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Dxj4NLInsOwXDlXB | maths | matrices-and-determinants | multiplication-of-matrices | If $$A = \left[ {\matrix{
a & b \cr
b & a \cr
} } \right]$$ and $${A^2} = \left[ {\matrix{
\alpha & \beta \cr
\beta & \alpha \cr
} } \right]$$, then | [{"identifier": "A", "content": "$$\\alpha = 2ab,\\,\\beta = {a^2} + {b^2}$$ "}, {"identifier": "B", "content": "$$\\alpha = {a^2} + {b^2},\\,\\beta = ab$$ "}, {"identifier": "C", "content": "$$\\alpha = {a^2} + {b^2},\\,\\beta = 2ab$$ "}, {"identifier": "D", "content": "$$\\alpha = {a^2} + {b^2},\\,\\beta = {a^2} - {b^2}$$ "}] | ["C"] | null | $${A^2} = \left[ {\matrix{
\alpha & \beta \cr
\beta & \alpha \cr
} } \right] = \left[ {\matrix{
a & b \cr
b & a \cr
} } \right]\left[ {\matrix{
a & b \cr
b & a \cr
} } \right]$$
<br><br>$$ = \left[ {\matrix{
{{a^2} + {b^2}} & {2ab} \cr
{2ab} & {{a^2} + {b^2}} \cr
} } \right]$$
<br><br>$$\alpha = {a^2} + {b^2};\,\,\beta = 2ab$$ | mcq | aieee-2003 |
KAstYuenUMEEwuNAkDNlk | maths | matrices-and-determinants | multiplication-of-matrices | Let A = $$\left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right]$$ and B = A<sup>20</sup>. Then the sum of the elements of the first column of B is : | [{"identifier": "A", "content": "210"}, {"identifier": "B", "content": "211"}, {"identifier": "C", "content": "231"}, {"identifier": "D", "content": "251"}] | ["C"] | null | A = $$\left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right]$$
<br><br>A<sup>2</sup> = A.A = $$\left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right] \times \left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
1 & 0 & 0 \cr
2 & 1 & 0 \cr
3 & 2 & 1 \cr
} } \right]$$
<br><br>A<sup>3</sup> = A<sup>2</sup>.A = $$\left[ {\matrix{
1 & 0 & 0 \cr
2 & 1 & 0 \cr
3 & 2 & 1 \cr
} } \right] \times \left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right]$$
<br><br>= $$\left[ {\matrix{
1 & 0 & 0 \cr
3 & 1 & 0 \cr
6 & 3 & 1 \cr
} } \right]$$
<br><br>Similarly
<br><br>A<sup>4</sup> = $$\left[ {\matrix{
1 & 0 & 0 \cr
4 & 1 & 0 \cr
{10} & 4 & 1 \cr
} } \right]$$
<br><br>From this we can say,
<br><br>A<sup>n</sup> = $$\left[ {\matrix{
1 & 0 & 0 \cr
n & 1 & 0 \cr
{{{n\left( {n + 1} \right)} \over 2}} & n & 1 \cr
} } \right]$$
<br><br>$$\therefore\,\,\,$$ A<sup>20</sup> = $$\left[ {\matrix{
1 & 0 & 0 \cr
{20} & 1 & 0 \cr
{210} & {20} & 1 \cr
} } \right]$$
<br><br>$$\therefore\,\,\,$$ Sum of the first column
<br><br>= 1 + 20 + 210
<br><br>= 231 | mcq | jee-main-2018-online-16th-april-morning-slot |
qXMCpQCnIQe4sYD3Q2jgy2xukf0ypwyy | maths | matrices-and-determinants | multiplication-of-matrices | Let A = $$\left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right]$$, x $$ \in $$ R and A<sup>4</sup> = [a<sub>ij</sub>].
<br/>If
a<sub>11</sub> = 109, then a<sub>22</sub> is equal to _______ . | [] | null | 10 | $${A^2} = \left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right]\left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right] = \left[ {\matrix{
{{x^2} + 1} & x \cr
x & 1 \cr
} } \right]$$<br><br>$${A^4} = \left[ {\matrix{
{{x^2} + 1} & x \cr
x & 1 \cr
} } \right]\left[ {\matrix{
{{x^2} + 1} & x \cr
x & 1 \cr
} } \right]$$<br><br>$$ = \left[ {\matrix{
{{{({x^2} + 1)}^2} + {x^2}} & {x({x^2} + 1) + x} \cr
{x({x^2} + 1) + x} & {{x^2} + 1} \cr
} } \right]$$<br><br>Given $${({x^2} + 1)^2} + {x^2} = 109$$<br><br>Let $${x^2} + 1$$ = t<br><br>$${t^2} + t - 1 = 109$$<br><br>$$ \Rightarrow $$ (t $$ - $$ 10) (t + 11) = 0<br><br>$$ \therefore $$ t = 10 = x<sup>2</sup> + 1 = a<sub>22</sub> | integer | jee-main-2020-online-3rd-september-morning-slot |
W3T58kXuIKp6eHtE5Ejgy2xukf8zff62 | maths | matrices-and-determinants | multiplication-of-matrices | If $$A = \left[ {\matrix{
{\cos \theta } & {i\sin \theta } \cr
{i\sin \theta } & {\cos \theta } \cr
} } \right]$$, $$\left( {\theta = {\pi \over {24}}} \right)$$<br/><br/>
and $${A^5} = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$, where $$i = \sqrt { - 1} $$ then which one of the following is
not true? | [{"identifier": "A", "content": "$$a$$<sup>2</sup> - $$c$$<sup>2</sup> = 1"}, {"identifier": "B", "content": "$$0 \\le {a^2} + {b^2} \\le 1$$"}, {"identifier": "C", "content": "$$ a$$<sup>2</sup> - $$d$$<sup>2</sup> = 0"}, {"identifier": "D", "content": "$${a^2} - {b^2} = {1 \\over 2}$$"}] | ["D"] | null | $$ \because $$ $$A = \left[ {\matrix{
{\cos \theta } & {i\sin \theta } \cr
{i\sin \theta } & {\cos \theta } \cr
} } \right]$$<br><br>$$ \therefore $$ $${A^n} = \left[ {\matrix{
{\cos \,n\theta } & {i\sin \,n\theta } \cr
{i\sin \,n\theta } & {\cos \,n\theta } \cr
} } \right],n \in N$$<br><br>$$ \therefore $$$${A^5} = \left[ {\matrix{
{\cos \,5\theta } & {i\sin \,5\theta } \cr
{i\sin \,5\theta } & {\cos \,5\theta } \cr
} } \right] = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$<br><br>$$ \therefore $$ $$a = \cos 5\theta ,\,b = i\sin 5\theta = c,\,d = \cos 5\theta $$<br><br>$$ \therefore $$ $${a^2} - {b^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$$<br><br>$${a^2} - {c^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$$<br><br>$${a^2} - {d^2} = {\cos ^2}5\theta - {\cos ^2}5\theta = 1$$<br><br>$${a^2} + {b^2} = {\cos ^2}5\theta - {\sin ^2}5\theta = \cos 10\theta = \cos {{10\pi } \over {24}}$$<br><br>and $$0 < \cos {{5\pi } \over {12}} < 1 $$
<br><br>$$\Rightarrow 0 \le {a^2} + {b^2} \le 1$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
PxfQPOh5QrObe8wYIi1kluy5ax1 | maths | matrices-and-determinants | multiplication-of-matrices | If the matrix $$A = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right]$$ satisfies the equation<br/><br/> $${A^{20}} + \alpha {A^{19}} + \beta A = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 4 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$ for some real numbers $$\alpha$$ and $$\beta$$, then $$\beta$$ $$-$$ $$\alpha$$ is equal to ___________. | [] | null | 4 | $${A^2} = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right]\left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right] = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 4 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$<br><br>$${A^3} = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 4 & 0 \cr
0 & 0 & 1 \cr
} } \right]\left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right] = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 8 & 0 \cr
3 & 0 & { - 1} \cr
} } \right]$$<br><br>$${A^4} = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 4 & 0 \cr
0 & 0 & 1 \cr
} } \right]\left[ {\matrix{
1 & 0 & 0 \cr
0 & 4 & 0 \cr
0 & 0 & 1 \cr
} } \right] = \left[ {\matrix{
1 & 0 & 0 \cr
0 & {16} & 0 \cr
0 & 0 & 1 \cr
} } \right]$$<br><br>$$\eqalign{
& . \cr
& . \cr
& . \cr
& . \cr
& . \cr
& . \cr} $$<br><br>$${A^{19}} = \left[ {\matrix{
1 & 0 & 0 \cr
0 & {{2^{19}}} & 0 \cr
3 & 0 & { - 1} \cr
} } \right],{A^{20}} = \left[ {\matrix{
1 & 0 & 0 \cr
0 & {{2^{20}}} & 0 \cr
0 & 0 & 1 \cr
} } \right]$$<br><br>$$L.H.S. = {A^{20}} + \alpha {A^{19}} + \beta A = $$
<br><br>$$\left[ {\matrix{
{1 + \alpha + \beta } & 0 & 0 \cr
0 & {{2^{20}} + \alpha {2^{19}} + 2\beta } & 0 \cr
{3\alpha + 3\beta } & 0 & {1 - \alpha - \beta } \cr
} } \right]$$<br><br>$$R.H.S. = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 4 & 0 \cr
0 & 0 & 1 \cr
} } \right] $$
<br><br>$$\Rightarrow \alpha + \beta = 0$$ and $${2^{20}} + \alpha {2^{19}} + 2\beta = 4$$<br><br>$$ \Rightarrow {2^{20}} + \alpha ({2^{19}} - 2) = 4$$<br><br>$$ \Rightarrow \alpha = {{4 - {2^{20}}} \over {{2^{19}} - 2}} = - 2$$<br><br>$$ \Rightarrow \beta = 2$$<br><br>$$ \therefore $$ $$\beta - \alpha = 4$$ | integer | jee-main-2021-online-26th-february-evening-slot |
KxeIvsh9BKSv7d6i4Q1kmknlkg7 | maths | matrices-and-determinants | multiplication-of-matrices | Let $$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$ and $$B = \left[ {\matrix{
\alpha \cr
\beta \cr
} } \right] \ne \left[ {\matrix{
0 \cr
0 \cr
} } \right]$$ such that AB = B and a + d = 2021, then the value of ad $$-$$ bc is equal to ___________. | [] | null | 2020 | $$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right],\,B = \left[ {\matrix{
\alpha \cr
\beta \cr
} } \right]$$<br><br>$$AB = B$$<br><br>$$\left[ {\matrix{
a & b \cr
c & d \cr
} } \right]\left[ {\matrix{
\alpha \cr
\beta \cr
} } \right] = \left[ {\matrix{
\alpha \cr
\beta \cr
} } \right]$$<br><br>$$\left[ {\matrix{
{a\alpha + b\beta } \cr
{c\alpha + d\beta } \cr
} } \right] = \left[ {\matrix{
\alpha \cr
\beta \cr
} } \right]$$$$ \Rightarrow $$ $$\eqalign{
& a\alpha + b\beta = \alpha \,......(1) \cr
& c\alpha + d\beta = \beta \,......(2) \cr} $$<br><br>$$\alpha (a - 1) = - b\beta $$ and $$c\alpha = \beta (1 - d)$$<br><br>$${\alpha \over \beta } = {{ - b} \over {a - 1}}$$ & $${\alpha \over \beta } = {{1 - d} \over c}$$<br><br>$$ \therefore $$ $${{ - b} \over {a - 1}} = {{1 - d} \over c}$$<br><br>$$ - bc = (a - 1)(1 - d)$$<br><br>$$ - bc = a - ad - 1 + d$$<br><br>$$ad - bc = a + d - 1$$<br><br>$$ = 2021 - 1 = 2020$$ | integer | jee-main-2021-online-17th-march-evening-shift |
1kruapkyd | maths | matrices-and-determinants | multiplication-of-matrices | Let $$A = \left[ {\matrix{
0 & 1 & 0 \cr
1 & 0 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$. Then the number of 3 $$\times$$ 3 matrices B with entries from the set {1, 2, 3, 4, 5} and satisfying AB = BA is ____________. | [] | null | 3125 | Let matrix $$B = \left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right]$$<br><br>$$\because$$ $$AB = BA$$<br><br>$$\left[ {\matrix{
0 & 1 & 0 \cr
1 & 0 & 0 \cr
0 & 0 & 1 \cr
} } \right]\left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right] = \left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right]\left[ {\matrix{
0 & 1 & 0 \cr
1 & 0 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$<br><br>$$\left[ {\matrix{
d & e & f \cr
a & b & c \cr
g & h & i \cr
} } \right] = \left[ {\matrix{
b & a & c \cr
e & d & f \cr
h & g & i \cr
} } \right]$$<br><br>$$ \Rightarrow d = b,e = a,f = c,g = h$$<br><br>$$\therefore$$ Matrix $$B = \left[ {\matrix{
a & b & c \cr
b & a & c \cr
g & g & i \cr
} } \right]$$<br><br>No. of ways of selecting a, b, c, g, i<br><br>$$ = 5 \times 5 \times 5 \times 5 \times 5$$<br><br>$$ = {5^5} = 3125$$<br><br>$$\therefore$$ No. of matrices B = 3125 | integer | jee-main-2021-online-22th-july-evening-shift |
1ktcy6d7j | maths | matrices-and-determinants | multiplication-of-matrices | Let $$A = \left( {\matrix{
1 & 0 & 0 \cr
0 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right)$$. Then A<sup>2025</sup> $$-$$ A<sup>2020</sup> is equal to : | [{"identifier": "A", "content": "A<sup>6</sup> $$-$$ A"}, {"identifier": "B", "content": "A<sup>5</sup>"}, {"identifier": "C", "content": "A<sup>5</sup> $$-$$ A"}, {"identifier": "D", "content": "A<sup>6</sup>"}] | ["A"] | null | $$A = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right] \Rightarrow {A^2} = \left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right]$$<br><br>$${A^3} = \left[ {\matrix{
1 & 0 & 0 \cr
2 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right] \Rightarrow {A^4} = \left[ {\matrix{
1 & 0 & 0 \cr
