5}$$ and $$A = \left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$. <br/><br/> If B = A + A<sup>4</sup>
, then det (B) : | [{"identifier": "A", "content": "lies in (1, 2)"}, {"identifier": "B", "content": "lies in (2, 3)."}, {"identifier": "C", "content": "is zero.\n"}, {"identifier": "D", "content": "is one."}] | ["A"] | null | $$A = \left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$
<br><br>A<sup>2</sup> = $$\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$$$\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$
<br><br>$$ \Rightarrow $$ A<sup>2</sup> = $$\left[ {\matrix{
{\cos 2\theta } & {\sin 2\theta } \cr
{ - \sin 2\theta } & {\cos 2\theta } \cr
} } \right]$$
<br><br>Similarly, A<sup>n</sup> = $$\left[ {\matrix{
{\cos n\theta } & {\sin n\theta } \cr
{ - \sin n\theta } & {\cos n\theta } \cr
} } \right]$$
<br><br>$$ \therefore $$ B = A + A<sup>4</sup>
<br><br>= $$\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$ + $$\left[ {\matrix{
{\cos 4\theta } & {\sin 4\theta } \cr
{ - \sin 4\theta } & {\cos 4\theta } \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
{\cos 4\theta + \cos \theta } & {\sin 4\theta + \sin \theta } \cr
{ - \sin 4\theta - \sin \theta } & {\cos 4\theta + \cos \theta } \cr
} } \right]$$
<br><br>detB = (cos4$$\theta $$ + cos$$\theta $$)<sup>2</sup>
+ (sin4$$\theta $$ + sin$$\theta $$)<sup>2</sup>
<br><br>= cos<sup>2</sup>4$$\theta $$ + cos<sup>2</sup>$$\theta $$ + 2cos4$$\theta $$ cos$$\theta $$
<br>+ sin<sup>2</sup>4$$\theta $$ + sin2$$\theta $$ + 2sin4$$\theta $$ –sin$$\theta $$
<br><br>= 2 + 2 ( cos4$$\theta $$ cos$$\theta $$ + sin4$$\theta $$ sin$$\theta $$)
<br><br>$$ \Rightarrow $$ detB = 2 + 2 cos3$$\theta $$
<br><br>at $$\theta $$ = $${\pi \over 5}$$
<br><br>detB = 2 + 2cos $${{3\pi } \over 5}$$
<br><br> = 2(1 - sin18)
<br><br>= 2(1 - $${{\sqrt 5 - 1} \over 4}$$)
<br><br>= 2$$\left( {{{5 - \sqrt 5 } \over 4}} \right)$$
<br><br>= $${{{5 - \sqrt 5 } \over 2}}$$ $$ \simeq $$ 1.385
<br><br>$$ \therefore $$ detB $$ \in $$ (1, 2)
| mcq | jee-main-2020-online-6th-september-evening-slot |
hqjDMffUsigfBYiX0jjgy2xukfuvg3xv | maths | matrices-and-determinants | expansion-of-determinant | Let m and M be respectively the minimum and maximum values of
<br/><br/>$$\left| {\matrix{
{{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr
{1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr
{{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr
} } \right|$$
<br/><br/>Then the ordered pair (m, M) is equal to : | [{"identifier": "A", "content": "(\u20133, \u20131)"}, {"identifier": "B", "content": "(\u20134, \u20131)"}, {"identifier": "C", "content": "(1, 3)"}, {"identifier": "D", "content": "(\u20133, 3)"}] | ["A"] | null | $$\left| {\matrix{
{{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr
{1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr
{{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr
} } \right|$$
<br><br>R<sub>1</sub> $$ \to $$ R<sub>1</sub> – R<sub>2</sub>, R<sub>2</sub> $$ \to $$ R<sub>2</sub> – R<sub>3</sub>
<br><br>$$\left| {\matrix{
{ - 1} & 1 & 0 \cr
1 & 0 & { - 1} \cr
{{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr
} } \right|$$
<br><br>= –1(sin<sup>2</sup>
x) – 1(1 + sin2x + cos<sup>2</sup>
x)
<br><br>= - sin2x - 2
<br><br>$$ \therefore $$ minimum value when sin2x = 1
<br><br>m = - 2 - 1 = -3
<br><br>$$ \therefore $$ Maximum value when sin2x = –1
<br><br> M = -2 + 1 = -1
<br><br>$$ \therefore $$ (m, M) = (–3, –1) | mcq | jee-main-2020-online-6th-september-morning-slot |
iTnghvwMsdwXFONHFgjgy2xukfg6qaw0 | maths | matrices-and-determinants | expansion-of-determinant | If the minimum and the maximum values of the function $$f:\left[ {{\pi \over 4},{\pi \over 2}} \right] \to R$$, defined by <br/>
$$f\left( \theta \right) = \left| {\matrix{
{ - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr
{ - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr
{12} & {10} & { - 2} \cr
} } \right|$$ are m and M respectively, then the ordered pair (m,M) is
equal to :
| [{"identifier": "A", "content": "$$\\left( {0,2\\sqrt 2 } \\right)$$"}, {"identifier": "B", "content": "(-4, 0)\n"}, {"identifier": "C", "content": "(-4, 4)"}, {"identifier": "D", "content": "(0, 4)"}] | ["B"] | null | Given <br>$$f\left( \theta \right) = \left| {\matrix{
{ - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr
{ - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr
{12} & {10} & { - 2} \cr
} } \right|$$
<br><br>C<sub>1</sub> $$ \to $$ C<sub>1</sub>
– C<sub>2</sub>
, C<sub>3</sub> $$ \to $$ C<sub>3</sub>
+ C<sub>2</sub>
<br><br>= $$\left| {\matrix{
1 & { - 1 - {{\sin }^2}\theta } & { - {{\sin }^2}\theta } \cr
1 & { - 1 - {{\cos }^2}\theta } & { - {{\cos }^2}\theta } \cr
2 & {10} & 8 \cr
} } \right|$$
<br><br>C<sub>2</sub> $$ \to $$ C<sub>2</sub>
– C<sub>3</sub>
<br><br>= $$\left| {\matrix{
1 & { - 1} & { - {{\sin }^2}\theta } \cr
1 & { - 1} & { - {{\cos }^2}\theta } \cr
2 & 2 & 8 \cr
} } \right|$$
<br><br>= 1(2cos<sup>2</sup>$$\theta $$ – 8) + (8 + 2cos<sup>2</sup>$$\theta $$) – 4sin<sup>2</sup>$$\theta $$
<br><br>= 4cos<sup>2</sup>$$\theta $$ - 4cos<sup>2</sup>$$\theta $$
<br><br>= 4 cos 2$$\theta $$
<br><br>$$\theta $$ $$ \in $$ $$\left[ {{\pi \over 4},{\pi \over 2}} \right]$$
<br><br>$$ \Rightarrow $$ 2$$\theta $$ $$ \in $$ $$\left[ {{\pi \over 2},{\pi }} \right]$$
<br><br>$$ \Rightarrow $$ cos 2$$\theta $$ $$ \in $$ [-1, 0]
<br><br>$$ \Rightarrow $$ 4cos 2$$\theta $$ $$ \in $$ [-4, 0]
<br><br>$$ \Rightarrow $$ $$f\left( \theta \right)$$ $$ \in $$ [-4, 0]
<br><br>$$ \therefore $$ (m, M) = (–4, 0) | mcq | jee-main-2020-online-5th-september-morning-slot |
ySPeAUbfr63EfeVlQdjgy2xukfqb5abj | maths | matrices-and-determinants | expansion-of-determinant | If a + x = b + y = c + z + 1, where a, b, c, x, y, z<br/>
are non-zero distinct real numbers, then
<br/>$$\left| {\matrix{
x & {a + y} & {x + a} \cr
y & {b + y} & {y + b} \cr
z & {c + y} & {z + c} \cr
} } \right|$$ is equal to : | [{"identifier": "A", "content": "y(b \u2013 a)"}, {"identifier": "B", "content": "y(a \u2013 b)"}, {"identifier": "C", "content": "y(a \u2013 c)"}, {"identifier": "D", "content": "0"}] | ["B"] | null | $$\left| {\matrix{
x & {a + y} & {x + a} \cr
y & {b + y} & {y + b} \cr
z & {c + y} & {z + c} \cr
} } \right|$$
<br><br>C<sub>3</sub> $$ \to $$ C<sub>3</sub> – C<sub>1</sub>
<br><br>= $$\left| {\matrix{
x & {a + y} & a \cr
y & {b + y} & b \cr
z & {c + y} & c \cr
} } \right|$$
<br><br>C<sub>2</sub> $$ \to $$ C<sub>2</sub> – C<sub>3</sub>
<br><br>= $$\left| {\matrix{
x & y & a \cr
y & y & b \cr
z & y & c \cr
} } \right|$$
<br><br>R<sub>3</sub> $$ \to $$ R<sub>3</sub> – R<sub>1</sub>, R<sub>2</sub> $$ \to $$ R<sub>2</sub> – R<sub>1</sub>
<br><br>= $$\left| {\matrix{
x & y & a \cr
{y - x} & 0 & {b - a} \cr
{z - x} & 0 & {c - a} \cr
} } \right|$$
<br><br>= (–y)[(y – x) (c – a) – (b – a) (z – x)]
<br><br>Given, a + x = b + y = c + z + 1
<br><br>= (–y)[(a – b) (c – a) + (a – b) (a – c – 1)]
<br><br>= (–y)[(a – b) (c – a) + (a – b) (a – c) + b – a)
<br><br>= –y(b – a) = y(a – b) | mcq | jee-main-2020-online-5th-september-evening-slot |
t8IVACMgbcUbnsFcWz1klt7w2kq | maths | matrices-and-determinants | expansion-of-determinant | Let A be a 3 $$\times$$ 3 matrix with det(A) = 4. Let R<sub>i</sub> denote the i<sup>th</sup> row of A. If a matrix B is obtained by performing the operation R<sub>2</sub> $$ \to $$ 2R<sub>2</sub> + 5R<sub>3</sub> on 2A, then det(B) is equal to : | [{"identifier": "A", "content": "64"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "128"}, {"identifier": "D", "content": "80"}] | ["A"] | null | $$A = \left[ {\matrix{