3 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right]$$<br><br>$${A^n} = \left[ {\matrix{
1 & 0 & 0 \cr
{n - 1} & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right]$$<br><br>$${A^{2025}} - {A^{2020}} = \left[ {\matrix{
0 & 0 & 0 \cr
5 & 0 & 0 \cr
0 & 0 & 0 \cr
} } \right]$$<br><br>$${A^6} - A = \left[ {\matrix{
0 & 0 & 0 \cr
5 & 0 & 0 \cr
0 & 0 & 0 \cr
} } \right]$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1l55j80fk | maths | matrices-and-determinants | multiplication-of-matrices | <p>Let $$A = \left( {\matrix{
{1 + i} & 1 \cr
{ - i} & 0 \cr
} } \right)$$ where $$i = \sqrt { - 1} $$. Then, the number of elements in the set { n $$\in$$ {1, 2, ......, 100} : A<sup>n</sup> = A } is ____________.</p> | [] | null | 25 | <p>$$\therefore$$ $${A^2} = \left[ {\matrix{
{1 + i} & 1 \cr
{ - i} & 0 \cr
} } \right]\left[ {\matrix{
{1 + i} & 1 \cr
{ - 1} & 0 \cr
} } \right] = \left[ {\matrix{
i & {1 + i} \cr
{1 - i} & { - i} \cr
} } \right]$$</p>
<p>$${A^4} = \left[ {\matrix{
i & {1 + i} \cr
{1 - i} & { - i} \cr
} } \right]\left[ {\matrix{
i & {1 + i} \cr
{1 - i} & { - i} \cr
} } \right] = I$$</p>
<p>So A<sup>5</sup> = A, A<sup>9</sup> = A and so on.</p>
<p>Clearly n = 1, 5, 9, ......, 97</p>
<p>Number of values of n = 25</p> | integer | jee-main-2022-online-28th-june-evening-shift |
1l6hxkaiy | maths | matrices-and-determinants | multiplication-of-matrices | <p>$$
\text { Let } A=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right] \text { and } B=\left[\begin{array}{ccc}
9^{2} & -10^{2} & 11^{2} \\
12^{2} & 13^{2} & -14^{2} \\
-15^{2} & 16^{2} & 17^{2}
\end{array}\right] \text {, then the value of } A^{\prime} B A \text { is: }
$$</p> | [{"identifier": "A", "content": "1224"}, {"identifier": "B", "content": "1042"}, {"identifier": "C", "content": "540"}, {"identifier": "D", "content": "539"}] | ["D"] | null | <p>$$A'BA = \left[ {\matrix{
1 & 1 & 1 \cr
} } \right]\left[ {\matrix{
{{9^2}} & { - {{10}^2}} & {{{11}^2}} \cr
{{{12}^2}} & {{{13}^2}} & { - {{14}^2}} \cr
{ - {{15}^2}} & {{{16}^2}} & {{{17}^2}} \cr
} } \right]A$$</p>
<p>$$ = \left[ {\matrix{
{{9^2} + {{12}^2} - {{15}^2}} & { - {{10}^2} + {{13}^2} + {{16}^2}} & {{{11}^2} - {{14}^2} + {{17}^2}} \cr
} } \right]\left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right]$$</p>
<p>$$ = \left[ {{9^2} + {{12}^2} - {{15}^2} - {{10}^2} + {{13}^2} + {{16}^2} + {{11}^2} - {{14}^2} + {{17}^2}} \right]$$</p>
<p>$$ = [({9^2} - {10^2}) + ({11^2} + {12^2}) + ({13^2} - {14^2}) + ({16^2} - {15^2}) + {17^2}]$$</p>
<p>$$ = [ - 19 + 265 + ( - 27) + 31 + 289]$$</p>
<p>$$ = [585 - 46] = [539]$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6jb5z9r | maths | matrices-and-determinants | multiplication-of-matrices | <p>Let $$A=\left(\begin{array}{cc}1 & 2 \\ -2 & -5\end{array}\right)$$. Let $$\alpha, \beta \in \mathbb{R}$$ be such that $$\alpha A^{2}+\beta A=2 I$$. Then $$\alpha+\beta$$ is equal to</p> | [{"identifier": "A", "content": "$$-$$10"}, {"identifier": "B", "content": "$$-$$6"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "10"}] | ["D"] | null | <p>$${A^2} = \left[ {\matrix{
1 & 2 \cr
{ - 2} & { - 5} \cr
} } \right]\left[ {\matrix{
1 & 2 \cr
{ - 2} & { - 5} \cr
} } \right] = \left[ {\matrix{
{ - 3} & { - 8} \cr
8 & {21} \cr
} } \right]$$</p>
<p>$$\alpha {A^2} + \beta A = \left[ {\matrix{
{ - 3\alpha } & { - 8\alpha } \cr
{8\alpha } & {21\alpha } \cr
} } \right] + \left[ {\matrix{
\beta & {2\beta } \cr
{ - 2\beta } & { - 5\beta } \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
{ - 3\alpha + \beta } & { - 8\alpha + 2\beta } \cr
{8\alpha - 2\beta } & {21\alpha - 5\beta } \cr
} } \right] = \left[ {\matrix{
2 & 0 \cr
0 & 2 \cr
} } \right]$$</p>
<p>On Comparing</p>
<p>$$8\alpha = 2\beta ,\, - 3\alpha + \beta = 2,\,21\alpha - 5\beta = 2$$</p>
<p>$$ \Rightarrow \alpha = 2,\,\beta = 8$$</p>
<p>So, $$\alpha + \beta = 10$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6m6njhu | maths | matrices-and-determinants | multiplication-of-matrices | <p>Let $$A=\left[\begin{array}{cc}1 & -1 \\ 2 & \alpha\end{array}\right]$$ and $$B=\left[\begin{array}{cc}\beta & 1 \\ 1 & 0\end{array}\right], \alpha, \beta \in \mathbf{R}$$. Let $$\alpha_{1}$$ be the value of $$\alpha$$ which satisfies $$(\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$$ and $$\alpha_{2}$$ be the value of $$\alpha$$ which satisfies $$(\mathrm{A}+\mathrm{B})^{2}=\mathrm{B}^{2}$$. Then $$\left|\alpha_{1}-\alpha_{2}\right|$$ is equal to ___________.</p> | [] | null | 2 | <p>$${(A + B)^2} = {A^2} + {B^2} + AB + BA$$</p>
<p>$$ = {A^2} + \left[ {\matrix{
2 & 2 \cr
2 & 2 \cr
} } \right]$$</p>
<p>$$\therefore$$ $${B^2} + AB + BA = \left[ {\matrix{
2 & 2 \cr
2 & 2 \cr
} } \right]$$ ..... (1)</p>
<p>$$AB = \left[ {\matrix{
1 & { - 1} \cr
2 & \alpha \cr
} } \right]\left[ {\matrix{
\beta & 1 \cr
1 & 0 \cr
} } \right] = \left[ {\matrix{
{\beta - 1} & 1 \cr
{\alpha + 2\beta } & 2 \cr
} } \right]$$</p>
<p>$$BA = \left[ {\matrix{
\beta & 1 \cr
1 & 0 \cr
} } \right]\left[ {\matrix{
1 & { - 1} \cr
2 & \alpha \cr
} } \right] = \left[ {\matrix{
{\beta + 2} & {\alpha - \beta } \cr
1 & { - 1} \cr
} } \right]$$</p>
<p>$${B^2} = \left[ {\matrix{
\beta & 1 \cr
1 & 0 \cr
} } \right]\left[ {\matrix{
\beta & 1 \cr
1 & 0 \cr
} } \right] = \left[ {\matrix{
{{\beta ^2} + 1} & \beta \cr
\beta & 1 \cr
} } \right]$$</p>
<p>By (1) we get</p>
<p>$$\left[ {\matrix{
{{\beta ^2} + 2\beta } + 2 & {\alpha + 1} \cr
{\alpha + 3\beta + 1} & 2 \cr
} } \right] = \left[ {\matrix{
2 & 2 \cr
2 & 2 \cr
} } \right]$$</p>
<p>$$\therefore$$ $$\alpha = 1\,\,\beta = 0\,\, \Rightarrow {\alpha _1} = 1$$</p>
<p>Similarly if $${A^2} + AB + BA = 0$$ then</p>
<p>$$\left( {{A^2} = \left[ {\matrix{
1 & { - 1} \cr
2 & \alpha \cr
} } \right]\left[ {\matrix{
1 & { - 1} \cr
2 & \alpha \cr
} } \right] = \left[ {\matrix{
{ - 1} & { - 1 - \alpha } \cr
{2 + 2\alpha } & {{\alpha ^2} - 2} \cr
} } \right]} \right)$$</p>
<p>$$\left[ {\matrix{
{2\beta } & {\alpha - \beta + 1 - 1 - \alpha } \cr
{\alpha + 2\beta + 1 + 2 + 2\alpha } & {{\alpha ^2} - 2 + 1} \cr
} } \right] = \left[ {\matrix{
0 & 0 \cr
0 & 0 \cr
} } \right]$$</p>
<p>$$ \Rightarrow \beta = 0$$ and $$\alpha = - 1\,\, \Rightarrow {\alpha _2} = - 1$$</p>
<p>$$\therefore$$ $$|{\alpha _1} - {\alpha _2}| = |2| = 2.$$</p> | integer | jee-main-2022-online-28th-july-morning-shift |
1l6rfk48l | maths | matrices-and-determinants | multiplication-of-matrices | <p>Let $$X=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$$ and $$A=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$$. For $$\mathrm{k} \in N$$, if $$X^{\prime} A^{k} X=33$$, then $$\mathrm{k}$$ is equal to _______.</p> | [] | null | 10 | Given $A=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$
<br/><br/>
$A^{2}=\left[\begin{array}{lll}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad A^{4}=\left[\begin{array}{ccc}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
<br/><br/>
$\Rightarrow A^{k}=\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
<br/><br/>
$\therefore \quad X^{\prime} A^{k} X=[111]\left[\begin{array}{ccc}1 & 0 & 3 k \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=[3 k+3]$
<br/><br/>
$\Rightarrow[3 k+3]=33$ (here it shall be [33] as matrix can't be equal to a scalar)
<br/><br/>
i.e. $[3 k+3]=33$
<br/><br/>
$3 k+3=[33] \quad \Rightarrow k=10$
<br/><br/>
If $k$ is odd and apply above process, we don't get odd value of $k$
<br/><br/>
$\therefore k=10$ | integer | jee-main-2022-online-29th-july-evening-shift |
1ldo6f9s1 | maths | matrices-and-determinants | multiplication-of-matrices | <p>If $$A = {1 \over 2}\left[ {\matrix{
1 & {\sqrt 3 } \cr
{ - \sqrt 3 } & 1 \cr
} } \right]$$, then :</p> | [{"identifier": "A", "content": "$$\\mathrm{A^{30}-A^{25}=2I}$$"}, {"identifier": "B", "content": "$$\\mathrm{A^{30}+A^{25}-A=I}$$"}, {"identifier": "C", "content": "$$\\mathrm{A^{30}=A^{25}}$$"}, {"identifier": "D", "content": "$$\\mathrm{A^{30}+A^{25}+A=I}$$"}] | ["B"] | null | $$A = {1 \over 2}\left[ {\matrix{
1 & {\sqrt 3 } \cr
{ - \sqrt 3 } & 1 \cr
} } \right]$$
<br/><br/>Let $\theta=\frac{\pi}{3}$
<br/><br/>$$
\begin{aligned}
A^2 & =\left[\begin{array}{cc}
\cos \theta & \sin \theta \\\\
-\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{cc}
\cos \theta & \sin \theta \\\\
-\sin \theta & \cos \theta
\end{array}\right] \\\\
& =\left[\begin{array}{cc}
\cos 2 \theta & \sin 2 \theta \\\\
-\sin 2 \theta & \cos 2 \theta
\end{array}\right] \\\\
A^3 & =\left[\begin{array}{cc}
\cos 2 \theta & \sin 2 \theta \\\\
-\sin 2 \theta & \cos 2 \theta
\end{array}\right]\left[\begin{array}{cc}
\cos \theta & \sin \theta \\\\
-\sin \theta & \cos \theta
\end{array}\right] \\\\
& =\left[\begin{array}{cc}
\cos 3 \theta & \sin 3 \theta \\\\
-\sin 3 \theta & \cos 3 \theta
\end{array}\right]
\end{aligned}
$$
<br/><br/>$\begin{aligned} & \therefore \quad A^{30}=\left[\begin{array}{cc}\cos 30 \theta & \sin 30 \theta \\\\ -\sin 30 \theta & \cos 30 \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\ & A^{25}=\left[\begin{array}{cc}\cos 25 \theta & \sin 25 \theta \\\\ -\sin 25 \theta & \cos 25 \theta\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right]=A \\\\ & \therefore \quad A^{30}+A^{25}-A=I\end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift |
1lgre7f3c | maths | matrices-and-determinants | multiplication-of-matrices | <p>Let $$A=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]$$. If $$\mathrm{B}=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] A\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$$, then the sum of all the elements of the matrix $$\sum_\limits{n=1}^{50} B^{n}$$ is equal to</p> | [{"identifier": "A", "content": "50"}, {"identifier": "B", "content": "75"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "125"}] | ["C"] | null | $$
\begin{aligned}
& \text { Let } C=\left[\begin{array}{cc}
1 & 2 \\
-1 & -1
\end{array}\right], \mathrm{D}=\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right] \\\\
& \mathrm{DC}=\left[\begin{array}{cc}
1 & 2 \\
-1 & -1
\end{array}\right]\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\mathrm{I}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \mathrm{B}=\mathrm{CAD} \\\\
& \mathrm{B}^{\mathrm{n}}=\underbrace{(\mathrm{CAD})(\mathrm{CAD})(\mathrm{CAD}) \ldots(\mathrm{CAD})}_{\text {n-times }}
\end{aligned}
$$
<br/><br/>$$
\Rightarrow \mathrm{B}^{\mathrm{n}}=\mathrm{CA}^{\mathrm{n}} \mathrm{D}
$$
<br/><br/>$$
\mathrm{A}^2=\left[\begin{array}{cc}
1 & \frac{1}{51} \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & \frac{1}{51} \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & \frac{2}{51} \\
0 & 1
\end{array}\right]
$$
<br/><br/>$$
\mathrm{A}^3=\left[\begin{array}{cc}
1 & \frac{1}{51} \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & \frac{2}{51} \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & \frac{3}{51} \\
0 & 1
\end{array}\right]
$$
<br/><br/>$$
\text { Similarly } A^{\mathrm{n}}=\left[\begin{array}{cc}
1 & \frac{\mathrm{n}}{51} \\
0 & 1
\end{array}\right]
$$
<br/><br/>$$
\begin{aligned}
\mathrm{B}^{\mathrm{n}} & =\left[\begin{array}{cc}
1 & 2 \\
-1 & -1
\end{array}\right]\left[\begin{array}{cc}
1 & \frac{\mathrm{n}}{51} \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right] \\\\
& =\left[\begin{array}{cc}
1 & \frac{\mathrm{n}}{51}+2 \\
-1 & -\frac{\mathrm{n}}{51}-1
\end{array}\right]\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right] \\\\
& =\left[\begin{array}{cc}
\frac{\mathrm{n}}{51}+1 & \frac{\mathrm{n}}{51} \\
-\frac{\mathrm{n}}{51} & 1-\frac{\mathrm{n}}{51}
\end{array}\right]
\end{aligned}
$$
<br/><br/>$$
\sum\limits_{n=1}^{50} B^n=\left[\begin{array}{cc}
50+\frac{50 \cdot 51}{2 \cdot 51} & \frac{50 \cdot 51}{2 \cdot 51} \\\\
\frac{-50 \cdot 51}{2 \cdot 51} & 50-\frac{50 \cdot 51}{2 \cdot 51}
\end{array}\right]=\left[\begin{array}{cc}
75 & 25 \\
-25 & 25
\end{array}\right]
$$
<br/><br/>$$ \therefore $$ Sum of the elements = 100 | mcq | jee-main-2023-online-12th-april-morning-shift |
lsan7qev | maths | matrices-and-determinants | multiplication-of-matrices | Let $A=I_2-2 M M^T$, where $M$ is a real matrix of order $2 \times 1$ such that the relation $M^T M=I_1$ holds. If $\lambda$ is a real number such that the relation $A X=\lambda X$ holds for some non-zero real matrix $X$ of order $2 \times 1$, then the sum of squares of all possible values of $\lambda$ is equal to __________. | [] | null | 2 | $\begin{aligned} & A=I_2-2 M^T \\\\ & A^2=\left(I_2-2 M M^T\right)\left(I_2-2 M^T\right) \\\\ & =I_2-2 M^T-2 M M^T+4 M^T M^T \\\\ & =I_2-4 M M^T+4 M M^T \\\\ & =I_2\end{aligned}$
<br/><br/>$\begin{aligned} & \mathrm{AX}=\lambda \mathrm{X} \\\\ & \mathrm{A}^2 \mathrm{X}=\lambda \mathrm{AX} \\\\ & \mathrm{X}=\lambda(\lambda \mathrm{X}) \\\\ & \mathrm{X}=\lambda^2 \mathrm{X} \\\\ & \mathrm{X}\left(\lambda^2-1\right)=0 \\\\ & \lambda^2=1 \\\\ & \lambda= \pm 1\end{aligned}$
<br/><br/>Sum of square of all possible values $=2$ | integer | jee-main-2024-online-1st-february-evening-shift |
lsblig15 | maths | matrices-and-determinants | multiplication-of-matrices | Let $A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right], B=\left[B_1, B_2, B_3\right]$, where $B_1, B_2, B_3$ are column matrics, and
<br/><br/>$$
\mathrm{AB}_1=\left[\begin{array}{l}
1 \\
0 \\
0
\end{array}\right], \mathrm{AB}_2=\left[\begin{array}{l}
2 \\
3 \\
0
\end{array}\right], \quad \mathrm{AB}_3=\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]
$$
<br/><br/>If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^3+\beta^3$ is equal to ____________. | [] | null | 28 | <p>$$\mathrm{A}=\left[\begin{array}{lll}
2 & 0 & 1 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{array}\right] \quad \mathrm{B}=\left[\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3\right]$$</p>
<p>$$\mathrm{B}_1=\left[\begin{array}{l}
\mathrm{x}_1 \\
\mathrm{y}_1 \\
\mathrm{z}_1
\end{array}\right], \quad \mathrm{B}_2=\left[\begin{array}{l}
\mathrm{x}_2 \\
\mathrm{y}_2 \\
\mathrm{z}_2
\end{array}\right], \quad \mathrm{B}_3=\left[\begin{array}{l}
\mathrm{x}_3 \\
\mathrm{y}_3 \\
\mathrm{z}_3
\end{array}\right]$$</p>
<p>$$\mathrm{AB}_1=\left[\begin{array}{lll}
2 & 0 & 1 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{x}_1 \\
\mathrm{y}_1 \\
\mathrm{z}_1
\end{array}\right]=\left[\begin{array}{l}
1 \\
0 \\
0
\end{array}\right]$$</p>
<p>$$\begin{gathered}
\mathrm{x}_1=1, \mathrm{y}_1=-1, \mathrm{z}_1=-1 \\
\mathrm{AB}_2=\left[\begin{array}{lll}
2 & 0 & 1 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{x}_2 \\
\mathrm{y}_2 \\
\mathrm{z}_2
\end{array}\right]=\left[\begin{array}{l}
2 \\
3 \\
0
\end{array}\right] \\
\mathrm{x}_2=2, \mathrm{y}_2=1, \mathrm{z}_2=-2 \\
\mathrm{AB}_3=\left[\begin{array}{lll}
2 & 0 & 1 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{x}_3 \\
\mathrm{y}_3 \\
\mathrm{z}_3
\end{array}\right]=\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right] \\
\mathrm{x}_3=2, \mathrm{y}_3=0, \mathrm{z}_3=-1 \\
\mathrm{~B}=\left[\begin{array}{ccc}
1 & 2 & 2 \\
-1 & 1 & 0 \\
-1 & -2 & -1
\end{array}\right] \\
\alpha=|\mathrm{B}|=3 \\
\beta=1 \\
\alpha^3+\beta^3=27+1=28
\end{gathered}$$</p> | integer | jee-main-2024-online-27th-january-morning-shift |
PhR7ljF2bx2QxzpZ | maths | matrices-and-determinants | operations-on-matrices | Let $$A = \left( {\matrix{
1 & 2 \cr
3 & 4 \cr
} } \right)$$ and $$B = \left( {\matrix{
a & 0 \cr
0 & b \cr
} } \right),a,b \in N.$$ Then | [{"identifier": "A", "content": "there cannot exist any $$B$$ such that $$AB=BA$$ "}, {"identifier": "B", "content": "there exist more then one but finite number of $$B'$$s such that $$AB=BA$$"}, {"identifier": "C", "content": "there exists exactly one $$B$$ such that $$AB=BA$$ "}, {"identifier": "D", "content": "there exist infinitely many $$B'$$s such that $$AB=BA$$"}] | ["D"] | null | $$A = \left[ {\matrix{
1 & 2 \cr
3 & 4 \cr
} } \right]\,\,\,\,B = \left[ {\matrix{
a & 0 \cr
0 & b \cr
} } \right]$$
<br><br>$$AB = \left[ {\matrix{
a & {2b} \cr
{3a} & {4b} \cr
} } \right]$$
<br><br>$$BA = \left[ {\matrix{
a & 0 \cr
0 & b \cr
} } \right]\left[ {\matrix{
1 & 2 \cr
3 & 4 \cr
} } \right] = \left[ {\matrix{
a & {2a} \cr
{3b} & {4b} \cr
} } \right]$$
<br><br>Hence, $$AB=BA$$ only when $$a=b$$
<br><br>$$\therefore$$ There can be infinitely many $$B's$$
<br><br>for which $$AB=BA$$ | mcq | aieee-2006 |
1ciXKjNjBu8EWyS3 | maths | matrices-and-determinants | operations-on-matrices | If $$A$$ and $$B$$ are square matrices of size $$n\, \times \,n$$ such that
<br/>$${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right),$$ then which of the following will be always true? | [{"identifier": "A", "content": "$$A=B$$ "}, {"identifier": "B", "content": "$$AB=BA$$ "}, {"identifier": "C", "content": "either of $$A$$ or $$B$$ is a zero matrix"}, {"identifier": "D", "content": "either of $$A$$ or $$B$$ is identity matrix"}] | ["B"] | null | $${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$$
<br><br>$${A^2} - {B^2} = {A^2} + AB - BA - {B^2}$$
<br><br>$$ \Rightarrow AB = BA$$ | mcq | aieee-2006 |
BVviBGTP0IGky423R1Ghg | maths | matrices-and-determinants | operations-on-matrices | Let P = $$\left[ {\matrix{
1 & 0 & 0 \cr
3 & 1 & 0 \cr
9 & 3 & 1 \cr
} } \right]$$ and Q = [q<sub>ij</sub>] be two 3 $$ \times $$ 3 matrices such that Q – P<sup>5</sup> = I<sub>3</sub>.
<br/><br/>Then $${{{q_{21}} + {q_{31}}} \over {{q_{32}}}}$$ is equal to : | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "135 "}, {"identifier": "D", "content": "10"}] | ["D"] | null | $$P = \left[ {\matrix{
1 & 0 & 0 \cr
3 & 1 & 0 \cr
9 & 3 & 1 \cr
} } \right]$$
<br><br>$${P^2} = \left[ {\matrix{
1 & 0 & 0 \cr
{3 + 3} & 1 & 0 \cr
{9 + 9 + 9} & {3 + 3} & 1 \cr
} } \right]$$
<br><br>$${P^3} = \left[ {\matrix{
1 & 0 & 0 \cr
{3 + 3 + 3} & 1 & 0 \cr
{6.9} & {3 + 3 + 3} & 1 \cr
} } \right]$$
<br><br>$${P^n} = \left[ {\matrix{
1 & 0 & 0 \cr
{3n} & 1 & 0 \cr
{{{n\left( {n + 1} \right)} \over 2}{3^2}} & {3n} & 1 \cr
} } \right]$$
<br><br>$${P^5} = \left[ {\matrix{
1 & 0 & 0 \cr
{5.3} & 1 & 0 \cr
{15.9} & {5.3} & 1 \cr
} } \right]$$
<br><br>$$Q = {P^5} + {{\rm I}_3}$$
<br><br>$$Q = \left[ {\matrix{
2 & 0 & 0 \cr
{15} & 2 & 0 \cr
{135} & {15} & 2 \cr
} } \right]$$
<br><br>$${{{q_{21}} + {q_{31}}} \over {{q_{32}}}} = {{15 + 135} \over {15}} = 10$$
<br><br>Aliter
<br><br>$$P = \left( {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right) + \left( {\matrix{
0 & 0 & 0 \cr
3 & 0 & 0 \cr
9 & 3 & 0 \cr
} } \right)$$
<br><br>$$P = {\rm I} + X$$
<br><br>$$X = \left( {\matrix{
0 & 0 & 0 \cr
3 & 0 & 0 \cr
9 & 3 & 0 \cr
} } \right)$$
<br><br>$${X^2} = \left( {\matrix{
0 & 0 & 0 \cr
0 & 0 & 0 \cr
9 & 0 & 0 \cr
} } \right)$$
<br><br>$${{X_3} = 0}$$
<br>$${{P^5} = {\rm I} + 5X + 10{X^2}}$$
<br>$${Q = {P^5} + {\rm I} = 2{\rm I} + 5X + 10{X^2}}$$
<br><br>$$Q = \left( {\matrix{
2 & 0 & 0 \cr
0 & 2 & 0 \cr
0 & 0 & 2 \cr
} } \right) + \left( {\matrix{
0 & 0 & 0 \cr
{15} & 0 & 0 \cr
{15} & {15} & 0 \cr
} } \right) + \left( {\matrix{
0 & 0 & 0 \cr
0 & 0 & 0 \cr
{90} & 0 & 0 \cr
} } \right)$$
<br><br>$$ \Rightarrow \,\,Q = \left( {\matrix{
2 & 0 & 0 \cr
{15} & 2 & 0 \cr
{135} & {15} & 2 \cr
} } \right)$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
Twer9dbwsJBwRwagoq2cN | maths | matrices-and-determinants | operations-on-matrices | Let $$A = \left( {\matrix{
{\cos \alpha } & { - \sin \alpha } \cr
{\sin \alpha } & {\cos \alpha } \cr
} } \right)$$, ($$\alpha $$ $$ \in $$ R)<br/> such that $${A^{32}} = \left( {\matrix{
0 & { - 1} \cr
1 & 0 \cr
} } \right)$$ then a value of $$\alpha $$ is | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${\\pi \\over {16}}$$"}, {"identifier": "C", "content": "$${\\pi \\over {32}}$$"}, {"identifier": "D", "content": "$${\\pi \\over {64}}$$"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265941/exam_images/vlkqlnh7isvrysclmlsm.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263666/exam_images/rf4qtzmkrumvqo5bw5x9.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266088/exam_images/bapofvunaei3ewpu2sex.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266055/exam_images/pn2fkxk2e7btjw3e87xb.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263483/exam_images/v60ksuiexs4xmoaxy2nv.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264884/exam_images/lwetrvd1yci6wistyder.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264935/exam_images/jrlvoj8gndiqxczs4gtp.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264644/exam_images/sd5znxchfmvbwrpympou.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Morning Slot Mathematics - Matrices and Determinants Question 239 English Explanation"></picture>
<br><br>From here sin 32$$\alpha $$ = 1 and cos 32$$\alpha $$ = 0
<br><br>$$ \therefore $$ 32$$\alpha $$ = 2n$$\pi $$ + $${\pi \over 2}$$
<br><br>$$ \Rightarrow $$ $$\alpha $$ = $${\pi \over {64}} + {{n\pi } \over {16}}$$
<br><br>Putting n = 0, $$\alpha $$ = $${\pi \over {64}}$$ | mcq | jee-main-2019-online-8th-april-morning-slot |
VLRXTfvhfMWdTaSlS37k9k2k5e2emb4 | maths | matrices-and-determinants | operations-on-matrices | Let $$\alpha $$ be a root of the equation x<sup>2</sup> + x + 1 = 0 and the <br/>matrix A = $${1 \over {\sqrt 3 }}\left[ {\matrix{
1 & 1 & 1 \cr
1 & \alpha & {{\alpha ^2}} \cr
1 & {{\alpha ^2}} & {{\alpha ^4}} \cr
} } \right]$$<br/><br/> then the matrix
A<sup>31</sup> is equal to
| [{"identifier": "A", "content": "A<sup>2</sup>"}, {"identifier": "B", "content": "A"}, {"identifier": "C", "content": "I<sub>3</sub>"}, {"identifier": "D", "content": "A<sup>3</sup>"}] | ["D"] | null | x<sup>2</sup> + x + 1 = 0
<br><br>$$ \Rightarrow $$ x = $${{ - 1 + i\sqrt 3 } \over 2}$$ = $$\omega $$ or $${{ - 1 - i\sqrt 3 } \over 2}$$ = $${\omega ^2}$$
<br><br>Let $$\alpha $$ = $$\omega $$
<br><br>$$ \therefore $$ A = $${1 \over {\sqrt 3 }}\left[ {\matrix{
1 & 1 & 1 \cr
1 & \omega & {{\omega ^2}} \cr
1 & {{\omega ^2}} & {{\omega ^4}} \cr
} } \right]$$
<br><br>A<sup>2</sup> = $${1 \over 3}\left[ {\matrix{
3 & 0 & 0 \cr
0 & 0 & 3 \cr
0 & 3 & 0 \cr
} } \right]$$ = $$\left[ {\matrix{
1 & 0 & 0 \cr
0 & 0 & 1 \cr
0 & 1 & 0 \cr
} } \right]$$
<br><br>Now A<sup>4</sup> = $$\left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$ = I
<br><br>$$ \therefore $$ A<sup>31</sup> = A<sup>28</sup>A<sup>3</sup> = A<sup>3</sup> | mcq | jee-main-2020-online-7th-january-morning-slot |
WRLXvWOxnX5Sxd5nGV7k9k2k5hjw6xo | maths | matrices-and-determinants | operations-on-matrices | If $$A = \left( {\matrix{