{{R_{11}}} & {{R_{12}}} & {{R_{13}}} \cr
{{R_{21}}} & {{R_{22}}} & {{R_{23}}} \cr
{{R_{31}}} & {{R_{32}}} & {{R_{33}}} \cr
} } \right]$$<br><br>$$2A = \left[ {\matrix{
{2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \cr
{2{R_{21}}} & {2{R_{22}}} & {2{R_{23}}} \cr
{2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \cr
} } \right]$$<br><br>$${R_2} \to 2{R_2} + 5{R_3}$$<br><br>$$B = \left[ {\matrix{
{2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \cr
{4{R_{21}} + 10{R_{31}}} & {4{R_{22}} + 10{R_{32}}} & {4{R_{23}} + 10{R_{33}}} \cr
{2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \cr
} } \right]$$<br><br>$${R_2} \to {R_2} - 5{R_3}$$<br><br>$$B = \left[ {\matrix{
{2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \cr
{4{R_{21}}} & {4{R_{22}}} & {4{R_{23}}} \cr
{2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \cr
} } \right]$$<br><br>$$\left| B \right| = \left[ {\matrix{
{2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \cr
{4{R_{21}}} & {4{R_{22}}} & {4{R_{23}}} \cr
{2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \cr
} } \right]$$<br><br>$$\left| B \right| = 2 \times 2 \times 4\left| {\matrix{
{{R_{11}}} & {{R_{12}}} & {{R_{13}}} \cr
{{R_{21}}} & {{R_{22}}} & {{R_{23}}} \cr
{{R_{31}}} & {{R_{32}}} & {{R_{33}}} \cr
} } \right|$$<br><br>$$ = 16 \times 4$$<br><br>$$ = 64$$ | mcq | jee-main-2021-online-25th-february-evening-slot |
5wV2M9p2j0ePQcEpT21klugwy2o | maths | matrices-and-determinants | expansion-of-determinant | The value of $$\left| {\matrix{
{(a + 1)(a + 2)} & {a + 2} & 1 \cr
{(a + 2)(a + 3)} & {a + 3} & 1 \cr
{(a + 3)(a + 4)} & {a + 4} & 1 \cr
} } \right|$$ is : | [{"identifier": "A", "content": "$$-$$2"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "(a + 2)(a + 3)(a + 4)"}, {"identifier": "D", "content": "(a + 1)(a + 2)(a + 3)"}] | ["A"] | null | Given, $$\Delta $$ = $$\left| {\matrix{
{(a + 1)(a + 2)} & {a + 2} & 1 \cr
{(a + 2)(a + 3)} & {a + 3} & 1 \cr
{(a + 3)(a + 4)} & {a + 4} & 1 \cr
} } \right|$$
<br><br>R<sub>2</sub> $$ \to $$ R<sub>2</sub> $$-$$ R<sub>1</sub> and R<sub>3</sub> $$ \to $$ R<sub>3</sub> $$-$$ R<sub>1</sub><br><br>$$\Delta$$ = $$\left| {\matrix{
{(a + 1)(a + 2)} & {a + 2} & 1 \cr
{(a + 2)(a + 3 - a - 1)} & 1 & 0 \cr
{{a^2} + 7a + 12 - {a^2} - 3a - 2} & 2 & 0 \cr
} } \right|$$<br><br>$$ = \left| {\matrix{
{{a^2} + 3a + 2} & {a + 2} & 1 \cr
{2(a + 2)} & 1 & 0 \cr
{4a + 10} & 2 & 0 \cr
} } \right|$$<br><br>$$ = 4(a + 2) - 4a - 10$$<br><br>$$ = 4a + 8 - 4a - 10 = - 2$$
| mcq | jee-main-2021-online-26th-february-morning-slot |
A4GUzTtASNdDS8C3KA1kmknd2bm | maths | matrices-and-determinants | expansion-of-determinant | If x, y, z are in arithmetic progression with common difference d, x $$\ne$$ 3d, and the determinant of the matrix $$\left[ {\matrix{
3 & {4\sqrt 2 } & x \cr
4 & {5\sqrt 2 } & y \cr
5 & k & z \cr
} } \right]$$ is zero, then the value of k<sup>2</sup> is : | [{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "6"}] | ["A"] | null | $$\left| {\matrix{
3 & {4\sqrt 2 } & x \cr
4 & {5\sqrt 2 } & y \cr
5 & k & z \cr
} } \right| = 0$$<br><br>$${R_1} \to {R_1} + {R_3} - 2{R_2}$$<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
0 & {4\sqrt 2 - k - 10\sqrt 2 } & 0 \cr
4 & {5\sqrt 2 } & y \cr
5 & k & z \cr
} } \right| = 0$$ { $$ \because $$ 2y = x + z}<br><br>$$ \Rightarrow (k - 6\sqrt 2 )(4z - 5y) = 0$$<br><br>$$ \Rightarrow $$ k = $$6\sqrt 2 $$ or 4z = 5y (Not possible $$ \because $$ x, y, z in A.P.)<br><br>So, k<sup>2</sup> = 72<br><br>$$ \therefore $$ Option (A) | mcq | jee-main-2021-online-17th-march-evening-shift |
J4wCNBxubchEpe31cN1kmko0m1s | maths | matrices-and-determinants | expansion-of-determinant | If 1, log<sub>10</sub>(4<sup>x</sup> $$-$$ 2) and log<sub>10</sub>$$\left( {{4^x} + {{18} \over 5}} \right)$$ are in arithmetic progression for a real number x, then the value of the determinant $$\left| {\matrix{
{2\left( {x - {1 \over 2}} \right)} & {x - 1} & {{x^2}} \cr
1 & 0 & x \cr
x & 1 & 0 \cr
} } \right|$$ is equal to : | [] | null | 2 | 1, $$lo{g_{10}}({4^x} - 2),\,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$$ in AP.<br><br>$$ \therefore $$ 2$$ \times $$$$lo{g_{10}}({4^x} - 2) = 1 + \,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$$ <br><br>$$lo{g_{10}}{({4^x} - 2)^2} = \,lo{g_{10}}\left( {10.\left( {{4^x} + {{18} \over 5}} \right)} \right)$$<br><br>$${({4^x} - 2)^2} = 10.\left( {{4^x} + {{18} \over 5}} \right)$$<br><br>$${({4^x})^2} + 4 - {4.4^x} = {10.4^x} + 36$$<br><br>$${({4^x})^2} - {14.4^x} - 32 = 0$$<br><br>$${({4^x})^2} + {2.4^x} - {16.4^x} - 32 = 0$$<br><br>$${4^x}({4^x} + 2) - 16.({4^x} + 2) = 0$$<br><br>$$({4^x} + 2)({4^x} - 16) = 0$$<br><br>4<sup>x</sup> = -2 (Not Possible)
<br><br>Or 4<sup>x</sup> = 16
<br><br>$$ \Rightarrow $$ x = 2<br><br>Therefore $$\left| {\matrix{
{2(x - 1/2)} & {x - 1} & {{x^2}} \cr
1 & 0 & x \cr
x & 1 & 0 \cr
} } \right|$$<br><br>$$ = \left| {\matrix{
3 & 1 & 4 \cr
1 & 0 & 2 \cr
2 & 1 & 0 \cr
} } \right|$$<br><br>$$ = 3( - 2) - 1(0 - 4) + 4(1 - 0)$$<br><br>$$ = - 6 + 4 + 4 = 2$$ | integer | jee-main-2021-online-17th-march-evening-shift |
0v6oV602XmCfWdoQV31kmli3lt0 | maths | matrices-and-determinants | expansion-of-determinant | The solutions of the equation $$\left| {\matrix{
{1 + {{\sin }^2}x} & {{{\sin }^2}x} & {{{\sin }^2}x} \cr
{{{\cos }^2}x} & {1 + {{\cos }^2}x} & {{{\cos }^2}x} \cr
{4\sin 2x} & {4\sin 2x} & {1 + 4\sin 2x} \cr
} } \right| = 0,(0 < x < \pi )$$, are | [{"identifier": "A", "content": "$${\\pi \\over {12}},{\\pi \\over 6}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6},{{5\\pi } \\over 6}$$"}, {"identifier": "C", "content": "$${{5\\pi } \\over {12}},{{7\\pi } \\over {12}}$$"}, {"identifier": "D", "content": "$${{7\\pi } \\over {12}},{{11\\pi } \\over {12}}$$"}] | ["D"] | null | By using C<sub>1</sub> $$ \to $$ C<sub>1</sub> $$-$$ C<sub>2</sub> and C<sub>3</sub> $$ \to $$ C<sub>3</sub> $$-$$ C<sub>2</sub> we get<br><br>$$\left| {\matrix{
1 & {{{\sin }^2}x} & 0 \cr
{ - 1} & {1 + {{\cos }^2}x} & { - 1} \cr
0 & {4\sin 2x} & 1 \cr
} } \right| = 0$$<br><br>Expanding by R<sub>1</sub> we get<br><br>$$1(1 + {\cos ^2}x + 4\sin 2x) - {\sin ^2}x( - 1) = 0$$<br><br>$$ \Rightarrow 2 + 4\sin 2x = 0$$<br><br>$$ \Rightarrow \sin 2x = {{ - 1} \over 2}$$<br><br>$$ \Rightarrow 2x = n\pi + {( - 1)^n}\left( {{{ - \pi } \over 6}} \right),n \in Z$$<br><br>$$ \therefore $$ $$2x = {{7\pi } \over 6},{{11\pi } \over 6} $$<br><br>$$\Rightarrow x = {{7\pi } \over {12}},{{11\pi } \over 2}$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
b29ZvX12aHiapPJUir1kmm46ruq | maths | matrices-and-determinants | expansion-of-determinant | Let I be an identity matrix of order 2 $$\times$$ 2 and P = $$\left[ {\matrix{
2 & { - 1} \cr
5 & { - 3} \cr
} } \right]$$. Then the value of n$$\in$$N for which P<sup>n</sup> = 5I $$-$$ 8P is equal to ____________. | [] | null | 6 | $$P = \left[ {\matrix{
2 & { - 1} \cr
5 & { - 3} \cr
} } \right]$$<br><br>$$\left| {\matrix{
{2 - \lambda } & { - 1} \cr
5 & { - 3 - \lambda } \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ $$\lambda$$<sup>2</sup> + $$\lambda$$ $$-$$ 1 = 0<br><br>$$ \Rightarrow $$ P<sup>2</sup> + P $$-$$ I = 0<br><br>$$ \Rightarrow $$ P<sup>2</sup> = I $$-$$ P<br><br>$$ \Rightarrow $$ P<sup>4</sup> = I + P<sup>2</sup> $$-$$ 2P<br><br>$$ \Rightarrow $$ P<sup>4</sup> = 2I $$-$$ 3P<br><br>Now, P<sup>4</sup> . P<sup>2</sup> = (2I $$-$$ 3P)(I $$-$$ P) = 2I $$-$$ 5P + 3P<sup>2</sup><br><br>$$ \Rightarrow $$ P<sup>6</sup> = 5I $$-$$ 8P<br><br>So n = 6 | integer | jee-main-2021-online-18th-march-evening-shift |