2 & 2 \cr
9 & 4 \cr
} } \right)$$ and $$I = \left( {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right)$$ then 10A<sup>–1</sup> is
equal to : | [{"identifier": "A", "content": "6I \u2013 A"}, {"identifier": "B", "content": "4I \u2013 A"}, {"identifier": "C", "content": "A \u2013 6I"}, {"identifier": "D", "content": "A \u2013 4I"}] | ["C"] | null | According to Cayley Hamilton equation
<br>|A – $$\lambda $$I| = 0
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
{2 - \lambda } & 2 \cr
9 & {4 - \lambda } \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ (2 – $$\lambda $$)(4 – $$\lambda $$) – 18 = 0
<br><br>$$ \Rightarrow $$ 8 – 2$$\lambda $$ – 4$$\lambda $$ + $$\lambda $$<sup>2</sup>
– 18 = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$<sup>2</sup>
– 6$$\lambda $$ – 10 = 0
<br><br>$$ \therefore $$ A<sup>2</sup>
– 6A– 10 = 0
<br><br>$$ \Rightarrow $$ A<sup>–1</sup>(A<sup>2</sup>) – 6A<sup>–1</sup>A – 10A<sup>–1</sup> = 0
<br><br>$$ \Rightarrow $$ A – 6I – 10A<sup>–1</sup> = 0
<br><br>$$ \Rightarrow $$ 10A<sup>–1</sup> = A – 6I | mcq | jee-main-2020-online-8th-january-evening-slot |
VQhiXPYmhIls9RqlD81kmizzhfm | maths | matrices-and-determinants | operations-on-matrices | Let $$A = \left[ {\matrix{
{{a_1}} \cr
{{a_2}} \cr
} } \right]$$ and $$B = \left[ {\matrix{
{{b_1}} \cr
{{b_2}} \cr
} } \right]$$ be two 2 $$\times$$ 1 matrices with real entries such that A = XB, where <br/><br/>$$X = {1 \over {\sqrt 3 }}\left[ {\matrix{
1 & { - 1} \cr
1 & k \cr
} } \right]$$, and k$$\in$$R. <br/><br/>If $$a_1^2$$ + $$a_2^2$$ = $${2 \over 3}$$(b$$_1^2$$ + b$$_2^2$$) and (k<sup>2</sup> + 1) b$$_2^2$$ $$\ne$$ $$-$$2b<sub>1</sub>b<sub>2</sub>, then the value of k is __________. | [] | null | 1 | $$XB = A$$
<br><br>$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}\left[ {\matrix{
1 & { - 1} \cr
1 & k \cr
} } \right]\left[ {\matrix{
{{b_1}} \cr
{{b_2}} \cr
} } \right] = \left[ {\matrix{
{{a_1}} \cr
{{a_2}} \cr
} } \right]$$
<br><br>$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}\left[ {\matrix{
{{b_1} - {b_2}} \cr
{{b_1} + k{b_2}} \cr
} } \right] = \left[ {\matrix{
{{a_1}} \cr
{{a_2}} \cr
} } \right]$$<br><br>$${b_1} - {b_2} = \sqrt 3 {a_1} \Rightarrow 3a_1^2 = b_1^2 + b_2^2 - 2{b_1}{b_2}$$<br><br>$${b_1} + k{b_2} = \sqrt 3 {a_2} \Rightarrow 3a_2^2 = b_1^2 + {k^2}b_2^2 + 2k{b_1}{b_2}$$<br><br>$$3\left( {a_1^2 + a_2^2} \right) = 2b_1^2 + \left( {{k^2} + 1} \right)b_2^2 + 2{b_1}{b_2}(k - 1)$$
<br><br>$$ \Rightarrow $$ $${a_1^2 + a_2^2} $$ = $${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$$
<br><br>Given $$a_1^2$$ + $$a_2^2$$ = $${2 \over 3}$$(b$$_1^2$$ + b$$_2^2$$)
<br><br>$$ \therefore $$ $${2 \over 3}$$(b$$_1^2$$ + b$$_2^2$$) = $${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$$
<br><br>$$ \Rightarrow $$ $${2 \over 3}b_2^2 = {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$$
<br><br>Comparing both sides, We get
<br><br>$${{\left( {{k^2} + 1} \right)} \over 3} = {2 \over 3}$$
<br><br>$$ \Rightarrow $$ k<sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ k = $$ \pm $$ 1 ......(1)
<br><br>and $${2 \over 3}\left( {k - 1} \right) = 0$$ $$ \Rightarrow $$ k = 1 ....(2)
<br><br>From (1) and (2),
<br><br>k = 1 | integer | jee-main-2021-online-16th-march-evening-shift |
1krq0aujn | maths | matrices-and-determinants | operations-on-matrices | Let $$A = \left( {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
0 & 0 & 1 \cr
} } \right)$$ and B = 7A<sup>20</sup> $$-$$ 20A<sup>7</sup> + 2I, where I is an identity matrix of order 3 $$\times$$ 3. If B = [b<sub>ij</sub>], then b<sub>13</sub>is equal to _____________. | [] | null | 910 | Let $$A = \left( {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
0 & 0 & 1 \cr
} } \right) = I + C$$<br><br>where, $$I = \left( {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right),C = \left( {\matrix{
0 & { - 1} & 0 \cr
0 & 0 & { - 1} \cr
0 & 0 & 0 \cr
} } \right)$$<br><br>$${C^2} = \left( {\matrix{
0 & 0 & 1 \cr
0 & 0 & 0 \cr
0 & 0 & 0 \cr
} } \right),$$<br><br>$${C^3} = \left( {\matrix{
0 & 0 & 0 \cr
0 & 0 & 0 \cr
0 & 0 & 0 \cr
} } \right) = {C^4} = {C^5} = ........$$<br><br>$$B = 7{A^{20}} - 20{A^7} + 2I$$<br><br>$$ = 7{(I + C)^{20}} + 20{(I + C)^7} + 2I$$<br><br>$$ = 7(I + 20C + {}^{20}{C_2}{C^2}) - 20(I + 7C + {}^7{C_2}{C^2}) + 2I$$<br><br> So<br><br>$${b_{13}} = 7 \times {}^{20}{C_2}{C^2} - 20 \times {}^7{C_2} = 910$$ | integer | jee-main-2021-online-20th-july-morning-shift |
1kru3wirg | maths | matrices-and-determinants | operations-on-matrices | Let A = [a<sub>ij</sub>] be a real matrix of order 3 $$\times$$ 3, such that a<sub>i1</sub> + a<sub>i2</sub> + a<sub>i3</sub> = 1, for i = 1, 2, 3. Then, the sum of all the entries of the matrix A<sup>3</sup> is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "9"}] | ["C"] | null | $$A = \left[ {\matrix{
{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right]$$<br><br>Let $$x = \left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right]$$<br><br>$$AX = \left[ {\matrix{
{{a_{11}} + {a_{12}} + {a_{13}}} \cr
{{a_{21}} + {a_{22}} + {a_{23}}} \cr
{{a_{31}} + {a_{32}} + {a_{33}}} \cr
} } \right] = \left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right]$$<br><br>$$\Rightarrow$$ AX = X<br><br>Replace X by AX<br><br>A<sup>2</sup>X = AX = X<br><br>Replace X by AX<br><br>A<sup>3</sup>X = AX = X<br><br>Let $${A^3} = \left[ {\matrix{
{{x_1}} & {{x_2}} & {{x_3}} \cr
{{y_1}} & {{y_2}} & {{y_3}} \cr
{{z_1}} & {{z_2}} & {{z_3}} \cr
} } \right]$$<br><br>$${A^3}\left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right] = \left[ {\matrix{
{{x_1}} & {{x_2}} & {{x_3}} \cr
{{y_1}} & {{y_2}} & {{y_3}} \cr
{{z_1}} & {{z_2}} & {{z_3}} \cr
} } \right] = \left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right]$$<br><br>Sum of all the element = 3 | mcq | jee-main-2021-online-22th-july-evening-shift |
1krygt7fm | maths | matrices-and-determinants | operations-on-matrices | If $$A = \left[ {\matrix{
1 & 1 & 1 \cr
0 & 1 & 1 \cr
0 & 0 & 1 \cr
} } \right]$$ and M = A + A<sup>2</sup> + A<sup>3</sup> + ....... + A<sup>20</sup>, then the sum of all the elements of the matrix M is equal to _____________. | [] | null | 2020 | $${A^n} = \left[ {\matrix{
1 & n & {{{{n^2} + n} \over 2}} \cr
0 & 1 & n \cr
0 & 0 & 1 \cr
} } \right]$$<br><br>So, required sum<br><br>$$ = 20 \times 3 + 2 \times \left( {{{20 \times 21} \over 2}} \right) + \sum\limits_{r = 1}^{20} {\left( {{{{r^2} + r} \over 2}} \right)} $$<br><br>$$ = 60 + 420 + 105 + 35 \times 41 = 2020$$ | integer | jee-main-2021-online-27th-july-evening-shift |
1krzn7q3c | maths | matrices-and-determinants | operations-on-matrices | If $$P = \left[ {\matrix{
1 & 0 \cr
{{1 \over 2}} & 1 \cr
} } \right]$$, then P<sup>50</sup> is : | [{"identifier": "A", "content": "$$\\left[ {\\matrix{\n 1 & 0 \\cr \n {25} & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {\\matrix{\n 1 & {50} \\cr \n 0 & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {\\matrix{\n 1 & {25} \\cr \n 0 & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {\\matrix{\n 1 & 0 \\cr \n {50} & 1 \\cr \n\n } } \\right]$$"}] | ["A"] | null | $$P = \left[ {\matrix{
1 & 0 \cr
{{1 \over 2}} & 1 \cr
} } \right]$$<br><br>$${P^2} = \left[ {\matrix{
1 & 0 \cr
{{1 \over 2}} & 1 \cr
} } \right]\left[ {\matrix{
1 & 0 \cr
{{1 \over 2}} & 1 \cr
} } \right] = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]$$<br><br>$${P^3} = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]\left[ {\matrix{
1 & 0 \cr
{{1 \over 2}} & 1 \cr
} } \right] = \left[ {\matrix{
1 & 0 \cr
{{3 \over 2}} & 1 \cr
} } \right]$$<br><br>$${P^4} = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]\left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right] = \left[ {\matrix{
1 & 0 \cr
2 & 1 \cr
} } \right]$$<br><br>$$ \vdots $$<br><br>$$\therefore$$ $${P^{50}} = \left[ {\matrix{
1 & 0 \cr
{25} & 1 \cr
} } \right]$$ | mcq | jee-main-2021-online-25th-july-evening-shift |
1ktbfkr1u | maths | matrices-and-determinants | operations-on-matrices | If $$A = \left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr
{{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)$$, $$B = \left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right)$$, $$i = \sqrt { - 1} $$, and Q = A<sup>T</sup>BA, then the inverse of the matrix A Q<sup>2021</sup> A<sup>T</sup> is equal to : | [{"identifier": "A", "content": "$$\\left( {\\matrix{\n {{1 \\over {\\sqrt 5 }}} & { - 2021} \\cr \n {2021} & {{1 \\over {\\sqrt 5 }}} \\cr \n\n } } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {\\matrix{\n 1 & 0 \\cr \n { - 2021i} & 1 \\cr \n\n } } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {\\matrix{\n 1 & 0 \\cr \n {2021i} & 1 \\cr \n\n } } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {\\matrix{\n 1 & { - 2021i} \\cr \n 0 & 1 \\cr \n\n } } \\right)$$"}] | ["B"] | null | $$A{A^T} = \left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr
{{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)\left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{{ - 2} \over {\sqrt 5 }}} \cr
{{2 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)$$<br><br>$$A{A^T} = \left( {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right) = I$$<br><br>$${Q^2} = {A^T}BA\,{A^T}BA = {A^T}BIBA$$<br><br>$$ \Rightarrow {Q^2} = {A^T}{B^2}A$$<br><br>$${Q^3} = {A^T}{B^2}A{A^T}BA \Rightarrow {Q^3} = {A^T}{B^3}A$$<br><br>Similarly : $${Q^{2021}} = {A^T}{B^{2021}}A$$<br><br>Now, $${B^2} = \left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right)\left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right) = \left( {\matrix{
1 & 0 \cr
{2i} & 1 \cr
} } \right)$$<br><br>$${B^3} = \left( {\matrix{
1 & 0 \cr
{2i} & 1 \cr
} } \right)\left( {\matrix{
1 & 0 \cr
i & 1 \cr
} } \right) \Rightarrow {B^3} = \left( {\matrix{
1 & 0 \cr
{3i} & 1 \cr
} } \right)$$<br><br>Similarly $${B^{2021}} = \left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)$$<br><br>$$\therefore$$ $$A{Q^{2021}} = {A^T} = A{A^T}{B^{2021}}\,A{A^T} = I{B^{2021}}I$$<br><br>$$ \Rightarrow A{Q^{2021}}\,{A^T} = {B^{2021}} = \left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)$$<br><br>$$\therefore$$ $${(A{Q^{2021}}\,{A^T})^{ - 1}} = {\left( {\matrix{
1 & 0 \cr
{2021i} & 1 \cr
} } \right)^{ - 1}} = \left( {\matrix{
1 & 0 \cr
{ - 2021i} & 1 \cr
} } \right)$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1kteid0ux | maths | matrices-and-determinants | operations-on-matrices | If the matrix $$A = \left( {\matrix{
0 & 2 \cr
K & { - 1} \cr
} } \right)$$ satisfies $$A({A^3} + 3I) = 2I$$, then the value of K is : | [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$$-$$$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "1"}] | ["A"] | null | Given matrix $$A = \left[ {\matrix{
0 & 2 \cr
k & { - 1} \cr
} } \right]$$<br><br>$${A^4} + 3IA = 2I$$<br><br>$$ \Rightarrow {A^4} = 2I - 3A$$<br><br>Also characteristic equation of A is $$|A - \lambda I|\, = 0$$<br><br>$$ \Rightarrow \left| {\matrix{
{0 - \lambda } & 2 \cr
k & { - 1 - \lambda } \cr
} } \right| = 0$$<br><br>$$ \Rightarrow \lambda + {\lambda ^2} - 2k = 0$$<br><br>$$ \Rightarrow A + {A^2} = 2K.I$$<br><br>$$ \Rightarrow {A^2} = 2KI - A$$<br><br>$$ \Rightarrow {A^4} = 4{K^2}I + {A^2} - 4AK$$<br><br>Put $${A^2} = 2KI - A$$<br><br>and $${A^4} = 2I - 3A$$<br><br>$$2I - 3A = 4{K^2}I + 2KI - A - 4AK$$<br><br>$$ \Rightarrow I(2 - 2K - 4{K^2}) = A(2 - 4K)$$<br><br>$$ \Rightarrow - 2I(2{K^2} + K - 1) = 2A(1 - 2K)$$<br><br>$$ \Rightarrow - 2I(2K - 1)(K + 1) = 2A(1 - 2K)$$<br><br>$$ \Rightarrow (2K - 1)(2A) - 2I(2K - 1)(K + 1) = 0$$<br><br>$$ \Rightarrow (2K - 1)[2A - 2I(K + 1)] = 0$$<br><br>$$ \Rightarrow K = {1 \over 2}$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktkekk3h | maths | matrices-and-determinants | operations-on-matrices | The number of elements in the set $$\left\{ {A = \left( {\matrix{
a & b \cr
0 & d \cr
} } \right):a,b,d \in \{ - 1,0,1\} \,and\,{{(I - A)}^3} = I - {A^3}} \right\}$$, where I is 2 $$\times$$ 2 identity matrix, is : | [] | null | 8 | $${(I - A)^3} = {I^3} - {A^3} - 3A(I - A) = I - {A^3}$$<br><br>$$ \Rightarrow 3A(I - A) = 0$$ or $${A^2} = A$$<br><br>$$ \Rightarrow \left[ {\matrix{
{{a^2}} & {ab + bd} \cr
0 & {{d^2}} \cr
} } \right] = \left[ {\matrix{
a & b \cr
0 & d \cr
} } \right]$$<br><br>$$ \Rightarrow {a^2} = a,b(a + d - 1) = 0,{d^2} = d$$<br><br>If b $$\ne$$ 0, a + d = 1 $$\Rightarrow$$ 4 ways<br><br>If b = 0, a = 0, 1 & d = 0, 1 $$\Rightarrow$$ 4 ways<br><br>$$\Rightarrow$$ Total 8 matrices | integer | jee-main-2021-online-31st-august-evening-shift |
1l5452x4x | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$A = [{a_{ij}}]$$ be a square matrix of order 3 such that $${a_{ij}} = {2^{j - i}}$$, for all i, j = 1, 2, 3. Then, the matrix A<sup>2</sup> + A<sup>3</sup> + ...... + A<sup>10</sup> is equal to :</p> | [{"identifier": "A", "content": "$$\\left( {{{{3^{10}} - 3} \\over 2}} \\right)A$$"}, {"identifier": "B", "content": "$$\\left( {{{{3^{10}} - 1} \\over 2}} \\right)A$$"}, {"identifier": "C", "content": "$$\\left( {{{{3^{10}} + 1} \\over 2}} \\right)A$$"}, {"identifier": "D", "content": "$$\\left( {{{{3^{10}} + 3} \\over 2}} \\right)A$$"}] | ["A"] | null | <p>Given, $${a_{ij}} = {2^{j - i}}$$</p>
<p>Now, $$A = \left[ {\matrix{
{{2^0}} & {{2^1}} & {{2^2}} \cr
{{2^{ - 1}}} & {{2^0}} & {{2^1}} \cr
{{2^{ - 2}}} & {{2^{ - 1}}} & {{2^0}} \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
1 & 2 & 4 \cr
{{1 \over 2}} & 1 & 2 \cr
{{1 \over 4}} & {{1 \over 2}} & 1 \cr
} } \right]$$</p>
<p>$${A^2} = \left[ {\matrix{
1 & 2 & 4 \cr
{{1 \over 2}} & 1 & 2 \cr
{{1 \over 4}} & {{1 \over 2}} & 1 \cr
} } \right]\left[ {\matrix{
1 & 2 & 4 \cr
{{1 \over 2}} & 1 & 2 \cr
{{1 \over 4}} & {{1 \over 2}} & 1 \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
{1 + 1 + 1} & {2 + 2 + 2} & {4 + 4 + 4} \cr
{{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} & {2 + 2 + 2} \cr
{{1 \over 4} + {1 \over 4} + {1 \over 4}} & {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
3 & 6 & {12} \cr
{{3 \over 2}} & 3 & 6 \cr
{{3 \over 4}} & {{3 \over 2}} & 3 \cr
} } \right]$$</p>
<p>$$ = 3\left[ {\matrix{
1 & 2 & 4 \cr
{{1 \over 2}} & 1 & 2 \cr
{{1 \over 4}} & {{1 \over 2}} & 1 \cr
} } \right]$$</p>
<p>$$ = 3A$$</p>
<p>Similarly, $${A^3} = {3^2}A$$</p>
<p>$${A^4} = {3^3}A$$</p>
<p>$$\therefore$$ $${A^2} + {A^3} + \,\,......\,\, + \,\,{A^{10}}$$</p>
<p>$$ = 3A + {3^2}A + {3^3}A + \,\,......\,\, + \,\,{3^9}A$$</p>
<p>$$ = A(3 + {3^2} + {3^3} + \,\,......\,\, + \,\,{3^9})$$</p>
<p>$$ = A\left( {{{3({3^9} - 1)} \over {3 - 1}}} \right) = {{3({3^9} - 1)} \over 2}A$$ = $$\left( {{{{3^{10}} - 3} \over 2}} \right)A$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54uepk6 | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$M = \left[ {\matrix{
0 & { - \alpha } \cr
\alpha & 0 \cr
} } \right]$$, where $$\alpha$$ is a non-zero real number an $$N = \sum\limits_{k = 1}^{49} {{M^{2k}}} $$. If $$(I - {M^2})N = - 2I$$, then the positive integral value of $$\alpha$$ is ____________.</p> | [] | null | 1 | $M=\left[\begin{array}{cc}0 & -\alpha \\ \alpha & 0\end{array}\right], M^{2}=\left[\begin{array}{cc}-\alpha^{2} & 0 \\ 0 & -\alpha^{2}\end{array}\right]=-\alpha^{2}$ I
<br/><br/>
$N=M^{2}+M^{4}+\ldots+M^{98}$
<br/><br/>
$=\left[-\alpha^{2}+\alpha^{4}-\alpha^{6}+\ldots\right] I$
<br/><br/>
$=\frac{-\alpha^{2}\left(1-\left(-\alpha^{2}\right)^{49}\right)}{1+\alpha^{2}} \cdot 1$
<br/><br/>
$I-M^{2}=\left(1+\alpha^{2}\right) I$
<br/><br/>
$\left(I-M^{2}\right) N=-\alpha^{2}\left(\alpha^{98}+1\right)=-2$
<br/><br/>
$\therefore \alpha=1$ | integer | jee-main-2022-online-29th-june-evening-shift |
1l59l4v93 | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$A = \left( {\matrix{
2 & { - 2} \cr
1 & { - 1} \cr
} } \right)$$ and $$B = \left( {\matrix{
{ - 1} & 2 \cr
{ - 1} & 2 \cr
} } \right)$$. Then the number of elements in the set {(n, m) : n, m $$\in$$ {1, 2, .........., 10} and nA<sup>n</sup> + mB<sup>m</sup> = I} is ____________.</p> | [] | null | 1 | <p>$${A^2} = \left[ {\matrix{
2 & { - 2} \cr
1 & { - 1} \cr
} } \right]\left[ {\matrix{
2 & { - 2} \cr
1 & { - 1} \cr
} } \right] = \left[ {\matrix{
2 & { - 2} \cr
1 & { - 1} \cr
} } \right] = A$$</p>
<p>$$ \Rightarrow {A^K} = A,\,K \in I$$</p>
<p>$${B^2} = \left[ {\matrix{
{ - 1} & 2 \cr
{ - 1} & 2 \cr
} } \right]\left[ {\matrix{
{ - 1} & 2 \cr
{ - 1} & 2 \cr
} } \right] = \left[ {\matrix{
{ - 1} & 2 \cr
{ - 1} & 2 \cr
} } \right] = B$$</p>
<p>So, $${B^K} = B,\,K \in I$$</p>
<p>$$n{A^n} + m{B^m} = nA + mB$$</p>
<p>$$ = \left[ {\matrix{
{2n - 2n} \cr
{n - n} \cr
} } \right] + \left[ {\matrix{
{ - m} & {2m} \cr
{ - m} & {2m} \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right]$$</p>
<p>So, $$2n - m = 1,\, - n + m = 0,\,2m - n = 1$$</p>
<p>So, $$(m,n) = (1,1)$$</p> | integer | jee-main-2022-online-25th-june-evening-shift |
1l5bb0wm7 | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$S = \left\{ {\left( {\matrix{
{ - 1} & a \cr
0 & b \cr
} } \right);a,b \in \{ 1,2,3,....100\} } \right\}$$ and let $${T_n} = \{ A \in S:{A^{n(n + 1)}} = I\} $$. Then the number of elements in $$\bigcap\limits_{n = 1}^{100} {{T_n}} $$ is ___________.</p> | [] | null | 100 | $$
\begin{aligned}
&\mathrm{A}=\left[\begin{array}{cc}
-1 & \mathrm{a} \\\\
0 & \mathrm{~b}
\end{array}\right] \\\\
&\mathrm{A}^2=\left[\begin{array}{cc}
-1 & \mathrm{a} \\\\
0 & \mathrm{~b}
\end{array}\right]\left[\begin{array}{cc}
-1 & \mathrm{a} \\\\
0 & \mathrm{~b}
\end{array}\right] \\\\
&=\left[\begin{array}{cc}
1 & -\mathrm{a}+\mathrm{ab} \\\\
0 & \mathrm{~b}^2
\end{array}\right] \\\\
&\therefore \mathrm{T}_{\mathrm{n}}=\left\{\mathrm{A} \in \mathrm{S} ; \mathrm{A}^{\mathrm{n}(\mathrm{n}+1)}=\mathrm{I}\right\}
\end{aligned}
$$<br/><br/>
$\therefore$ b must be equal to 1<br/><br/>
$\therefore$ In this case $\mathrm{A}^2$ will become identity matrix and a can take any value from 1 to 100<br/><br/>
$\therefore$ Total number of common element will be 100 . | integer | jee-main-2022-online-24th-june-evening-shift |
1l6dwytee | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$A=\left(\begin{array}{rrr}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{array}\right)$$ and $$B=A-I$$. If $$\omega=\frac{\sqrt{3} i-1}{2}$$, then the number of elements in the $$\operatorname{set}\left\{n \in\{1,2, \ldots, 100\}: A^{n}+(\omega B)^{n}=A+B\right\}$$ is equal to ____________.</p> | [] | null | 17 | Here $A=\left(\begin{array}{ccc}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{array}\right)$
<br/><br/>
We get $A^{2}=A$ and similarly for
<br/><br/>
$$
B=A-I=\left[\begin{array}{lll}
1 & -1 & -1 \\
1 & -1 & -1 \\
1 & -1 & -1
\end{array}\right]
$$
<br/><br/>
We get $B^{2}=-B \Rightarrow B^{3}=B$
<br/><br/>
$$
\therefore A^{n}+(\omega B)^{n}=A+(\omega B)^{n} \quad \text { for } n \in \mathrm{N}
$$
<br/><br/>
For $\omega^{n}$ to be unity $n$ shall be multiple of 3 and for $B^{n}$ to be $B . n$ shell be $3,5,7, \ldots 99$
<br/><br/>
$\therefore n=\{3,9,15, \ldots . .99\}$
<br/><br/>
Number of elements $=17$ | integer | jee-main-2022-online-25th-july-morning-shift |
1l6rdsrjt | maths | matrices-and-determinants | operations-on-matrices | <p>Which of the following matrices can NOT be obtained from the matrix $$\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$$ by a single elementary row operation ?</p> | [{"identifier": "A", "content": "$$\\left[\\begin{array}{cc}0 & 1 \\\\ 1 & -1\\end{array}\\right]$$"}, {"identifier": "B", "content": "$$\\left[\\begin{array}{cc}1 & -1 \\\\ -1 & 2\\end{array}\\right]$$"}, {"identifier": "C", "content": "$$\\left[\\begin{array}{rr}-1 & 2 \\\\ -2 & 7\\end{array}\\right]$$"}, {"identifier": "D", "content": "$$\\left[\\begin{array}{ll}-1 & 2 \\\\ -1 & 3\\end{array}\\right]$$"}] | ["C"] | null | <p>Given matrix $$A = \left[ {\matrix{
{ - 1} & 2 \cr
1 & { - 1} \cr
} } \right]$$</p>
<p>For option A :</p>
<p>$${R_1} \to {R_1} + {R_2}$$</p>
<p>$$A = \left[ {\matrix{
0 & 1 \cr
1 & { - 1} \cr
} } \right]$$</p>
<p>$$\therefore$$ Option A can be obtained.</p>
<p>For option B :</p>
<p>$${R_1} \leftrightarrow {R_2}$$</p>
<p>$$A = \left[ {\matrix{
1 & { - 1} \cr
{ - 1} & 2 \cr
} } \right]$$</p>
<p>$$\therefore$$ Option B can be obtained.</p>
<p>Option C :</p>
<p>Not possible by a single elementary row operation.</p>
<p>Option D :</p>
<p>$${R_2} \to {R_2} + 2{R_1}$$</p>
<p>$$A = \left[ {\matrix{
{ - 1} & 2 \cr
{ - 1} & 3 \cr
} } \right]$$</p>
<p>$$\therefore$$ Option D can be obtained.</p> | mcq | jee-main-2022-online-29th-july-evening-shift |
1ldsuoc5l | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$\alpha$$ and $$\beta$$ be real numbers. Consider a 3 $$\times$$ 3 matrix A such that $$A^2=3A+\alpha I$$. If $$A^4=21A+\beta I$$, then</p> | [{"identifier": "A", "content": "$$\\alpha=1$$"}, {"identifier": "B", "content": "$$\\alpha=4$$"}, {"identifier": "C", "content": "$$\\beta=8$$"}, {"identifier": "D", "content": "$$\\beta=-8$$"}] | ["D"] | null | $\mathrm{A}^{2}=3 \mathrm{~A}+\alpha \mathrm{I}$
<br/><br/>
$A^{3}=3 A^{2}+\alpha A$
<br/><br/>
$\mathrm{A}^{3}=3(3 \mathrm{~A}+\alpha \mathrm{I})+\alpha \mathrm{A}$
<br/><br/>
$\mathrm{A}^{3}=9 \mathrm{~A}+\alpha \mathrm{A}+3 \alpha \mathrm{I}$
<br/><br/>
$\mathrm{A}^{4}=(9+\alpha) \mathrm{A}^{2}+3 \alpha \mathrm{A}$
<br/><br/>
$=(9+\alpha)(3 \mathrm{~A}+\alpha \mathrm{I})+3 \alpha \mathrm{A}$
<br/><br/>
$=\mathrm{A}(27+6 \alpha)+\alpha(9+\alpha)$
<br/><br/>
$\Rightarrow 27+6 \alpha=21 \Rightarrow \alpha=-1$
<br/><br/>
$\Rightarrow \beta=\alpha(9+\alpha)=-8$ | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldybe3tm | maths | matrices-and-determinants | operations-on-matrices | <p>If A and B are two non-zero n $$\times$$ n matrices such that $$\mathrm{A^2+B=A^2B}$$, then :</p> | [{"identifier": "A", "content": "$$\\mathrm{A^2B=I}$$"}, {"identifier": "B", "content": "$$\\mathrm{A^2=I}$$ or $$\\mathrm{B=I}$$"}, {"identifier": "C", "content": "$$\\mathrm{A^2B=BA^2}$$"}, {"identifier": "D", "content": "$$\\mathrm{AB=I}$$"}] | ["C"] | null | Given : $A^{2}+B=A^{2} B\quad...(i)$
<br/><br/>
$\Rightarrow A^{2}+B-I=A^{2} B-I$
<br/><br/>
$\Rightarrow A^{2} B-A^{2}-B+I=I$
<br/><br/>
$\Rightarrow A^{2}(B-I)-I(B-I)=I$
<br/><br/>
$\Rightarrow\left(A^{2}-I\right)(B-I)=I$
<br/><br/>
$\therefore A^{2}-I$ is the inverse matrix of $B-I$ and vice versa.