1krq1a1sm | maths | matrices-and-determinants | expansion-of-determinant | Let a, b, c, d in arithmetic progression with common difference $$\lambda$$. If $$\left| {\matrix{
{x + a - c} & {x + b} & {x + a} \cr
{x - 1} & {x + c} & {x + b} \cr
{x - b + d} & {x + d} & {x + c} \cr
} } \right| = 2$$, then value of $$\lambda$$<sup>2</sup> is equal to ________________. | [] | null | 1 | $$\left| {\matrix{
{x + a - c} & {x + b} & {x + a} \cr
{x - 1} & {x + c} & {x + b} \cr
{x - b + d} & {x + d} & {x + c} \cr
} } \right| = 2$$<br><br>$${C_2} \to {C_2} - {C_3}$$<br><br>$$ \Rightarrow \left| {\matrix{
{x - 2\lambda } & \lambda & {x + a} \cr
{x - 1} & \lambda & {x + b} \cr
{x + 2\lambda } & \lambda & {x + c} \cr
} } \right| = 2$$<br><br>$${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$$<br><br>$$ \Rightarrow \left| {\matrix{
{x - 2\lambda } & 1 & {x + a} \cr
{2\lambda - 1} & 0 & \lambda \cr
{4\lambda } & 0 & {2\lambda } \cr
} } \right| = 2$$<br><br>$$ \Rightarrow 1(4{\lambda ^2} - 4{\lambda ^2} + 2\lambda ) = 2$$<br><br>$$ \Rightarrow {\lambda ^2} = 1$$ | integer | jee-main-2021-online-20th-july-morning-shift |
1krzmszlz | maths | matrices-and-determinants | expansion-of-determinant | The number of distinct real roots <br/><br/>of $$\left| {\matrix{
{\sin x} & {\cos x} & {\cos x} \cr
{\cos x} & {\sin x} & {\cos x} \cr
{\cos x} & {\cos x} & {\sin x} \cr
} } \right| = 0$$ in the interval $$ - {\pi \over 4} \le x \le {\pi \over 4}$$ is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["B"] | null | $$\left| {\matrix{
{\sin x} & {\cos x} & {\cos x} \cr
{\cos x} & {\sin x} & {\cos x} \cr
{\cos x} & {\cos x} & {\sin x} \cr
} } \right| = 0, - {\pi \over 4} \le x \le {\pi \over 4}$$<br><br>Apply : $${R_1} \to {R_1} - {R_2}$$ & $${R_2} \to {R_2} - {R_3}$$<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
{\sin x - \cos x} & {\cos x - \sin x} & 0 \cr
0 & {\sin x - \cos x} & {\cos x - \sin x} \cr
{\cos x} & {\cos x} & {\sin x} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ $${(\sin x - \cos x)^2}\left| {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
{\cos x} & {\cos x} & {\sin x} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow $$ $${(\sin x - \cos x)^2}(\sin x + 2\cos x) = 0$$<br><br>$$\therefore$$ $$x = {\pi \over 4}$$ | mcq | jee-main-2021-online-25th-july-evening-shift |
1ks0ch3dj | maths | matrices-and-determinants | expansion-of-determinant | Let $$f(x) = \left| {\matrix{
{{{\sin }^2}x} & { - 2 + {{\cos }^2}x} & {\cos 2x} \cr
{2 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \cr
} } \right|,x \in [0,\pi ]$$. Then the maximum value of f(x) is equal to ______________. | [] | null | 6 | $$\left| {\matrix{
{ - 2} & { - 2} & 0 \cr
2 & 0 & { - 1} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \cr
} } \right|\left( \matrix{
{R_1} \to {R_1} - {R_2} \hfill \cr
\& \,{R_2} \to {R_2} - {R_3} \hfill \cr} \right)$$<br><br>= $$ - 2({\cos ^2}x) + 2(2 + 2\cos 2x + {\sin ^2}x)$$<br><br>= $$4 + 4\cos 2x - 2({\cos ^2}x - {\sin ^2}x)$$<br><br>$$ \therefore $$ $$f(x) = 4 + \underbrace {2\cos 2x}_{\max = 1}$$<br><br>$$ \Rightarrow $$ $$f{(x)_{\max }} = 4 + 2 = 6$$ | integer | jee-main-2021-online-27th-july-morning-shift |
1ktfw40qs | maths | matrices-and-determinants | expansion-of-determinant | Let $$A = \left( {\matrix{
{[x + 1]} & {[x + 2]} & {[x + 3]} \cr
{[x]} & {[x + 3]} & {[x + 3]} \cr
{[x]} & {[x + 2]} & {[x + 4]} \cr
} } \right)$$, where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval : | [{"identifier": "A", "content": "[68, 69)"}, {"identifier": "B", "content": "[62, 63)"}, {"identifier": "C", "content": "[65, 66)"}, {"identifier": "D", "content": "[60, 61)"}] | ["B"] | null | $$\left| {\matrix{
{[x + 1]} & {[x + 2]} & {[x + 3]} \cr
{[x]} & {[x + 3]} & {[x + 3]} \cr
{[x]} & {[x + 2]} & {[x + 4]} \cr
} } \right| = 192$$<br><br>R<sub>1</sub> $$\to$$ R<sub>1</sub> $$-$$ R<sub>3</sub> & R<sub>2</sub> $$\to$$ R<sub>2</sub> $$-$$ R<sub>3</sub><br><br>$$\left[ {\matrix{
1 & 0 & { - 1} \cr
0 & 1 & { - 1} \cr
{[x]} & {[x] + 2} & {[x] + 4} \cr
} } \right] = 192$$<br><br>$$2[x] + 6 + [x] = 192 \Rightarrow [x] = 62$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktirfw8s | maths | matrices-and-determinants | expansion-of-determinant | If $${a_r} = \cos {{2r\pi } \over 9} + i\sin {{2r\pi } \over 9}$$, r = 1, 2, 3, ....., i = $$\sqrt { - 1} $$, then<br/> the determinant $$\left| {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right|$$ is equal to : | [{"identifier": "A", "content": "a<sub>2</sub>a<sub>6</sub> $$-$$ a<sub>4</sub>a<sub>8</sub>"}, {"identifier": "B", "content": "a<sub>9</sub>"}, {"identifier": "C", "content": "a<sub>1</sub>a<sub>9</sub> $$-$$ a<sub>3</sub>a<sub>7</sub>"}, {"identifier": "D", "content": "a<sub>5</sub>"}] | ["C"] | null | $${a_r} = {e^{{{i2\pi r} \over 9}}}$$, r = 1, 2, 3, ......, a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... are in G.P.<br><br>$$\left| {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_n}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right| = \left| {\matrix{
{{a_1}} & {a_1^2} & {a_1^3} \cr
{a_1^4} & {a_1^5} & {a_1^6} \cr
{a_1^7} & {a_1^8} & {a_1^9} \cr
} } \right| $$
<br><br>$$= {a_1}\,.\,a_1^4\,.\,a_1^7\left| {\matrix{
1 & {{a_1}} & {a_1^2} \cr
1 & {{a_1}} & {a_1^2} \cr
1 & {{a_1}} & {a_1^2} \cr
} } \right| = 0$$<br><br>Now, $${a_1}{a_9} - {a_3}{a_7} = {a_1}^{10} - {a_1}^{10} = 0$$ | mcq | jee-main-2021-online-31st-august-morning-shift |
1l5vzbklz | maths | matrices-and-determinants | expansion-of-determinant | <p>Let $$A = \left[ {\matrix{
1 & { - 2} & \alpha \cr
\alpha & 2 & { - 1} \cr
} } \right]$$ and $$B = \left[ {\matrix{
2 & \alpha \cr
{ - 1} & 2 \cr
4 & { - 5} \cr
} } \right],\,\alpha \in C$$. Then the absolute value of the sum of all values of $$\alpha$$ for which det(AB) = 0 is :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "5"}] | ["A"] | null | <p>Given,</p>
<p>$$A = \left[ {\matrix{
1 & { - 2} & \alpha \cr
\alpha & 2 & { - 1} \cr
} } \right]$$</p>
<p>and $$B = \left[ {\matrix{
2 & \alpha \cr
{ - 1} & 2 \cr
4 & { - 5} \cr
} } \right]$$</p>
<p>$$AB = \left[ {\matrix{
1 & { - 2} & \alpha \cr
\alpha & 2 & { - 1} \cr
} } \right]\left[ {\matrix{
2 & \alpha \cr
{ - 1} & 2 \cr
4 & { - 5} \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
{4 + 4\alpha } & { - 4\alpha - 4} \cr
{2\alpha - 6} & {{\alpha ^2} + 9} \cr
} } \right]$$</p>
<p>Given,</p>
<p>$$|AB| = 0$$</p>
<p>$$\therefore$$ $$\left| {\matrix{
{4 + 4\alpha } & { - 4\alpha - 4} \cr
{2\alpha - 6} & {{\alpha ^2} + 9} \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow (4\alpha + 4)\left| {\matrix{
1 & { - 1} \cr
{2\alpha - 6} & {{\alpha ^2} + 9} \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow (4\alpha + 4)({\alpha ^2} + 9 + 2\alpha - 6) = 0$$</p>
<p>$$ \Rightarrow (4\alpha + 4)({\alpha ^2} + 2\alpha + 3) = 0$$</p>
<p>$$\therefore$$ $$\alpha - = - 1$$</p>
<p>or $${\alpha ^2} + 2\alpha + 3 = 0$$</p>
<p>$${\alpha _1} + {\alpha _2} = - 2$$</p>
<p>$$\therefore$$ Sum of all values of $$\alpha = - 1 - 2 = - 3$$</p>
<p>$$\therefore$$ Absolute value of $$\alpha = | - 3| = 3$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6p3o4ow | maths | matrices-and-determinants | expansion-of-determinant | <p>Let p and p + 2 be prime numbers and let</p>
<p>$$
\Delta=\left|\begin{array}{ccc}
\mathrm{p} ! & (\mathrm{p}+1) ! & (\mathrm{p}+2) ! \\
(\mathrm{p}+1) ! & (\mathrm{p}+2) ! & (\mathrm{p}+3) ! \\
(\mathrm{p}+2) ! & (\mathrm{p}+3) ! & (\mathrm{p}+4) !