<br/><br/>
So, $(B-I)\left(A^{2}-I\right)=I$
<br/><br/>
$\Rightarrow B A^{2}-B-A^{2}+I=I$
<br/><br/>
$\therefore A^{2}+B=B A^{2} \quad...(ii)$
<br/><br/>
So, by (i) and (ii)
<br/><br/>
$A^{2} B=B A^{2}$ | mcq | jee-main-2023-online-24th-january-morning-shift |
1lguwckg8 | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$A=\left[\begin{array}{lll}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{array}\right]$$, where $$a, c \in \mathbb{R}$$. If $$A^{3}=A$$ and the positive value of $$a$$ belongs to the interval $$(n-1, n]$$, where $$n \in \mathbb{N}$$, then $$n$$ is equal to ___________.</p> | [] | null | 2 | $$
\text { We have, } A=\left[\begin{array}{lll}
0 & 1 & 2 \\
a & 0 & 3 \\
1 & c & 0
\end{array}\right] \text {, where } a, c \in R
$$
<br/><br/>$$
\begin{aligned}
A^2 & =\left[\begin{array}{lll}
0 & 1 & 2 \\
a & 0 & 3 \\
1 & c & 0
\end{array}\right]\left[\begin{array}{lll}
0 & 1 & 2 \\
a & 0 & 3 \\
1 & c & 0
\end{array}\right] \\\\
& =\left[\begin{array}{ccc}
a+2 & 2 c & 3 \\
3 & a+3 c & 2 a \\
a c & 1 & 2+3 c
\end{array}\right]
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
A^3 & =\left[\begin{array}{ccc}
a+2 & 2 c & 3 \\
3 & a+3 c & 2 a \\
a c & 1 & 2+3 c
\end{array}\right]\left[\begin{array}{ccc}
0 & 1 & 2 \\
a & 0 & 3 \\
1 & c & 0
\end{array}\right] \\\\
& =\left[\begin{array}{ccc}
2 a c+3 & a+2+3 c & 2 a+4+6 c \\
a(a+3 c)+2 a & 3+2 a c & 6+3 a+9 c \\
a+2+3 c & a c+2 c+3 c^2 & 2 a c+3
\end{array}\right]
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& A^3 =A [Given]\\\\
& 2 a c+3= 0 \text { and } a+2+3 c=1 \\\\
& a^2+2 a+3 a c =a \\\\
& \Rightarrow a^2 +a+3\left(-\frac{3}{2}\right)=0\\\\
& \Rightarrow 2 a^2+2 a-9=0
\end{aligned}
$$
<br/><br/>When, $a=1,2 a^2+2 a-9<0$ and
<br/><br/>When, $a=2,2 a^2+2 a-9>0$
<br/><br/>$\therefore$ Positive value of $a \in(1,2]$
<br/><br/>Hence, $n=2$ | integer | jee-main-2023-online-11th-april-morning-shift |
1lh21bstu | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{2 \times 2}$$, where $$\mathrm{a}_{\mathrm{ij}} \neq 0$$ for all $$\mathrm{i}, \mathrm{j}$$ and $$\mathrm{A}^{2}=\mathrm{I}$$. Let a be the sum of all diagonal elements of $$\mathrm{A}$$ and $$\mathrm{b}=|\mathrm{A}|$$. Then $$3 a^{2}+4 b^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "14"}, {"identifier": "D", "content": "7"}] | ["A"] | null | Given, $A^2=I$
<br/><br/>and $b=|A|$
<br/><br/>Let
$$
A=\left[\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right]
$$
<br/><br/>$$
\begin{aligned}
\therefore \quad A^2 & =\left[\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right]\left[\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right] \\\\
& =\left[\begin{array}{cc}
a_1^2+b_1 a_2 & a_1 b_1+b_1 b_2 \\
a_1 a_2+a_2 b_2 & b_1 a_2+b_2^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \left(\because A^2=I\right)
\end{aligned}
$$
<br/><br/>By equality of matrices, we get
<br/><br/>$$
\begin{aligned}
& a_1 a_2+a_2 b_2=0 \text { and } a_1 b_1+b_1 b_2=0 \\\\
&\Rightarrow a_2\left(a_1+b_2\right)=0 \text { and } b_1\left(a_1+b_2\right)=0 \\\\
&\Rightarrow a_1+b_2=0 \text { (since, } b_1, a_2 \neq 0 \text { ) }\\\\
&\Rightarrow \text { Sum of diagonal elements }=0 \\\\
&\Rightarrow a=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Now, } A^2=I \\\\
& \Rightarrow \left|A^2\right|=1 \\\\
& \Rightarrow |A|^2=1 ~~~~~~~\left(\because\left|A^n\right|=|A|^n\right)\\\\
& \Rightarrow b^2=1 ~~~~~~~~~~~(\because|A|=b)\\\\
& \therefore 3 a^2+4 b^2=3(0)+4(1)=4
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2xzk3z | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$P$$ be a square matrix such that $$P^{2}=I-P$$. For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$P^{\alpha}+P^{\beta}=\gamma I-29 P$$ and $$P^{\alpha}-P^{\beta}=\delta I-13 P$$, then $$\alpha+\beta+\gamma-\delta$$ is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "40"}] | ["C"] | null | We have, $P^2=I-P$
<br/><br/>$$
\begin{aligned}
\Rightarrow P^4 & =(I-P)^2=I+P^2-2 P \\\\
& =2 I-3 P \text { [Using Eq. (i)] }
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\Rightarrow P^8 & =(2 I-3 P)^2 \\\\
& =4 I+9 P^2-12 P \\\\
& =13 I-21 P \text { [Using Eq. (i)] }
\end{aligned}
$$
<br/><br/>and
<br/><br/>$$
\begin{aligned}
P^6 & =(I-P)(2 I-3 P) \\\\
& =2 I-3 P-2 P+3 P^2 \\\\
& =5 I-8 P \text { [Using Eq. (i)] }
\end{aligned}
$$
<br/><br/>$\begin{aligned} & \text { Now, } P^8+P^6=18 I-29 P \\\\ & \text { and } P^8-P^6=8 I-13 P \\\\ & \therefore \alpha=8, \beta=6, \gamma=18, \delta=8 \\\\ & \therefore \alpha+\beta+\gamma-\delta=8+6+18-8=24\end{aligned}$ | mcq | jee-main-2023-online-6th-april-evening-shift |
jaoe38c1lsfkynmn | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$A=\left[\begin{array}{ccc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right]$$ and $$P=\left[\begin{array}{lll}1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5\end{array}\right]$$. The sum of the prime factors of $$\left|P^{-1} A P-2 I\right|$$ is equal to</p> | [{"identifier": "A", "content": "66"}, {"identifier": "B", "content": "27"}, {"identifier": "C", "content": "23"}, {"identifier": "D", "content": "26"}] | ["D"] | null | <p>$$\begin{aligned}
\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right| & =\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right| \\
& =\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right| \\
& =\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}| \\
& =|\mathrm{A}-2 \mathrm{I}| \\
& =\left|\begin{array}{ccc}
0 & 1 & 2 \\
6 & 0 & 11 \\
3 & 3 & 0
\end{array}\right|=69
\end{aligned}$$</p>
<p>So, Prime factor of 69 is 3 & 23</p>
<p>So, sum = 26</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
lv5grw76 | maths | matrices-and-determinants | operations-on-matrices | <p>Let $$A=\left[\begin{array}{lll}2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b\end{array}\right]$$. If $$A^3=4 A^2-A-21 I$$, where $$I$$ is the identity matrix of order $$3 \times 3$$, then $$2 a+3 b$$ is equal to</p> | [{"identifier": "A", "content": "$$-10$$\n"}, {"identifier": "B", "content": "$$-12$$\n"}, {"identifier": "C", "content": "$$-13$$\n"}, {"identifier": "D", "content": "$$-9$$"}] | ["C"] | null | <p>$$\begin{aligned}
& |A-\lambda I|=0 \\
& \left|\begin{array}{ccc}
2-\lambda & a & 0 \\
1 & 3-\lambda & 1 \\
0 & 5 & b-\lambda
\end{array}\right|=0 \\
& (2-\lambda)[(3-\lambda)(b-\lambda)-5]-a[b-\lambda-0]+0=0 \\
& (2-\lambda)\left[3 b-3 \lambda-b \lambda+\lambda^2-5\right]-a b+a \lambda=0 \\
& \lambda^3-(b+5) \lambda^2+(1-a+5 b) \lambda+(10-6 b+a b)=0 \\
& A^3-(b+5) A^2+(1-a+5 b) A+(10-6 b+a b) I=0 \\
& \Rightarrow \mathrm{b}+5=4,1-a+5 b=1,10-6 b+a b=21 \\
& \Rightarrow a=-5, b=-1 \\
& \Rightarrow 2 a+3 b=-13
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
YSvNld4KWlRuXhMh | maths | matrices-and-determinants | properties-of-determinants | Let $$A = \left| {\matrix{
5 & {5\alpha } & \alpha \cr
0 & \alpha & {5\alpha } \cr
0 & 0 & 5 \cr
} } \right|.$$ If $$\,\,\left| {{A^2}} \right| = 25,$$ then $$\,\left| \alpha \right|$$ equals | [{"identifier": "A", "content": "$$1/5$$ "}, {"identifier": "B", "content": "$$5$$"}, {"identifier": "C", "content": "$${5^2}$$ "}, {"identifier": "D", "content": "$$1$$"}] | ["A"] | null | $$\left| {{A^2}} \right| = 25 \Rightarrow {\left| A \right|^2} = 25$$
<br><br>$$ \Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}$$ | mcq | aieee-2007 |
XrzcXWoZKBi1HBgV | maths | matrices-and-determinants | properties-of-determinants | Let $$A$$ be a square matrix all of whose entries are integers.
<br/>Then which one of the following is true? | [{"identifier": "A", "content": "If det $$A = \\pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are not necessarily integers"}, {"identifier": "B", "content": "If det $$A \\ne \\pm 1,$$ then $${A^{ - 1}}$$ exists and all its entries are non integers"}, {"identifier": "C", "content": "If det $$A = \\pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are integers"}, {"identifier": "D", "content": "If det $$A = \\pm 1,$$ then $${A^{ - 1}}$$ need not exists "}] | ["C"] | null | As all entries of square matrix $$A$$ are integers, therefore all co-factors should also be integers.
<br><br>If det $$A = \pm 1\,\,$$ then $${A^{ - 1}}\,\,$$ exists. Also all entries of $${A^{ - 1}}$$ are integers. | mcq | aieee-2008 |
8TSmM1Tn5g2DomPz | maths | matrices-and-determinants | properties-of-determinants | Let $$A$$ be a $$\,2 \times 2$$ matrix
<br/><b>Statement - 1 :</b> $$adj\left( {adj\,A} \right) = A$$
<br/><b>Statement - 2 :</b>$$\left| {adj\,A} \right| = \left| A \right|$$ | [{"identifier": "A", "content": "statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1. "}, {"identifier": "B", "content": "statement - 1 is true, statement - 2 is false. "}, {"identifier": "C", "content": "statement - 1 is false, statement -2 is true "}, {"identifier": "D", "content": "statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1. "}] | ["D"] | null | We know that $$\left| {adj\left( {adj\,\,A} \right)} \right| = {\left| {Adj\,\,A} \right|^{2 - 1}}$$b
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left| A \right|^{2 - 1}} = \left| A \right|$$
<br><br>$$\therefore$$ $$\,\,\,\,\,\,$$ Both the statements are true and statement $$-2$$ is a correct explanation for statement - $$1.$$ | mcq | aieee-2009 |
5Mqcuw2868tUWChI | maths | matrices-and-determinants | properties-of-determinants | Let $$P$$ and $$Q$$ be $$3 \times 3$$ matrices $$P \ne Q.$$ If $${P^3} = {Q^3}$$ and
<br/> $${P^2}Q = {Q^2}P$$ then determinant of $$\left( {{P^2} + {Q^2}} \right)$$ is equal to : | [{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$0$$ "}, {"identifier": "D", "content": "$$-1$$"}] | ["C"] | null | Given
<br><br>$${P^3} = {q^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$${P^2}Q = {Q^2}p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>Subtracting $$(1)$$ and $$(2)$$, we get
<br><br>$${P^3} - {P^2}Q = {Q^3} - {Q^2}P$$
<br><br>$$ \Rightarrow {P^2}\left( {P - Q} \right) + {Q^2}\left( {P - Q} \right) = 0$$
<br><br>$$ \Rightarrow \left( {{P^2} + {Q^2}} \right)\left( {P - Q} \right) = 0$$
<br><br>$$ \Rightarrow \left| {{p^2} + {Q^2}} \right| = 0$$
<br><br>as $$P \ne Q$$
| mcq | aieee-2012 |
seFgM9t0D88KoRqm | maths | matrices-and-determinants | properties-of-determinants | Let $$A = \left( {\matrix{
1 & 0 & 0 \cr
2 & 1 & 0 \cr
3 & 2 & 1 \cr
} } \right)$$. If $${u_1}$$ and $${u_2}$$ are column matrices such
<br/> that $$A{u_1} = \left( {\matrix{
1 \cr
0 \cr
0 \cr
} } \right)$$ and $$A{u_2} = \left( {\matrix{
0 \cr
1 \cr
0 \cr
} } \right),$$ then $${u_1} + {u_2}$$ is equal to : | [{"identifier": "A", "content": "$$\\left( {\\matrix{\n -1 \\cr \n 1 \\cr \n 0 \\cr \n\n } } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {\\matrix{\n -1 \\cr \n 1 \\cr \n -1 \\cr \n\n } } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {\\matrix{\n -1 \\cr \n -1 \\cr \n 0 \\cr \n\n } } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {\\matrix{\n 1 \\cr \n -1 \\cr \n -1 \\cr \n\n } } \\right)$$"}] | ["D"] | null | Let $$A{u_1} = \left( {\matrix{
1 \cr
0 \cr
0 \cr
} } \right)\,\,\,\,\,\,A{u_2} = \left( {\matrix{
0 \cr
1 \cr
0 \cr
} } \right)$$
<br><br>Then, $$A{u_1} + A{u_2} = \left( {\matrix{
1 \cr
0 \cr
0 \cr
} } \right) + \left( {\matrix{
0 \cr
1 \cr
0 \cr
} } \right)$$
<br><br>$$ \Rightarrow A\left( {{u_1} + {u_2}} \right) = \left( {\matrix{
1 \cr
1 \cr
0 \cr
} } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>Also, $$A = \left( {\matrix{
1 & 0 & 0 \cr
2 & 1 & 0 \cr
3 & 2 & 1 \cr
} } \right)$$
<br><br>$$ \Rightarrow \left| A \right| = 1\left( 1 \right) - 0\left( 2 \right) + 0\left( {4 - 3} \right) = 1$$
<br><br>We know,
<br><br>$${A^{ - 1}} = {1 \over {\left| A \right|}}\,adjA \Rightarrow {A^{ - 1}} = adj\left( A \right)$$
<br><br>( as $$\left| A \right| = 1$$ )
<br><br>Now, from equation $$(1)$$, we have
<br><br>$${u_1} + {u_2} = {A^{ - 1}}\left( {\matrix{
1 \cr
1 \cr
0 \cr
} } \right)$$
<br><br>$$ = \left[ {\matrix{
1 & 0 & 0 \cr
{ - 2} & 1 & 0 \cr
1 & { - 2} & 1 \cr
} } \right]\left( {\matrix{
1 \cr
1 \cr
0 \cr
} } \right)$$
<br><br>$$ = \left[ {\matrix{
1 \cr
{ - 1} \cr
{ - 1} \cr
} } \right]$$ | mcq | aieee-2012 |
tQwnUOEU3VBd7zli | maths | matrices-and-determinants | properties-of-determinants | If $$P = \left[ {\matrix{
1 & \alpha & 3 \cr
1 & 3 & 3 \cr
2 & 4 & 4 \cr
} } \right]$$ is the adjoint of a $$3 \times 3$$ matrix $$A$$ and
<br/>$$\left| A \right| = 4,$$ then $$\alpha $$ is equal to : | [{"identifier": "A", "content": "$$4$$ "}, {"identifier": "B", "content": "$$11$$ "}, {"identifier": "C", "content": "$$5$$ "}, {"identifier": "D", "content": "$$0$$"}] | ["B"] | null | $$\left| P \right| = 1\left( {12 - 12} \right) - \alpha \left( {4 - 6} \right) + $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,3\left( {4 - 6} \right) = 2\alpha - 6$$
<br><br>Now, $$adj\,\,A = P\,$$ $$\,\,\,\,\,\,\,\, \Rightarrow \left| {adj\,A} \right| = \left| P \right|$$
<br><br>$$ \Rightarrow {\left| A \right|^2} = \left| P \right|$$
<br><br>$$ \Rightarrow \left| P \right| = 16$$
<br><br>$$ \Rightarrow 2\alpha - 6 = 16$$
<br><br>$$ \Rightarrow \alpha = 11$$ | mcq | jee-main-2013-offline |
9cnuwNKo0TcaRlwfihNII | maths | matrices-and-determinants | properties-of-determinants | Let A and B be two invertible matrices of order 3 $$ \times $$ 3. If det(ABA<sup>T</sup>) = 8 and det(AB<sup>–1</sup>) = 8,
<br/>then det (BA<sup>–1</sup> B<sup>T</sup>) is equal to :
| [{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "$${1 \\over {16}}$$"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $${\left| A \right|^2}.\left| B \right| = 8$$
<br><br>and $${{\left| A \right|} \over {\left| B \right|}} = 8 \Rightarrow \left| A \right| = 4$$
<br><br>and $$\left| B \right| = {1 \over 2}$$
<br><br>$$ \therefore $$ det(BA<sup>$$-$$1</sup>. B<sup>T</sup>) $$ = {1 \over 4} \times {1 \over 4} = {1 \over {16}}$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
RZJCYBV6H95GfEUmuY7k9k2k5fm0w1j | maths | matrices-and-determinants | properties-of-determinants | Let A = [a<sub>ij</sub>] and B = [b<sub>ij</sub>] be two 3 × 3 real matrices such that b<sub>ij</sub> = (3)<sup>(i+j-2)</sup>a<sub>ji</sub>, where i, j = 1, 2, 3.
If the determinant of B is 81, then the determinant of A is:
| [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${1 \\over 9}$$"}, {"identifier": "D", "content": "$${1 \\over {81}}$$"}] | ["C"] | null | |B| = $$\left| {\matrix{
{{b_{11}}} & {{b_{12}}} & {{b_{13}}} \cr
{{b_{21}}} & {{b_{22}}} & {{b_{23}}} \cr
{{b_{31}}} & {{b_{32}}} & {{b_{33}}} \cr
} } \right|$$
<br><br>= $$\left| {\matrix{
{{3^0}{a_{11}}} & {{3^1}{a_{12}}} & {{3^2}{a_{13}}} \cr
{{3^1}{a_{21}}} & {{3^2}{a_{22}}} & {{3^3}{a_{23}}} \cr
{{3^2}{a_{31}}} & {{3^3}{a_{32}}} & {{3^4}{a_{33}}} \cr
} } \right|$$
<br><br>= $${3.3^2}\left| {\matrix{
{{a_{11}}} & {{3^1}{a_{12}}} & {{3^2}{a_{13}}} \cr
{{a_{21}}} & {{3^1}{a_{22}}} & {{3^2}{a_{23}}} \cr
{{a_{31}}} & {{3^1}{a_{32}}} & {{3^2}{a_{33}}} \cr
} } \right|$$
<br><br>= $${3.3^2}{.3.3^2}\left| {\matrix{
{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right|$$
<br><br>= 3<sup>6</sup>.|A|
<br><br>$$ \therefore $$ 3<sup>6</sup>.|A| = 81
<br><br>$$ \Rightarrow $$ |A| = $${1 \over 9}$$ | mcq | jee-main-2020-online-7th-january-evening-slot |
YYS2DGHZEoLw2DMefI7k9k2k5iqzn1t | maths | matrices-and-determinants | properties-of-determinants | If the matrices A = $$\left[ {\matrix{
1 & 1 & 2 \cr
1 & 3 & 4 \cr
1 & { - 1} & 3 \cr
} } \right]$$,
<br/><br/>B = adjA and
C = 3A, then $${{\left| {adjB} \right|} \over {\left| C \right|}}$$ is equal to : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "16"}] | ["A"] | null | A = $$\left[ {\matrix{
1 & 1 & 2 \cr
1 & 3 & 4 \cr
1 & { - 1} & 3 \cr
} } \right]$$
<br><br>$$ \Rightarrow $$ |A| = 6
<br><br>$${{\left| {adjB} \right|} \over {\left| C \right|}}$$
<br><br>= $${{\left| {adj\left( {adjA} \right)} \right|} \over {\left| {3A} \right|}}$$
<br><br>= $${{{{\left| A \right|}^4}} \over {{3^3}\left| A \right|}}$$
<br><br>= $${{{{\left| A \right|}^3}} \over {{3^3}}}$$
<br><br>= $${{{6^3}} \over {{3^3}}}$$ = 8 | mcq | jee-main-2020-online-9th-january-morning-slot |
8DDQ9pLxEGNcY3sEdujgy2xukf4552qe | maths | matrices-and-determinants | properties-of-determinants | Let A be a 3 $$ \times $$ 3 matrix such that
<br/>adj A = $$\left[ {\matrix{
2 & { - 1} & 1 \cr
{ - 1} & 0 & 2 \cr
1 & { - 2} & { - 1} \cr
} } \right]$$ and B = adj(adj A).
<br/><br/>If |A| = $$\lambda $$ and |(B<sup>-1</sup>)<sup>T</sup>| = $$\mu $$ , then the ordered pair,
<br/>(|$$\lambda $$|, $$\mu $$) is equal to :
| [{"identifier": "A", "content": "(3, 81)"}, {"identifier": "B", "content": "$$\\left( {9,{1 \\over 9}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {3,{1 \\over {81}}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {9,{1 \\over {81}}} \\right)$$"}] | ["C"] | null | $$adj\,A = \left[ {\matrix{
2 & { - 1} & 1 \cr
{ - 1} & 0 & 2 \cr
1 & { - 2} & { - 1} \cr
} } \right]$$<br><br>$$B = adj\,(adj\,A)$$<br><br>$$ = |A{|^{n - 2}}A$$<br><br>$$ = |A{|^{3 - 2}}.A$$ [As here n = 3]<br><br>$$ = |A|.A$$ .....(1)<br><br>Now, $$|adj\,A| = \left[ {\matrix{
2 & { - 1} & 1 \cr
{ - 1} & 0 & 2 \cr
1 & { - 2} & { - 1} \cr
} } \right]$$<br><br>$$ = + 2(0 + 4) + 1(1 - 2) + 1(2 - 0)$$<br><br>$$ = 8 - 1 + 2$$<br><br>$$ = 9$$<br><br>Also we know, |adj A| = |A|<sup>n$$-$$1</sup><br><br>$$ \therefore $$ Here |adj A| = |A|<sup>3 $$-$$ 1</sup> = |A|<sup>2</sup><br><br>$$ \therefore $$ |A|<sup>2</sup> = 9<br><br>$$ \Rightarrow $$ |A| = $$ \pm $$3<br><br>Given, |A| = $$\lambda $$<br><br>$$ \therefore $$ $$\lambda $$ = $$ \pm $$3<br><br>$$ \Rightarrow $$ |$$\lambda $$| = 3<br><br>From (1)<br><br>B = ($$ \pm $$3)A<br><br>Given, |(B<sup>$$-$$1</sup>)<sup>T</sup>| = $$\mu $$<br><br>$$ \Rightarrow $$ |(B<sup>T</sup>)<sup>$$-$$1</sup>| = $$\mu $$<br><br>$$ \Rightarrow $$ $${1 \over {|{B^T}|}} = \mu $$<br><br>$$ \Rightarrow $$ $${1 \over {|B|}} = \mu $$ [As |B<sup>T</sup>| = |B|]
<br><br>$$ \Rightarrow $$ $${1 \over {| \pm 3A|}} = \mu $$
<br><br>$$ \Rightarrow $$ $${1 \over {{{\left( { \pm 3} \right)}^3}\left| A \right|}} = \mu $$ [As |KA| = K<sup>n</sup> |A|]<br><br>$$ \Rightarrow {1 \over {( \pm 27)|A|}} = \mu $$<br><br>$$ \Rightarrow \mu = {1 \over {( \pm 27)( \pm 3)}} = \pm {1 \over {81}}$$<br><br>$$ \therefore $$ $$(|\lambda |,\,\mu ) = \left( {3,\, \pm {1 \over {81}}} \right)$$<br><br>Here correct option will be $$\left( {3,\,{1 \over {81}}} \right)$$ | mcq | jee-main-2020-online-3rd-september-evening-slot |
BLrpVgxd8BhsNuoTtj1kls5lrip | maths | matrices-and-determinants | properties-of-determinants | Let $$A = \left[ {\matrix{
x & y & z \cr
y & z & x \cr
z & x & y \cr
} } \right]$$, where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If $${A^2} = {I_3}$$, then the value of $${x^3} + {y^3} + {z^3}$$ is ____________. | [] | null | 7 | $$A = \left[ {\matrix{
x & y & z \cr
y & z & x \cr
z & x & y \cr
} } \right]$$
<br><br>$$ \therefore $$ $$|A| = \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$$<br><br>Given $${A^2} = {I_3}$$<br><br>$$|{A^2}| = 1$$<br><br>$$ \therefore $$ $${({x^3} + {y^3} + {z^3} - 3xyz)^2} = 1$$<br><br>$$ \Rightarrow {x^3} + {y^3} + {z^3} - 3xyz = 1$$ only as $$(x + y + z > 0)$$<br><br>$$ \Rightarrow {x^3} + {y^3} + {z^3} = 6 + 1 = 7$$ | integer | jee-main-2021-online-25th-february-morning-slot |
mjhoVcgyX2dU8iVWBG1kmhzoe5g | maths | matrices-and-determinants | properties-of-determinants | Let $$P = \left[ {\matrix{
{ - 30} & {20} & {56} \cr
{90} & {140} & {112} \cr
{120} & {60} & {14} \cr
} } \right]$$ and<br/><br/> $$A = \left[ {\matrix{
2 & 7 & {{\omega ^2}} \cr
{ - 1} & { - \omega } & 1 \cr
0 & { - \omega } & { - \omega + 1} \cr
} } \right]$$ where <br/><br/>$$\omega = {{ - 1 + i\sqrt 3 } \over 2}$$, and I<sub>3</sub> be the identity matrix of order 3. If the <br/>determinant of the matrix (P<sup>$$-$$1</sup>AP$$-$$I<sub>3</sub>)<sup>2</sup> is $$\alpha$$$$\omega$$<sup>2</sup>, then the value of $$\alpha$$ is equal to ______________. | [] | null | 36 | $$|{P^{ - 1}}AP - I{|^2}$$<br><br>$$ = |({P^{ - 1}}AP - I){({P^{ - 1}}AP - 1)^2}|$$<br><br>$$ = |{P^{ - 1}}AP{P^{ - 1}}AP - 2{P^{ - 1}}AP + I|$$<br><br>$$ = |{P^{ - 1}}{A^2}P - 2{P^{ - 1}}AP + {P^{ - 1}}IP|$$<br><br>$$ = |{P^{ - 1}}({A^2} - 2A + I)P|$$<br><br>$$ = |{P^{ - 1}}{(A - I)^2}P|$$<br><br>$$ = |{P^{ - 1}}||A - I{|^2}|P|$$<br><br>$$ = |A - I{|^2}$$<br><br>$$ = \left| {\matrix{
1 & 7 & {{\omega ^2}} \cr
{ - 1} & { - \omega - 1} & 1 \cr
0 & { - \omega } & { - \omega } \cr
} } \right|$$<br><br>$$ = {(1(\omega (\omega + 1) + \omega ) - 7\omega + {\omega ^2}.\omega )^2}$$<br><br>$$ = {({\omega ^2} + 2\omega - 7\omega + 1)^2}$$<br><br>$$ = {({\omega ^2} - 5\omega + 1)^2}$$<br><br>$$ = {( - 6\omega )^2}$$<br><br>$$ = 36{\omega ^2} $$
<br><br>$$ \therefore $$ $$\alpha$$$$\omega$$<sup>2</sup> = $$36{\omega ^2} $$
<br><br>$$\Rightarrow \alpha = 36$$ | integer | jee-main-2021-online-16th-march-morning-shift |
PdrFkT0rspxM55uW6h1kmjbdnb6 | maths | matrices-and-determinants | properties-of-determinants | If $$A = \left( {\matrix{
0 & {\sin \alpha } \cr
{\sin \alpha } & 0 \cr
} } \right)$$ and $$\det \left( {{A^2} - {1 \over 2}I} \right) = 0$$, then a possible value of $$\alpha$$ is : | [{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6}$$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$"}, {"identifier": "D", "content": "$${\\pi \\over 3}$$"}] | ["A"] | null | $${A^2} = \left[ {\matrix{
0 & {\sin \alpha } \cr
{\sin \alpha } & 0 \cr
} } \right]\left[ {\matrix{
0 & {\sin \alpha } \cr
{\sin \alpha } & 0 \cr
} } \right] = \left[ {\matrix{
{{{\sin }^2}\alpha } & 0 \cr
0 & {{{\sin }^2}\alpha } \cr
} } \right]$$<br><br>$${A^2} - {1 \over 2}I = \left[ {\matrix{
{{{\sin }^2}\alpha } & 0 \cr
0 & {{{\sin }^2}\alpha } \cr
} } \right] - \left[ {\matrix{
{{1 \over 2}} & 0 \cr
0 & {{1 \over 2}} \cr
} } \right] = \left[ {\matrix{
{{{\sin }^2}\alpha - {1 \over 2}} & 0 \cr
0 & {{{\sin }^2}\alpha - {1 \over 2}} \cr
} } \right]$$<br><br>Given, $$\left| {{A^2} - {1 \over 2}I} \right| = 0$$<br><br>$$ \Rightarrow \left| {\matrix{
{{{\sin }^2}\alpha - {1 \over 2}} & 0 \cr
0 & {{{\sin }^2}\alpha - {1 \over 2}} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow {\left( {{{\sin }^2}\alpha - {1 \over 2}} \right)^2} = 0 $$<br><br>$$\Rightarrow {\sin ^2}\alpha = {1 \over 2} \Rightarrow \sin \alpha = {1 \over {\sqrt 2 }}, - {1 \over {\sqrt 2 }}$$<br><br>$$ \therefore $$ $$\alpha = {\pi \over 4}$$ | mcq | jee-main-2021-online-17th-march-morning-shift |
6D7g0xG8HmUmFaTFHj1kmjbrw8g | maths | matrices-and-determinants | properties-of-determinants | If $$A = \left[ {\matrix{
2 & 3 \cr
0 & { - 1} \cr
} } \right]$$, then the value of det(A<sup>4</sup>) + det(A<sup>10</sup> $$-$$ (Adj(2A))<sup>10</sup>) is equal to _____________. | [] | null | 16 | $$A = \left[ {\matrix{
2 & 3 \cr
0 & { - 1} \cr
} } \right]$$
<br><br>$$|A|\, = - 2 \Rightarrow |A{|^4} = 16$$
<br><br>$${A^2} = \left[ {\matrix{
4 & 3 \cr
0 & 1 \cr
} } \right]$$
<br><br>$${A^3} = \left[ {\matrix{
8 & 9 \cr
0 & { - 1} \cr
} } \right]$$
<br><br>$$ \therefore $$ $${A^{10}} = \left[ {\matrix{
{{2^{10}}} & {{2^{10}} - 1} \cr
0 & 1 \cr
} } \right] = \left[ {\matrix{
{1024} & {1023} \cr
0 & 1 \cr
} } \right]$$<br><br>$$2A = \left[ {\matrix{
4 & 6 \cr
0 & { - 2} \cr
} } \right]$$<br><br>$$adj(2A) = \left[ {\matrix{
{ - 2} & { - 6} \cr
0 & 4 \cr
} } \right]$$<br><br>$$adj(2A) = - 2\left[ {\matrix{
1 & 3 \cr
0 & { - 2} \cr
} } \right]$$<br><br>$${(adj(2A))^{10}} = {2^{10}}{\left[ {\matrix{
1 & 3 \cr
0 & { - 2} \cr
} } \right]^{10}}$$<br><br>$$ = {2^{10}}\left[ {\matrix{
1 & { - ({2^{10}} - 1)} \cr
0 & {{2^{10}}} \cr
} } \right]$$<br><br>$$ = {2^{10}}\left[ {\matrix{
1 & { - 1023} \cr
0 & {1024} \cr
} } \right]$$<br><br>$${A^{10}} - {(adj(2A))^{10}} = \left[ {\matrix{
0 & {{2^{11}} \times 1023} \cr
0 & {1 - {{(1024)}^2}} \cr
} } \right]$$<br><br>$$|{A^{10}} - adj{(2A)^{10}}| = 0$$