\end{array}\right|
$$</p>
<p>Then the sum of the maximum values of $$\alpha$$ and $$\beta$$, such that $$\mathrm{p}^{\alpha}$$ and $$(\mathrm{p}+2)^{\beta}$$ divide $$\Delta$$, is __________.</p> | [] | null | 4 | <p>$$\Delta = \left| {\matrix{
{p!} & {(p + 1)!} & {(p + 2)!} \cr
{(p + 1)!} & {(p + 2)!} & {(p + 3)!} \cr
{(p + 2)!} & {(p + 3)!} & {(p + 4)!} \cr
} } \right|$$</p>
<p>$$ = p!\,.\,(p + 1)!\,.\,(p + 2)!\left| {\matrix{
1 & {(p + 1)} & {(p + 1)(p + 2)} \cr
1 & {(p + 2)} & {(p + 2)(p + 3)} \cr
1 & {(p + 3)} & {(p + 3)(p + 4)} \cr
} } \right|$$</p>
<p>$$ = p!\,.\,(p + 1)!\,.\,(p + 2)!\left| {\matrix{
1 & {p + 1} & {{p^2} + 3p + 2} \cr
0 & 1 & {2p + 4} \cr
0 & 1 & {2p + 6} \cr
} } \right|$$</p>
<p>$$ = 2(p!)\,.\,\left( {(p + 1)!} \right)\,.\,\left( {(p + 2)!} \right)$$</p>
<p>$$ = 2(p + 1)\,.\,{(p!)^2}\,.\,\left( {(p + 2)!} \right)$$</p>
<p>$$ = 2{(p + 1)^2}\,.\,{(p!)^3}\,.\,\left( {(p + 2)!} \right)$$</p>
<p>$$\therefore$$ Maximum value of $$\alpha$$ is 3 and $$\beta$$ is 1.</p>
<p>$$\therefore$$ $$\alpha + \beta = 4$$</p> | integer | jee-main-2022-online-29th-july-morning-shift |
1ldv34755 | maths | matrices-and-determinants | expansion-of-determinant | <p>Let $$\mathrm{A_1,A_2,A_3}$$ be the three A.P. with the same common difference d and having their first terms as $$\mathrm{A,A+1,A+2}$$, respectively. Let a, b, c be the $$\mathrm{7^{th},9^{th},17^{th}}$$ terms of $$\mathrm{A_1,A_2,A_3}$$, respective such that $$\left| {\matrix{
a & 7 & 1 \cr
{2b} & {17} & 1 \cr
c & {17} & 1 \cr
} } \right| + 70 = 0$$.</p>
<p>If $$a=29$$, then the sum of first 20 terms of an AP whose first term is $$c-a-b$$ and common difference is $$\frac{d}{12}$$, is equal to ___________.</p> | [] | null | 495 | $a=A+6 d$
<br/><br/>
$$
\begin{aligned}
& b=A+8 d+1 \\\\
& c=A+16 d+2 \\\\
& \left|\begin{array}{ccc}
a & 7 & 1 \\
26 & 17 & 1 \\
c & 17 & 1
\end{array}\right|=-70 \\\\
& \Rightarrow\left|\begin{array}{ccc}
A+6 d & 7 & 1 \\
2 A+16 d+2 & 17 & 1 \\
A+16 d+2 & 17 & 1
\end{array}\right|=-70 \\\\
& R_{3} \rightarrow R_{3}-R_{2}, \quad R_{2} \rightarrow R_{2}-R_{1} \\\\
& \Rightarrow\left|\begin{array}{ccc}
A+6 d & 7 & 1 \\
A+10 d+2 & 10 & 0 \\
-A & 0 & 0
\end{array}\right|=-70
\end{aligned}
$$
<br/><br/>
$$
\begin{aligned}
\Rightarrow \quad & A=-7 \\\\
& a=A+6 d=29 \Rightarrow d=6 \\\\
& b=-7+48+1=42 \\\\
& c=-7+96+2=91 \\\\
& c-a-b=91-29-42=20 \\\\
& \text { Sum }=\frac{20}{2}\left[2 \times 20+19 \times \frac{6}{12}\right]=10\left[40+\frac{19}{2}\right]=495
\end{aligned}
$$ | integer | jee-main-2023-online-25th-january-morning-shift |
1lgrghmb3 | maths | matrices-and-determinants | expansion-of-determinant | <p>Let $$\mathrm{D}_{\mathrm{k}}=\left|\begin{array}{ccc}1 & 2 k & 2 k-1 \\
n & n^{2}+n+2 & n^{2} \\
n & n^{2}+n & n^{2}+n+2\end{array}\right|$$. If $$\sum_\limits{k=1}^{n} \mathrm{D}_{\mathrm{k}}=96$$, then $$n$$ is equal to _____________.</p> | [] | null | 6 | $$
\begin{aligned}
& \sum_{k=1}^n D_k=\left|\begin{array}{ccc}
\sum 1 & 2 \sum k & 2 \sum k-\sum 1 \\
n & n^2+n+2 & n^2 \\
n & n^2+n & n^2+n+2
\end{array}\right| \\\\
& =\left|\begin{array}{ccc}
n & n(n+1) & n^2 \\
n & n^2+n+2 & n^2 \\
n & n^2+n & n^2+n+2
\end{array}\right| \\\\
& =\left|\begin{array}{ccc}
0 & -2 & 0 \\
0 & 2 & -n-2 \\
n & n^2+n & n^2+n+2
\end{array}\right| \\\\
& =2((-n)(-n-2))=96 \\\\
& \Rightarrow n^2+2 n=48 \\\\
& \Rightarrow n=6,-8 \\\\
& \Rightarrow n=6
\end{aligned}
$$ | integer | jee-main-2023-online-12th-april-morning-shift |
1lgsverwb | maths | matrices-and-determinants | expansion-of-determinant | <p>$$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$$, then $$\lambda, \frac{\lambda}{3}$$ are the roots of the equation :</p> | [{"identifier": "A", "content": "$$4 x^{2}+24 x-27=0$$"}, {"identifier": "B", "content": "$$4 x^{2}-24 x+27=0$$"}, {"identifier": "C", "content": "$$4 x^{2}-24 x-27=0$$"}, {"identifier": "D", "content": "$$4 x^{2}+24 x+27=0$$"}] | ["B"] | null | $$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$$
<br/><br/>Put $x=0$
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{ccc}
1 & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda^2
\end{array}\right|=\frac{9}{8} \times 81 \\\\
& \lambda^3=\frac{3^6}{2^3}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \lambda=\frac{9}{2} \\\\
& \Rightarrow \frac{\lambda}{3}=\frac{3}{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Equation: } x^2-\left(\frac{9}{2}+\frac{3}{2}\right) x+\frac{9}{2} \times \frac{3}{2}=0 \\\\
& \Rightarrow 4 x^2-24 x+27=0
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-evening-shift |
jaoe38c1lscntluu | maths | matrices-and-determinants | expansion-of-determinant | <p>The values of $$\alpha$$, for which $$\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$$, lie in the interval</p> | [{"identifier": "A", "content": "$$(-2,1)$$\n"}, {"identifier": "B", "content": "$$\\left(-\\frac{3}{2}, \\frac{3}{2}\\right)$$\n"}, {"identifier": "C", "content": "$$(-3,0)$$\n"}, {"identifier": "D", "content": "$$(0,3)$$"}] | ["C"] | null | <p>$$\left|\begin{array}{ccc}
1 & \frac{3}{2} & \alpha+\frac{3}{2} \\
1 & \frac{1}{3} & \alpha+\frac{1}{3} \\
2 \alpha+3 & 3 \alpha+1 & 0
\end{array}\right|=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0 \\
& \Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0 \\
& \Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0 \\
& \Rightarrow 2 \alpha^2+6 \alpha+1=0 \\
& \Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2}
\end{aligned}$$</p>
<p>Hence option (C) is correct.</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lseyky4c | maths | matrices-and-determinants | expansion-of-determinant | <p>$$\text { Let } A=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & \alpha & \beta \\
0 & \beta & \alpha
\end{array}\right] \text { and }|2 \mathrm{~A}|^3=2^{21} \text { where } \alpha, \beta \in Z \text {, Then a value of } \alpha \text { is }$$</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "17"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["D"] | null | <p>$$\begin{aligned}
& |\mathrm{A}|=\alpha^2-\beta^2 \\
& |2 \mathrm{~A}|^3=2^{21} \Rightarrow|\mathrm{A}|=2^4 \\
& \alpha^2-\beta^2=16 \\
& (\alpha+\beta)(\alpha-\beta)=16 \Rightarrow \alpha=4 \text { or } 5
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
lvc57b13 | maths | matrices-and-determinants | expansion-of-determinant | <p>For $$\alpha, \beta \in \mathbb{R}$$ and a natural number $$n$$, let $$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$$. Then $$2 A_{10}-A_8$$ is</p> | [{"identifier": "A", "content": "$$4 \\alpha+2 \\beta$$\n"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$2 n$$\n"}, {"identifier": "D", "content": "$$2 \\alpha+4 \\beta$$"}] | ["A"] | null | <p>$$A_r=\left|\begin{array}{ccc}
r & 1 & \frac{n^2}{2}+\alpha \\
2 r & 2 & n^2-\beta \\
3 r-2 & 3 & \frac{n(3 n-1)}{2}
\end{array}\right|$$</p>
<p>$${A_r} = 2\left| {\matrix{
r & 1 & {{{{n^2}} \over 2} + \alpha } \cr
{2r} & 2 & {{{{n^2}} \over 2} - \beta } \cr
{3r - 2} & 3 & {{{n(3n - 1)} \over 2}} \cr
} } \right|$$</p>
<p>$$R_1 \rightarrow R_1-R_2$$</p>
<p>$$ = 2\left| {\matrix{
0 & 1 & {\alpha + {\beta \over 2}} \cr
r & 2 & {{{{n^2}} \over 2} - \beta } \cr
{3r - 2} & 3 & {{{n(3n - 1)} \over 2}} \cr
} } \right|$$</p>
<p>$$\begin{aligned}
& =2\left(\alpha+\frac{\beta}{2}\right)(3 r-3 r+2) \\
& A_r=4 \alpha+2 \beta \\
& 2 A_{10}-A_8=4 \alpha+2 \beta
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
L5OkUKHe1Wo61y8B | maths | matrices-and-determinants | inverse-of-a-matrix | Let $$A = \left( {\matrix{
1 & { - 1} & 1 \cr
2 & 1 & { - 3} \cr
1 & 1 & 1 \cr
} } \right).$$ and $$10$$ $$B = \left( {\matrix{
4 & 2 & 2 \cr
{ - 5} & 0 & \alpha \cr
1 & { - 2} & 3 \cr
} } \right)$$. if $$B$$ is
<p>the inverse of matrix $$A$$, then $$\alpha $$ is</p> | [{"identifier": "A", "content": "$$5$$"}, {"identifier": "B", "content": "$$-1$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$-2$$"}] | ["A"] | null | Given that $$10B$$ $$\,\,\, = \left[ {\matrix{
4 & 2 & 2 \cr
{ - 5} & 0 & \alpha \cr
1 & { - 2} & 3 \cr
} } \right]$$
<br><br>$$ \Rightarrow B = {1 \over {10}}\left[ {\matrix{
4 & 2 & 2 \cr
{ - 5} & 0 & \alpha \cr
1 & { - 2} & 3 \cr
} } \right]$$
<br><br>Also since, $$B = {A^{ - 1}} \Rightarrow AB = I$$
<br><br>$$ \Rightarrow {1 \over {10}}\left[ {\matrix{
1 & { - 1} & 1 \cr
2 & 1 & { - 3} \cr
1 & 1 & 1 \cr
} } \right]\left[ {\matrix{
4 & 2 & 2 \cr
{ - 5} & 0 & \alpha \cr
1 & { - 2} & 3 \cr
} } \right] = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$
<br><br>$$ \Rightarrow {1 \over {10}}\left[ {\matrix{
{10} & 0 & {5 - 2} \cr
0 & {10} & { - 5 + \alpha } \cr
0 & 0 & {5 + \alpha } \cr
} } \right] = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$
<br><br>$$ \Rightarrow {{5 - \alpha } \over {10}} = 0$$
<br><br>$$ \Rightarrow \alpha = 5$$ | mcq | aieee-2004 |
BrIRgABjKyBYb3DI | maths | matrices-and-determinants | inverse-of-a-matrix | Let $$A = \left( {\matrix{
0 & 0 & { - 1} \cr
0 & { - 1} & 0 \cr
{ - 1} & 0 & 0 \cr
} } \right)$$. The only correct
<p>statement about the matrix $$A$$ is</p> | [{"identifier": "A", "content": "$${A^2} = 1$$ "}, {"identifier": "B", "content": "$$A=(-1)I,$$ where $$I$$ is a unit matrix "}, {"identifier": "C", "content": "$${A^{ - 1}}$$ does not exist "}, {"identifier": "D", "content": "$$A$$ is a zero matrix"}] | ["A"] | null | $$A = \left[ {\matrix{
0 & 0 & { - 1} \cr
0 & { - 1} & 0 \cr
{ - 1} & 0 & 0 \cr
} } \right]$$
<br><br>clearly $$\,\,\,A \ne 0.\,$$ Also $$\,\,\left| A \right| = - 1 \ne 0$$
<br><br>$$\therefore$$ $${A^{ - 1}}\,\,$$ exists, further
<br><br>$$\left( { - 1} \right)I = \left[ {\matrix{
{ - 1} & 0 & 0 \cr
0 & { - 1} & 0 \cr
0 & 0 & { - 1} \cr
} } \right] \ne A$$
<br><br>Also $${A^2} = \left[ {\matrix{
0 & 0 & { - 1} \cr
0 & { - 1} & 0 \cr
{ - 1} & 0 & 0 \cr
} } \right]\left[ {\matrix{
0 & 0 & { - 1} \cr
0 & { - 1} & 0 \cr
{ - 1} & 0 & 0 \cr
} } \right]$$
<br><br>$$ = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right] = I$$ | mcq | aieee-2004 |
gVRbAj7S0qPH4tAt | maths | matrices-and-determinants | inverse-of-a-matrix | If $${A^2} - A + 1 = 0$$, then the inverse of $$A$$ is : | [{"identifier": "A", "content": "$$A+I$$ "}, {"identifier": "B", "content": "$$A$$ "}, {"identifier": "C", "content": "$$A-I$$ "}, {"identifier": "D", "content": "$$I-A$$"}] | ["D"] | null | Given $${A^2} - A + I = 0$$
<br><br>$${A^{ - 1}}{A^2} - {A^{ - 1}}A + {A^{ - 1}}.I = {A^{ - 1}}.0$$
<br><br>(Multiplying $$\,\,\,{A^{ - 1}}$$ on both sides)
<br><br>$$ \Rightarrow A - 1 + {A^{ - 1}} = 0$$
<br><br>or $${A^{ - 1}} = 1 - A$$ | mcq | aieee-2005 |
NZE3QvqIskeDsv0i | maths | matrices-and-determinants | inverse-of-a-matrix | If $$A$$ is a $$3 \times 3$$ non-singular matrix such that $$AA'=A'A$$ and
<br/>$$B = {A^{ - 1}}A',$$ then $$BB'$$ equals: | [{"identifier": "A", "content": "$${B^{ - 1}}$$ "}, {"identifier": "B", "content": "$$\\left( {{B^{ - 1}}} \\right)'$$"}, {"identifier": "C", "content": "$$I+B$$ "}, {"identifier": "D", "content": "$$I$$ "}] | ["D"] | null | $$BB' = B\left( {{A^{ - 1}}A'} \right)' = B\left( {A'} \right)'\left( {{A^{ - 1}}} \right)' = BA\left( {{A^{ - 1}}} \right)'$$
<br><br>$$ = \left( {{A^{ - 1}}A'} \right)\left( {A\left( {{A^{ - 1}}} \right)'} \right)$$
<br><br>$$ = {A^{ - 1}}A.A'.\left( {{A^{ - 1}}} \right)'\,\,\,\,\,\,$$ $$\left\{ {} \right.$$ as $$\,\,\,\,\,\,$$ $$AA' = A'A$$ $$\left. \, \right\}$$
<br><br>$$ = I\left( {{A^{ - 1}}A} \right)'$$
<br><br>$$ = I.I = {I^2} = I$$ | mcq | jee-main-2014-offline |
WdAPCKhiSFPJAR2URgL8E | maths | matrices-and-determinants | inverse-of-a-matrix | Let A be a 3 $$ \times $$ 3 matrix such that A<sup>2</sup> $$-$$ 5A + 7I = 0
<br/><br/><b>Statement - I :</b>
<br/><br/>A<sup>$$-$$1</sup> = $${1 \over 7}$$ (5I $$-$$ A).
<br/><br/><b>Statement - II :</b>
<br/><br/>The polynomial A<sup>3</sup> $$-$$ 2A<sup>2</sup> $$-$$ 3A + I can be reduced to 5(A $$-$$ 4I).
<br/><br/>Then : | [{"identifier": "A", "content": "Statement-I is true, but Statement-II is false."}, {"identifier": "B", "content": "Statement-I is false, but Statement-II is true."}, {"identifier": "C", "content": "Both the statements are true."}, {"identifier": "D", "content": "Both the statements are false"}] | ["C"] | null | Given,
<br><br>A<sup>2</sup> $$-$$ 5A + 7I = 0
<br><br>$$ \Rightarrow $$ A<sup>2</sup> $$-$$ 5A = $$-$$ 7I
<br><br>$$ \Rightarrow $$ AAA<sup>$$-$$1</sup> $$-$$ 5AA<sup>$$-$$1</sup> = $$-$$ 7IA<sup>$$-$$1</sup>
<br><br>$$ \Rightarrow $$ AI $$-$$ 5I = $$-$$ 7A<sup>$$-$$1</sup>
<br><br>$$ \Rightarrow $$ A $$-$$ 5I = $$-$$ 7A<sup>$$-$$1</sup>
<br><br>$$ \Rightarrow $$ A<sup>$$-$$1</sup> = $${1 \over 7}$$(5I $$-$$ A)
<br><br>Hence, statement 1 is true.
<br><br>Now A<sup>3</sup> $$-$$ 2A<sup>2</sup> $$-$$ 3A + I
<br><br>= A(A<sup>2</sup>) $$-$$ 2A<sup>2</sup> $$-$$ 3A + I
<br><br>= A(5A $$-$$ 7I) $$-$$ 2A<sup>2</sup> $$-$$ 3A + I
<br><br>= 5A<sup>2</sup> $$-$$ 7A $$-$$ 2A<sup>2</sup> $$-$$ 3A + I
<br><br>= 3A<sup>2</sup> $$-$$ 10A + I
<br><br>= 3(5A $$-$$ 7I) $$-$$ 10A + I
<br><br>= 15A $$-$$ 21A $$-$$ 10A + I
<br><br>= 5A $$-$$ 20I
<br><br>= 5(A $$-$$ 4I)
<br><br>So, statement 2 is also correct. | mcq | jee-main-2016-online-10th-april-morning-slot |
G13VIQ4UxGankd46IX9Mu | maths | matrices-and-determinants | inverse-of-a-matrix | Suppose A is any 3$$ \times $$ 3 non-singular matrix and ( A $$-$$ 3I) (A $$-$$ 5I) = O where I = I<sub>3</sub> and O = O<sub>3</sub>. If $$\alpha $$A + $$\beta $$A<sup>-1</sup> = 4I, then $$\alpha $$ + $$\beta $$ is equal to : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "12"}] | ["A"] | null | Given,
<br><br>( A $$-$$ 3I) (A $$-$$ 5I) = O
<br><br>$$ \Rightarrow $$ A<sup>2</sup> - 8A + 15I = O
<br><br>Multiplying both sides by A<sup>- 1</sup>, we get,
<br><br>A<sup>- 1</sup>A.A - 8A<sup>- 1</sup>A + 15A<sup>- 1</sup>I = A<sup>- 1</sup>O
<br><br>$$ \Rightarrow $$ A - 8I + 15A<sup>- 1</sup> = O
<br><br>$$ \Rightarrow $$ A + 15A<sup>- 1</sup> = 8I
<br><br>$$ \Rightarrow $$$${A \over 2} + {{15{A^{ - 1}}} \over 2} = 4I$$
<br><br>Comparing with the equation $$\alpha $$A + $$\beta $$A<sup>-1</sup> = 4I, we get
<br><br>$$\alpha $$ = $${1 \over 2}$$ and $$\beta $$ = $${15 \over 2}$$
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $${1 \over 2}$$ + $${15 \over 2}$$ = $${16 \over 2}$$ = 8 | mcq | jee-main-2018-online-15th-april-evening-slot |
1dHMda3zohOtFwklZhLEe | maths | matrices-and-determinants | inverse-of-a-matrix | If $$A = \left[ {\matrix{
{\cos \theta } & { - \sin \theta } \cr
{\sin \theta } & {\cos \theta } \cr
} } \right]$$, then the matrix A<sup>–50</sup> when $$\theta $$ = $$\pi \over 12$$, is equal to :
| [{"identifier": "A", "content": "$$\\left[ {\\matrix{\n { {{\\sqrt 3 } \\over 2}} & { - {1 \\over 2}} \\cr \n {{{ 1} \\over 2}} & {{{\\sqrt 3 } \\over 2}} \\cr \n\n } } \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {\\matrix{\n {{1 \\over 2}} & -{{{\\sqrt 3 } \\over 2}} \\cr \n {{{\\sqrt 3 } \\over 2}} & {{{ - 1} \\over 2}} \\cr \n\n } } \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {\\matrix{\n {{{\\sqrt 3 } \\over 2}} & {{1 \\over 2}} \\cr \n -{{1 \\over 2}} & {{{\\sqrt 3 } \\over 2}} \\cr \n\n } } \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {\\matrix{\n {{1 \\over 2}} & {{{\\sqrt 3 } \\over 2}} \\cr \n {-{{\\sqrt 3 } \\over 2}} & {{{ 1} \\over 2}} \\cr \n\n } } \\right]$$"}] | ["C"] | null | (A<sup>$$-$$50</sup>) = (A<sup>$$-$$1</sup>)<sup>50</sup>
<br><br>We know,
<br><br>A<sup>$$-$$1</sup> = $${{adjA} \over {\left| A \right|}}$$
<br><br>$$\left| A \right|$$ = cos<sup>2</sup>$$\theta $$ + sin<sup>2</sup>$$\theta $$ = 1
<br><br>cofactor of A = $$\left[ {\matrix{
{\cos \theta } & { - \sin \theta } \cr
{\sin \theta } & {\cos \theta } \cr
} } \right]$$
<br><br>Adjoint of A = Transpose of cofactor matrix
<br><br>$$ \therefore $$ Adj A = $$\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$
<br><br>$$ \therefore $$ A<sup>$$-$$1</sup> = $$\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$
<br><br>$$ \therefore $$ (A<sup>$$-$$1</sup>)<sup>2</sup> = $$\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]\left[ {\matrix{
{\cos \theta } & {\sin \theta } \cr
{ - \sin \theta } & {\cos \theta } \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
{\cos 2\theta } & {\sin 2\theta } \cr
{ - \sin 2\theta } & {\cos 2\theta } \cr
} } \right]$$
<br><br>Similarly,
<br><br>(A<sup>$$-$$1</sup>)<sup>3</sup> = $$\left[ {\matrix{
{\cos 3\theta } & {\sin 3\theta } \cr
{ - \sin 3\theta } & {\cos 3\theta } \cr
} } \right]$$
<br><br>:
<br><br>:
<br><br>:
<br><br>(A<sup>$$-$$1</sup>)<sup>50</sup> = $$\left[ {\matrix{
{\cos 50\theta } & {\sin 50\theta } \cr
{ - \sin 50\theta } & {\cos 50\theta } \cr
} } \right]$$
<br><br>when $$\theta $$ = $${\pi \over {12}}$$ then
<br><br>$${A^{ - 50}}$$ = $$\left[ {\matrix{
{\cos {{25\pi } \over 6}} & {\sin {{25\pi } \over 6}} \cr
{ - \sin {{25\pi } \over 6}} & {\cos {{25\pi } \over 6}} \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
{{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr
{ - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr
} } \right]$$
<br><br><u>Note:</u>
<br><br>$$\cos {{25\pi } \over 6} = \cos \left( {4\pi + {\pi \over 6}} \right) = \cos {\pi \over 6} = {{\sqrt 3 } \over 2}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
FDEz8mcROZKQr1vuUT18hoxe66ijvwpuwxp | maths | matrices-and-determinants | inverse-of-a-matrix | If $$\left[ {\matrix{
1 & 1 \cr
0 & 1 \cr
} } \right]\left[ {\matrix{
1 & 2 \cr
0 & 1 \cr
} } \right]$$$$\left[ {\matrix{
1 & 3 \cr
0 & 1 \cr
} } \right]$$....$$\left[ {\matrix{
1 & {n - 1} \cr
0 & 1 \cr
} } \right] = \left[ {\matrix{
1 & {78} \cr
0 & 1 \cr
} } \right]$$,
<br/><br/>then the inverse of $$\left[ {\matrix{
1 & n \cr
0 & 1 \cr
} } \right]$$ is | [{"identifier": "A", "content": "$$\\left[ {\\matrix{\n 1 & { 0} \\cr \n {12} & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {\\matrix{\n 1 & { 0} \\cr \n {13} & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {\\matrix{\n 1 & { - 13} \\cr \n 0 & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {\\matrix{\n 1 & { - 12} \\cr \n 0 & 1 \\cr \n\n } } \\right]$$"}] | ["C"] | null | Given<br><br>
$$\left[ {\matrix{
1 & 1 \cr
0 & 1 \cr
} } \right]\left[ {\matrix{
1 & 2 \cr
0 & 1 \cr
} } \right]$$$$\left[ {\matrix{
1 & 3 \cr
0 & 1 \cr
} } \right]$$....$$\left[ {\matrix{
1 & {n - 1} \cr
0 & 1 \cr
} } \right] = \left[ {\matrix{
1 & {78} \cr
0 & 1 \cr
} } \right]$$
<br><br>$$ \Rightarrow $$ $$\left[ {\matrix{
1 & 3 \cr
0 & 1 \cr
} } \right]\left[ {\matrix{
1 & 3 \cr
0 & 1 \cr
} } \right].....\left[ {\matrix{
1 & {n - 1} \cr
0 & 1 \cr
} } \right]$$ = $$\left[ {\matrix{
1 & {78} \cr
0 & 1 \cr
} } \right]$$
<br><br>$$ \Rightarrow $$ $$\left[ {\matrix{
1 & 6 \cr
0 & 1 \cr
} } \right].....\left[ {\matrix{
1 & {n - 1} \cr
0 & 1 \cr
} } \right]$$ = $$\left[ {\matrix{
1 & {78} \cr
0 & 1 \cr
} } \right]$$
<br><br>$$ \Rightarrow $$ $$\left[ {\matrix{
1 & {1 + 2 + 3} \cr
0 & 1 \cr
} } \right].....\left[ {\matrix{
1 & {n - 1} \cr
0 & 1 \cr
} } \right]$$ = $$\left[ {\matrix{
1 & {78} \cr
0 & 1 \cr
} } \right]$$
<br>.
<br>.
<br>.
<br>.
<br>$$ \Rightarrow $$ $$\left[ {\matrix{
1 & {1 + 2 + 3 + .... + \left( {n - 1} \right)} \cr
0 & 1 \cr
} } \right]$$ = $$\left[ {\matrix{
1 & {78} \cr
0 & 1 \cr
} } \right]$$
<br><br>By comparing both sides we get,
<br><br>1 + 2 + 3 + ........+ (n - 1) = 78
<br><br>$$ \Rightarrow $$ $${{n\left( {n - 1} \right)} \over 2}$$ = 78
<br><br>$$ \Rightarrow $$ n = 13, - 12(not possible)
<br><br>$$ \therefore $$ The inverse of $$\left[ {\matrix{
1 & 13 \cr
0 & 1 \cr
} } \right]$$ = $$\left[ {\matrix{
1 & -13 \cr
0 & 1 \cr
} } \right]$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
rvBAr0Vg54rfbOxUni1klrhw2ua | maths | matrices-and-determinants | inverse-of-a-matrix | Let P = $$\left[ {\matrix{
3 & { - 1} & { - 2} \cr
2 & 0 & \alpha \cr
3 & { - 5} & 0 \cr
} } \right]$$, where $$\alpha $$ $$ \in $$ R. Suppose Q = [ q<sub>ij</sub>] is a matrix satisfying PQ = kl<sub>3</sub> for some non-zero k $$ \in $$ R. <br/>If q<sub>23</sub> = $$ - {k \over 8}$$
and |Q| = $${{{k^2}} \over 2}$$, then a<sup>2</sup> + k<sup>2</sup> is equal to ______. | [] | null | 17 | As $$PQ = kI \Rightarrow Q = k{P^{ - 1}}I$$<br><br>now $$Q = {k \over {|P|}}(adjP)I $$
<br><br>$$\Rightarrow Q = {k \over {(20 + 12\alpha )}}\left[ {\matrix{
- & - & - \cr
- & - & {( - 3\alpha - 4)} \cr
- & - & - \cr
} } \right]\left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$<br><br>$$ \because $$ $${q_{23}} = {{ - k} \over 8} $$
<br><br>$$\Rightarrow {k \over {(20 + 12\alpha )}}( - 3\alpha - 4) = {{ - k} \over 8} $$
<br><br>$$\Rightarrow 2(3\alpha + 4) = 5 + 3\alpha $$<br><br>$$3\alpha = - 3 \Rightarrow \alpha = - 1$$<br><br>also $$|Q| = {{{k^3}|I|} \over {|P|}} \Rightarrow {{{k^2}} \over 2} = {{{k^3}} \over {(20 + 12\alpha )}}$$<br><br>$$ \Rightarrow $$ $$(20 + 12\alpha ) = 2k \Rightarrow 8 = 2k \Rightarrow k = 4$$ | integer | jee-main-2021-online-24th-february-morning-slot |
vTciYbdTPPnzWP11Sg1kls5jz0i | maths | matrices-and-determinants | inverse-of-a-matrix | If $$A = \left[ {\matrix{
0 & { - \tan \left( {{\theta \over 2}} \right)} \cr
{\tan \left( {{\theta \over 2}} \right)} & 0 \cr
} } \right]$$ and <br/>$$({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{
a & { - b} \cr
b & a \cr
} } \right]$$, then $$13({a^2} + {b^2})$$ is equal to | [] | null | 13 | $$A = \left[ {\matrix{
0 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 0 \cr
} } \right]$$<br><br>$$ \Rightarrow I + A = \left[ {\matrix{
1 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 1 \cr
} } \right]$$<br><br>$$ \Rightarrow I - A = \left[ {\matrix{
1 & {\tan {\theta \over 2}} \cr
{ - \tan {\theta \over 2}} & 1 \cr
} } \right]$$ { $$\therefore$$ $$\left| {I - A} \right| = {\sec ^2}\theta /2$$}<br><br>$$ \Rightarrow {(I - A)^{ - 1}} = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{
1 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 1 \cr
} } \right]$$<br><br>$$ \Rightarrow (1 + A){(I - A)^{ - 1}} $$
<br><br>$$= {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{
1 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 1 \cr
} } \right]\left[ {\matrix{
1 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 1 \cr
} } \right]$$<br><br>$$ = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{
{1 - {{\tan }^2}{\theta \over 2}} & { - 2\tan {\theta \over 2}} \cr
{2\tan {\theta \over 2}} & {1 - {{\tan }^2}{\theta \over 2}} \cr
} } \right]$$<br><br>$$a = {{1 - {{\tan }^2}{\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$$<br><br>$$b = {{2\tan {\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$$<br><br>$$\therefore$$ $${a^2} + {b^2} = 1$$
<br><br>$$ \Rightarrow $$ $$13({a^2} + {b^2})$$ = 13 | integer | jee-main-2021-online-25th-february-morning-slot |
1ks088t6b | maths | matrices-and-determinants | inverse-of-a-matrix | Let $$A = \left[ {\matrix{
1 & 2 \cr
{ - 1} & 4 \cr
} } \right]$$. If A<sup>$$-$$1</sup> = $$\alpha$$I + $$\beta$$A, $$\alpha$$, $$\beta$$ $$\in$$ R, I is a 2 $$\times$$ 2 identity matrix then 4($$\alpha$$ $$-$$ $$\beta$$) is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "$${8 \\over 3}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["D"] | null | $$A = \left[ {\matrix{
1 & 2 \cr
{ - 1} & 4 \cr
} } \right],|A| = 6$$<br><br>$${A^{ - 1}} = {{adjA} \over {|A|}} = {1 \over 6}\left[ {\matrix{
4 & { - 2} \cr
1 & 1 \cr
} } \right] = \left[ {\matrix{
{{2 \over 3}} & { - {1 \over 3}} \cr
{{1 \over 6}} & {{1 \over 6}} \cr
} } \right]$$<br><br>$$\left[ {\matrix{
{{2 \over 3}} & { - {1 \over 3}} \cr
{{1 \over 6}} & {{1 \over 6}} \cr
} } \right] = \left[ {\matrix{
\alpha & 0 \cr
0 & \alpha \cr
} } \right] + \left[ {\matrix{
\beta & {2\beta } \cr
{ - \beta } & {4\beta } \cr
} } \right]$$<br><br>$$\left. \matrix{
\alpha + \beta = {2 \over 3} \hfill \cr
\beta = - {1 \over 6} \hfill \cr} \right\} \Rightarrow \alpha = {2 \over 3} + {1 \over 6} = {5 \over 6}$$<br><br>$$ \therefore $$ $$4(\alpha - \beta ) = 4(1) = 4$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1l58gzk5u | maths | matrices-and-determinants | inverse-of-a-matrix | <p>Let $$X = \left[ {\matrix{
0 & 1 & 0 \cr
0 & 0 & 1 \cr
0 & 0 & 0 \cr
} } \right],\,Y = \alpha I + \beta X + \gamma {X^2}$$ and $$Z = {\alpha ^2}I - \alpha \beta X + ({\beta ^2} - \alpha \gamma ){X^2}$$, $$\alpha$$, $$\beta$$, $$\gamma$$ $$\in$$ R. If $${Y^{ - 1}} = \left[ {\matrix{
{{1 \over 5}} & {{{ - 2} \over 5}} & {{1 \over 5}} \cr
0 & {{1 \over 5}} & {{{ - 2} \over 5}} \cr
0 & 0 & {{1 \over 5}} \cr
} } \right]$$, then ($$\alpha$$ $$-$$ $$\beta$$ + $$\gamma$$)<sup>2</sup> is equal to ____________.</p> | [] | null | 100 | <p>$$\because$$ $$X = \left[ {\matrix{
0 & 1 & 0 \cr
0 & 0 & 1 \cr
0 & 0 & 0 \cr
} } \right]$$</p>
<p>$$\therefore$$ $${X^2} = \left[ {\matrix{
0 & 0 & 1 \cr
0 & 0 & 0 \cr
0 & 0 & 0 \cr
} } \right]$$</p>
<p>$$\therefore$$ $$Y = \alpha I + \beta X + \gamma {X^2}\left[ {\matrix{
\alpha & \beta & \gamma \cr
0 & \alpha & \beta \cr
0 & 0 & \alpha \cr
} } \right]$$</p>
<p>$$\because$$ $$Y\,.\,{Y^{ - 1}} = I$$</p>
<p>$$\therefore$$ $$\left[ {\matrix{
\alpha & \beta & \gamma \cr
0 & \alpha & \beta \cr
0 & 0 & \alpha \cr
} } \right]\left[ {\matrix{
{{1 \over 5}} & {{{ - 2} \over 5}} & {{1 \over 5}} \cr
0 & {{1 \over 5}} & {{{ - 2} \over 5}} \cr
0 & 0 & {{1 \over 5}} \cr
} } \right]\left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$</p>
<p>$$\therefore$$ $$\left[ {\matrix{
{{\alpha \over 5}} & {{{\beta - 2\alpha } \over 5}} & {{{\alpha - 2\beta + \gamma } \over 5}} \cr
0 & {{\alpha \over 5}} & {{{\beta - 2\alpha } \over 5}} \cr
0 & 0 & {{\alpha \over 5}} \cr
} } \right] = \left[ {\matrix{
0 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$</p>
<p>$$\therefore$$ $$\alpha$$ = 5, $$\beta$$ = 10, $$\gamma$$ =15</p>
<p>$$\therefore$$ ($$\alpha$$ $$-$$ $$\beta$$ + $$\gamma$$)<sup>2</sup> = 100</p> | integer | jee-main-2022-online-26th-june-evening-shift |
1l6i06330 | maths | matrices-and-determinants | inverse-of-a-matrix | <p>The number of matrices $$A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$$, where $$a, b, c, d \in\{-1,0,1,2,3, \ldots \ldots, 10\}$$, such that $$A=A^{-1}$$, is ___________.</p> | [] | null | 50 | <p>$$\because$$ $$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$ then $${A^2} = \left[ {\matrix{
{{a^2} + bc} & {b(a + d)} \cr
{c(a + d)} & {bc + {d^2}} \cr
} } \right]$$</p>
<p>For A<sup>$$-$$1</sup> must exist $$ad - bc \ne 0$$ ...... (i)</p>
<p>and $$A = {A^{ - 1}} \Rightarrow {A^2} = I$$</p>
<p>$$\therefore$$ $${a^2} + bc = {d^2} + bc = 1$$ ...... (ii)</p>
<p>and $$b(a + d) = c(a + d) = 0$$ ...... (iii)</p>
<p>Case I : When a = d = 0, then possible values of (b, c) are (1, 1), ($$-$$1, 1) and (1, $$-$$1) and ($$-$$1, 1).</p>
<p>Total four matrices are possible.</p>
<p>Case II : When a = $$-$$d then (a, d) be (1, $$-$$1) or ($$-$$1, 1).</p>
<p>Then total possible values of (b, c) are $$(12 + 11) \times 2 = 46$$.</p>
<p>$$\therefore$$ Total possible matrices $$= 46 + 4 = 50$$.</p> | integer | jee-main-2022-online-26th-july-evening-shift |
1ldu5socp | maths | matrices-and-determinants | inverse-of-a-matrix | <p>Let $$A = \left[ {\matrix{
{{1 \over {\sqrt {10} }}} & {{3 \over {\sqrt {10} }}} \cr
{{{ - 3} \over {\sqrt {10} }}} & {{1 \over {\sqrt {10} }}} \cr
} } \right]$$ and $$B = \left[ {\matrix{
1 & { - i} \cr
0 & 1 \cr
} } \right]$$, where $$i = \sqrt { - 1} $$. If $$\mathrm{M=A^T B A}$$, then the inverse of the matrix $$\mathrm{AM^{2023}A^T}$$ is</p> | [{"identifier": "A", "content": "$$\\left[ {\\matrix{\n 1 & { - 2023i} \\cr \n 0 & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {\\matrix{\n 1 & 0 \\cr \n {2023i} & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {\\matrix{\n 1 & {2023i} \\cr \n 0 & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {\\matrix{\n 1 & 0 \\cr \n { - 2023i} & 1 \\cr \n\n } } \\right]$$"}] | ["C"] | null | $$
\begin{aligned}
& \mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{cc}
\frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\
\frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}
\end{array}\right]\left[\begin{array}{cc}
\frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\
\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}}
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] = I \\\\\
& \mathrm{B}^2=\left[\begin{array}{cc}
1 & -\mathrm{i} \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & -\mathrm{i} \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & -2 \mathrm{i} \\
0 & 1
\end{array}\right] \\\\
& \mathrm{B}^3=\left[\begin{array}{cc}
1 & -3 \mathrm{i} \\
0 & 1
\end{array}\right]
\end{aligned}
$$
<br/><br/>Similarly,
<br/><br/>$$
\begin{aligned}
& \mathrm{B}^{2023}=\left[\begin{array}{cc}
1 & -2023 \mathrm{i} \\
0 & 1
\end{array}\right] \\\\
& \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \\\\
& \mathrm{M}^2=\mathrm{M} \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \mathrm{A}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{~A} \\\\
& \mathrm{M}^3=\mathrm{M}^2 \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{AA}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^3 \mathrm{~A}
\end{aligned}
$$
<br/><br/>Similarly,
<br/><br/>$$
\begin{aligned}
& \mathrm{M}^{2023}= \mathrm{A}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{~A} \\\\
& \mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}=\mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{AA}^{\mathrm{T}} = \mathrm{I}.\mathrm{B}^{2023}.\mathrm{I}=\mathrm{B}^{2023} \\\\
& =\left[\begin{array}{cc}
1 & -2023 \mathrm{i} \\
0 & 1
\end{array}\right]
\end{aligned}
$$
<br/><br/>$$ \therefore $$ $$
\text { Inverse of }\left(\mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}\right) \text { is }\left[\begin{array}{cc}
1 & 2023 i \\
0 & 1
\end{array}\right]
$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1lguucigm | maths | matrices-and-determinants | inverse-of-a-matrix | <p>Let $$\mathrm{A}$$ be a $$2 \times 2$$ matrix with real entries such that $$\mathrm{A}'=\alpha \mathrm{A}+\mathrm{I}$$, where $$\alpha \in \mathbb{R}-\{-1,1\}$$. If $$\operatorname{det}\left(A^{2}-A\right)=4$$, then the sum of all possible values of $$\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$\\frac{3}{2}$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$\\frac{5}{2}$$"}] | ["D"] | null | We have, $A^T=\alpha A+I$, where $A$ is $2 \times 2$ matrix and $\alpha \in R-\{-1,1\}$
<br/><br/>$$
\begin{aligned}
\left(A^T\right)^T & =\alpha A^T+I \\\\
A & =\alpha A^T+I \\\\
A & =\alpha(\alpha A+I)+I \left[\because A^T=\alpha A+I\right]\\\\
A & =\alpha^2 A+(\alpha+1) I \\\\
A & \left(1-\alpha^2\right)=(\alpha+1) I \\\\
A & =\frac{(\alpha+1)}{(1-\alpha)(1+\alpha)} I
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
A & =\frac{1}{1-\alpha} I ..........(i)\\\\
|A| & =\frac{1}{(1-\alpha)^2} ...........(ii)
\end{aligned}
$$
<br/><br/>Also,
<br/><br/>$$
\begin{aligned}
\left|A^2-A\right| & =4 \\\\
|A||A-I| & =4 \\\\
\frac{1}{(1-\alpha)^2}\left|\left(\frac{1}{1-\alpha}-1\right) I\right| & =4 \\\\
\frac{1}{(1-\alpha)^2}\left|\left(\frac{\alpha}{1-\alpha}\right) I\right| & =4 \\\\
\frac{1}{(1-\alpha)^2} \times \frac{\alpha^2}{(1-\alpha)^2} & =4 \\\\
\frac{\alpha^2}{(1-\alpha)^4} & =4 \\\\
\frac{\alpha}{(1-\alpha)^2} & = \pm 2 \\\\
2(1-\alpha)^2 & = \pm \alpha
\end{aligned}
$$
<br/><br/>If $2(1-\alpha)^2=\alpha$, then $2 \alpha^2-5 \alpha+2=0$
<br/><br/>Sum of value of $\alpha=\frac{5}{2} \quad\left[\because\right.$ Sum of zero $\left.=\frac{\text {-Coefficient of } x}{\text { Coefficient of } x^2}\right]$
<br/><br/>If $2(1-\alpha)^2=-\alpha$, then $2 \alpha^2-3 \alpha+2=0$
<br/><br/>$\therefore$ No real value of $\alpha$
<br/><br/>Hence, sum of all values of $\alpha=\frac{5}{2}$ | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgylvto4 | maths | matrices-and-determinants | inverse-of-a-matrix | <p>If $$A=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right], \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$$ and $$\alpha+\beta=-2$$, then $$4 \alpha^{2}+\beta^{2}+\lambda^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "19"}, {"identifier": "D", "content": "14"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{A}=\left[\begin{array}{cc}
1 & 5 \\
\lambda & 10
\end{array}\right] \\\\
& \Rightarrow|\mathrm{A}-x \mathrm{I}|=0 \\\\
& \Rightarrow\left|\begin{array}{cc}
1-x & 5 \\
\lambda & 10-x
\end{array}\right|=0 \\\\
& \Rightarrow(1-x)(10-x)-5 \lambda=0 \\\\
& \Rightarrow 10-11 x+x^2-5 \lambda=0
\end{aligned}
$$
<br/><br/>Also, $\Rightarrow \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$
<br/><br/>$$
\Rightarrow \alpha A^2+\beta A-I=0
$$
<br/><br/>and $A^2-11 A+(10-5 \lambda) I=0$
<br/><br/>On solving, we get
<br/><br/>$$
\alpha=\frac{1}{5}, \beta=-\frac{11}{5}
$$
<br/><br/>$$
\begin{aligned}
& \text { So, } 5 \lambda-10=5 \Rightarrow \lambda=3 \\\\
& \therefore 4 \alpha^2+\beta^2+\lambda^2 \\\\
& =4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9 \\\\
& =\frac{125}{25}+9=14
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
lsbku1aw | maths | matrices-and-determinants | inverse-of-a-matrix | Consider the matrix $f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$.
<br/><br/>Given below are two statements :
<br/><br/>Statement I : $ f(-x)$ is the inverse of the matrix $f(x)$.
<br/><br/>Statement II : $f(x) f(y)=f(x+y)$.
<br/><br/>In the light of the above statements, choose the correct answer from the options given below : | [{"identifier": "A", "content": "Statement I is false but Statement II is true"}, {"identifier": "B", "content": "Both Statement I and Statement II are false"}, {"identifier": "C", "content": "Both Statement I and Statement II are true"}, {"identifier": "D", "content": "Statement I is true but Statement II is false"}] | ["C"] | null | <p>$$\begin{aligned}
& f(-x)=\left[\begin{array}{ccc}
\cos x & \sin x & 0 \\
-\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right] \\
& f(x) \cdot f(-x)=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=I
\end{aligned}$$</p>
<p>Hence statement- I is correct</p>
<p>Now, checking statement II</p>
<p>$$\begin{aligned}
& f(y)=\left[\begin{array}{ccc}
\cos y & -\sin y & 0 \\
\sin y & \cos y & 0 \\
0 & 0 & 1
\end{array}\right] \\
& f(x) \cdot f(y)=\left[\begin{array}{ccc}
\cos (x+y) & -\sin (x+y) & 0 \\
\sin (x+y) & \cos (x+y) & 0 \\
0 & 0 & 1
\end{array}\right] \\
& \Rightarrow f(x) \cdot f(y)=f(x+y)
\end{aligned}$$</p>
<p>Hence statement-II is also correct.</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
luxwcxug | maths | matrices-and-determinants | inverse-of-a-matrix | <p>Let $$B=\left[\begin{array}{ll}1 & 3 \\ 1 & 5\end{array}\right]$$ and $$A$$ be a $$2 \times 2$$ matrix such that $$A B^{-1}=A^{-1}$$. If $$B C B^{-1}=A$$ and $$C^4+\alpha C^2+\beta I=O$$, then $$2 \beta-\alpha$$ is equal to</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "2"}] | ["B"] | null | <p>$$\begin{aligned}
& B=\left[\begin{array}{ll}
1 & 3 \\
1 & 5
\end{array}\right] \\
& A B^{-1}=A^{-1} \\
& \Rightarrow A^2=B
\end{aligned}$$</p>
<p>Also, $$B C B^{-1}=A$$</p>
<p>$$\begin{aligned}
\Rightarrow C & =B^{-1} A B \\
\Rightarrow C^4 & =\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right) \\
& =B^{-1} A^4 B \\
& =B^{-1} B^2 B \\
\Rightarrow \quad C^4 & =B^2
\end{aligned}$$</p>
<p>Also, $$C^2=\left(B^{-1} A B\right)\left(B^{-1} A B\right)$$</p>
<p>$$\begin{aligned}
& =B^{-1} A^2 B \\
& =B^{-1} B B
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow C^2=B \\
& \Rightarrow C^4+\alpha C^2+\beta I=0 \\
& \Rightarrow B^2+\alpha B+\beta I=0 \\
& B^2=\left[\begin{array}{ll}
1 & 3 \\
1 & 5
\end{array}\right]\left[\begin{array}{ll}
1 & 3 \\
1 & 5
\end{array}\right]=\left[\begin{array}{ll}
4 & 18 \\
6 & 28
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
4 & 18 \\
6 & 28
\end{array}\right]+\alpha\left[\begin{array}{ll}
1 & 3 \\
1 & 5
\end{array}\right]+\left[\begin{array}{ll}
\beta & 0 \\
0 & \beta
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& \Rightarrow 4+\alpha+\beta=0
\end{aligned}$$</p>
<p>and $$18+3 \alpha=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow \alpha=-6 \\
& \Rightarrow \beta=2 \\
& \Rightarrow 2 \beta-\alpha=10
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
lv2eqxv2 | maths | matrices-and-determinants | inverse-of-a-matrix | <p>Let $$A$$ be a $$2 \times 2$$ symmetric matrix such that $$A\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right]$$ and the determinant of $$A$$ be 1 . If $$A^{-1}=\alpha A+\beta I$$, where $$I$$ is an identity matrix of order $$2 \times 2$$, then $$\alpha+\beta$$ equals _________.</p> | [] | null | 5 | <p>Let $$A=\left[\begin{array}{ll}a & b \\ b & c\end{array}\right]$$</p>
<p>$$|A|=1 \Rightarrow a c-b^2=0 \quad \text{... (i)}$$</p>
<p>$$\text { Given }\left[\begin{array}{ll}
a & b \\
b & c
\end{array}\right]\left[\begin{array}{l}
1 \\
1
\end{array}\right]=\left[\begin{array}{l}
3 \\
7
\end{array}\right]$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad a+b=3 \quad \text{... (ii)}\\
& \text { and } b+c=7 \quad \text{... (iii)}
\end{aligned}$$</p>
<p>from (i), (ii) and (iii) $$a=1, b=2, c=5$$</p>
<p>$$\begin{aligned}
\Rightarrow & A=\left[\begin{array}{ll}
1 & 2 \\
2 & 5
\end{array}\right] \Rightarrow A^{-1}=\left[\begin{array}{cc}
5 & -2 \\
-2 & 1
\end{array}\right] \\
& \text { Given } A^{-1}=\alpha A+\beta I \\
\Rightarrow & {\left[\begin{array}{cc}
5 & -2 \\
-2 & 1
\end{array}\right]=\alpha\left[\begin{array}{ll}
1 & 2 \\
2 & 5
\end{array}\right]+\beta\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] } \\
\Rightarrow & \alpha=-1 \text { and } \beta=6 \\
& \alpha+\beta=5
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-evening-shift |
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