<br><br>$$ \therefore $$
det(A<sup>4</sup>) + det(A<sup>10</sup> $$-$$ (Adj(2A))<sup>10</sup>)
<br><br> = 16 + 0 = 16 | integer | jee-main-2021-online-17th-march-morning-shift |
1krrv7h3p | maths | matrices-and-determinants | properties-of-determinants | Let $$A = \{ {a_{ij}}\} $$ be a 3 $$\times$$ 3 matrix, <br/><br/>where $${a_{ij}} = \left\{ {\matrix{
{{{( - 1)}^{j - i}}} & {if} & {i < j,} \cr
2 & {if} & {i = j,} \cr
{{{( - 1)}^{i + j}}} & {if} & {i > j} \cr
} } \right.$$ <br/><br/>then $$\det (3Adj(2{A^{ - 1}}))$$ is equal to _____________. | [] | null | 108 | $$A = \left[ {\matrix{
2 & { - 1} & 1 \cr
{ - 1} & 2 & { - 1} \cr
1 & { - 1} & 2 \cr
} } \right]$$<br><br>$$|A| = 4$$<br><br>$$\det (3adj(2{A^{ - 1}}))$$<br><br>$$ = {3^3}\left| {adj(2{a^{ - 1}})} \right|$$<br><br>$$ = {3^2}{\left| {2{A^{ - 1}}} \right|^2}$$<br><br>$$ = {3^3}{.2^2}|{A^{ - 1}}{|^2} = {3^3}{.2^2}.{1 \over {|A{|^2}}} = {3^2}{.2^2}.{1 \over {{4^2}}} = 108$$ | integer | jee-main-2021-online-20th-july-evening-shift |
1krw2sssh | maths | matrices-and-determinants | properties-of-determinants | Let $$M = \left\{ {A = \left( {\matrix{
a & b \cr
c & d \cr
} } \right):a,b,c,d \in \{ \pm 3, \pm 2, \pm 1,0\} } \right\}$$. Define f : M $$\to$$ Z, as f(A) = det(A), for all A$$\in$$M, where z is set of all integers. Then the number of A$$\in$$M such that f(A) = 15 is equal to _____________. | [] | null | 16 | | A | = ad $$-$$ bc = 15<br><br>where $${a,b,c,d \in \{ \pm 3, \pm 2, \pm 1,0\} }$$<br><br>Case I ad = 9 & bc = $$-$$6<br><br>For ad possible pairs are (3, 3), ($$-$$3, $$-$$3)<br><br>For bc possible pairs are (3, $$-$$2), ($$-$$3, 2), ($$-$$2, 3), (2, $$-$$3)<br><br>So total matrix = 2 $$\times$$ 4 = 8<br><br>Case II ad = 6 & bc = $$-$$9<br><br>Similarly total matrix = 2 $$\times$$ 4 = 8<br><br>$$\Rightarrow$$ Total such matrices are = 16 | integer | jee-main-2021-online-25th-july-morning-shift |
1kryf4lkx | maths | matrices-and-determinants | properties-of-determinants | Let A and B be two 3 $$\times$$ 3 real matrices such that (A<sup>2</sup> $$-$$ B<sup>2</sup>) is invertible matrix. If A<sup>5</sup> = B<sup>5</sup> and A<sup>3</sup>B<sup>2</sup> = A<sup>2</sup>B<sup>3</sup>, then the value of the determinant of the matrix A<sup>3</sup> + B<sup>3</sup> is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "0"}] | ["D"] | null | C = A<sup>2</sup> $$-$$ B<sup>2</sup>; | C | $$\ne$$ 0<br><br>A<sup>2</sup> = B<sup>5</sup> and A<sup>3</sup>B<sup>2</sup> = A<sup>2</sup>B<sup>2</sup><br><br>Now, A<sup>5</sup> $$-$$ A<sup>3</sup>B<sup>2</sup> = B<sup>5</sup> $$-$$ A<sup>2</sup>B<sup>3</sup><br><br>$$\Rightarrow$$ A<sup>3</sup> (A<sup>2</sup> $$-$$ B<sup>2</sup>) + B<sup>3</sup> (A<sup>2</sup> $$-$$ B<sup>2</sup>) = 0<br><br>$$\Rightarrow$$ (A<sup>3</sup> + B<sup>3</sup>(A<sup>2</sup> $$-$$ B<sup>2</sup>) = 0<br><br>$$ \Rightarrow $$ A<sup>3</sup> + B<sup>3</sup> = 0 $$\left( \because{\left| {{A^2} - {B^2} \ne 0} \right|} \right)$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1ktd4vbhk | maths | matrices-and-determinants | properties-of-determinants | Let A be a 3 $$\times$$ 3 real matrix. If det(2Adj(2 Adj(Adj(2A)))) = 2<sup>41</sup>, then the value of det(A<sup>2</sup>) equal __________. | [] | null | 4 | adj (2A) = 2<sup>2</sup> adjA<br><br>$$\Rightarrow$$ adj(adj (2A)) = adj(4 adjA) = 16 adj (adj A)<br><br>= 16 | A | A<br><br>$$\Rightarrow$$ adj (32 | A | A) = (32 | A |)<sup>2</sup> adj A<br><br>12(32| A |)<sup>2</sup> |adj A | = 2<sup>3</sup> (32 | A |)<sup>6</sup> | adj A |<br><br>2<sup>3</sup> . 2<sup>30</sup> | A |<sup>6</sup> . | A |<sup>2</sup> = 2<sup>41</sup><br><br>| A |<sup>8</sup> = 2<sup>8</sup> $$\Rightarrow$$ | A | = $$\pm$$2<br><br>| A |<sup>2</sup> = | A |<sup>2</sup> = 4 | integer | jee-main-2021-online-26th-august-evening-shift |
1ktg05qbp | maths | matrices-and-determinants | properties-of-determinants | <sub></sub>Let A(a, 0), B(b, 2b + 1) and C(0, b), b $$\ne$$ 0, |b| $$\ne$$ 1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is : | [{"identifier": "A", "content": "$${{ - 2b} \\over {b + 1}}$$"}, {"identifier": "B", "content": "$${{2b} \\over {b + 1}}$$"}, {"identifier": "C", "content": "$${{2{b^2}} \\over {b + 1}}$$"}, {"identifier": "D", "content": "$${{ - 2{b^2}} \\over {b + 1}}$$"}] | ["D"] | null | $$\left| {{1 \over 2}\left| {\matrix{
a & 0 & 1 \cr
b & {2b + 1} & 1 \cr
0 & b & 1 \cr
} } \right|} \right| = 1$$<br><br>$$ \Rightarrow \left| {\matrix{
a & 0 & 1 \cr
b & {2b + 1} & 1 \cr
0 & b & 1 \cr
} } \right| = \pm \,2$$<br><br>$$ \Rightarrow a(2b + 1 - b) - 0 + 1({b^2} - 0) = \pm \,2$$<br><br>$$ \Rightarrow a = {{ \pm \,2 - {b^2}} \over {b + 1}}$$<br><br>$$\therefore$$ $$a = {{2 - {b^2}} \over {b + 1}}$$ and $$a = {{ - 2 - {b^2}} \over {b + 1}}$$<br><br>Sum of possible values of 'a' is <br><br>$$ = {{ - 2{b^2}} \over {a + 1}}$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1l54anlfc | maths | matrices-and-determinants | properties-of-determinants | <p>Let $$A = \left( {\matrix{
2 & { - 1} \cr
0 & 2 \cr
} } \right)$$. If $$B = I - {}^5{C_1}(adj\,A) + {}^5{C_2}{(adj\,A)^2} - \,\,.....\,\, - {}^5{C_5}{(adj\,A)^5}$$, then the sum of all elements of the matrix B is</p> | [{"identifier": "A", "content": "$$-$$5"}, {"identifier": "B", "content": "$$-$$6"}, {"identifier": "C", "content": "$$-$$7"}, {"identifier": "D", "content": "$$-$$8"}] | ["C"] | null | <p>Given $$A = \left[ {\matrix{
2 & { - 1} \cr
0 & 2 \cr
} } \right]$$</p>
<p>and</p>
<p>$$B = I - {5_{{C_1}}}(adj\,A) + {5_{{C_2}}}{(adj\,A)^2} - {5_{{C_3}}}{(adj\,A)^3} + {5_{{C_4}}}{(adj\,A)^4} - {5_{{C_5}}}{(adj\,A)^5}$$</p>
<p>$$ = {\left( {I - (adj\,A)} \right)^5}$$</p>
<p>Cofactor of $$A = \left[ {\matrix{
{{{( - 1)}^{1 + 1}}\,.\,2} & {{{( - 1)}^{1 + 2}}\,.\,0} \cr
{{{( - 1)}^{2 + 1}}\,.\,( - 1)} & {{{( - 1)}^{2 + 2}}\,.\,2} \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
2 & 0 \cr
1 & 2 \cr
} } \right]$$</p>
<p>Transpose of cofactor of $$A = \left[ {\matrix{
2 & 1 \cr
0 & 2 \cr
} } \right]$$</p>
<p>$$\therefore$$ $$adj\,A = \left[ {\matrix{
2 & 1 \cr
0 & 2 \cr
} } \right]$$</p>
<p>Now, $$I - adj\,A$$</p>
<p>$$ = \left[ {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right] - \left[ {\matrix{
2 & 1 \cr
0 & 2 \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
{ - 1} & { - 1} \cr
0 & { - 1} \cr
} } \right]$$</p>
<p>Now let,</p>
<p>$$P = I - adj\,A = \left[ {\matrix{
{ - 1} & { - 1} \cr
0 & { - 1} \cr
} } \right]$$</p>
<p>$$\therefore$$ $${P^2} = \left[ {\matrix{
{ - 1} & { - 1} \cr
0 & { - 1} \cr
} } \right]\left[ {\matrix{
{ - 1} & { - 1} \cr
0 & { - 1} \cr
} } \right]$$</p>
<p>$$ = \left[ {\matrix{
1 & 2 \cr
0 & 1 \cr
} } \right]$$</p>
<p>$${P^4} = {P^2}\,.\,{P^2} = \left[ {\matrix{
1 & 2 \cr
0 & 1 \cr
} } \right]\left[ {\matrix{
1 & 2 \cr
0 & 1 \cr
} } \right] = \left[ {\matrix{
1 & 4 \cr
0 & 1 \cr
} } \right]$$</p>
<p>$${P^5} = {P^4}\,.\,P = \left[ {\matrix{
1 & 4 \cr
0 & 1 \cr
} } \right]\left[ {\matrix{
{ - 1} & { - 1} \cr
0 & { - 1} \cr
} } \right] = \left[ {\matrix{
{ - 1} & { - 5} \cr
0 & { - 1} \cr
} } \right]$$</p>
<p>$$\therefore$$ $$B = \left[ {\matrix{
{ - 1} & { - 5} \cr
0 & { - 1} \cr
} } \right]$$</p>
<p>Now sum of elements $$ = - 1 - 5 - 1 + 0 = - 7$$</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l5667sum | maths | matrices-and-determinants | properties-of-determinants | <p>Let A be a matrix of order 3 $$\times$$ 3 and det (A) = 2. Then det (det (A) adj (5 adj (A<sup>3</sup>))) is equal to _____________.</p> | [{"identifier": "A", "content": "512 $$\\times$$ 10<sup>6</sup>"}, {"identifier": "B", "content": "256 $$\\times$$ 10<sup>6</sup>"}, {"identifier": "C", "content": "1024 $$\\times$$ 10<sup>6</sup>"}, {"identifier": "D", "content": "256 $$\\times$$ 10<sup>11</sup>"}] | ["A"] | null | <p>$$|A| = 2$$</p>
<p>$$||A| = adj\,(5\,adj\,{A^3})|$$</p>
<p>$$ = |25|A|adj\,(adj\,{A^3})|$$</p>
<p>$$ = {25^3}|A{|^3}\,.\,|adj\,{A^3}{|^2}$$</p>
<p>$$ = {25^3}\,.\,{2^3}\,.\,|{A^3}{|^4}$$</p>
<p>$$ = {25^3}\,.\,{2^3}\,.\,{2^{12}} = {10^6}\,.\,512$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l56q3mwj | maths | matrices-and-determinants | properties-of-determinants | <p>Let $$f(x) = \left| {\matrix{
a & { - 1} & 0 \cr
{ax} & a & { - 1} \cr
{a{x^2}} & {ax} & a \cr
} } \right|,\,a \in R$$. Then the sum of the squares of all the values of a, for which $$2f'(10) - f'(5) + 100 = 0$$, is</p> | [{"identifier": "A", "content": "117"}, {"identifier": "B", "content": "106"}, {"identifier": "C", "content": "125"}, {"identifier": "D", "content": "136"}] | ["C"] | null | <p>$$f(x) = \left| {\matrix{
a & { - 1} & 0 \cr
{ax} & a & { - 1} \cr
{a{x^2}} & {ax} & a \cr
} } \right|,\,a \in R$$</p>
<p>$$f(x) = a({a^2} + ax) + 1({a^2}x + a{x^2})$$</p>
<p>$$ = a{(x + a)^2}$$</p>
<p>$$f'(x) = 2a(x + a)$$</p>
<p>Now, $$2f'(10) - f'(5) + 100 = 0$$</p>
<p>$$ \Rightarrow 2.\,2a(10 + a) - 2a(5 + a) + 100 = 0$$</p>
<p>$$ \Rightarrow 2a(a + 15) + 100 = 0$$</p>
<p>$$ \Rightarrow {a^2} + 15a + 50 = 0$$</p>
<p>$$ \Rightarrow a = - 10,\, - 5$$</p>
<p>$$\therefore$$ Sum of squares of values of a = 125.</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l56q60a3 | maths | matrices-and-determinants | properties-of-determinants | <p>Let A and B be two 3 $$\times$$ 3 matrices such that $$AB = I$$ and $$|A| = {1 \over 8}$$. Then $$|adj\,(B\,adj(2A))|$$ is equal to</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "64"}, {"identifier": "D", "content": "128"}] | ["C"] | null | <p>A and B are two matrices of order 3 $$\times$$ 3.</p>
<p>and $$AB = I$$,</p>
<p>$$|A| = {1 \over 8}$$</p>
<p>Now, $$|A||B| = 1$$</p>
<p>$$|B| = 8$$</p>
<p>$$\therefore$$ $$|adj(B(adj(2A))| = |B(adj(2A)){|^2}$$</p>
<p>$$ = |B{|^2}|adj(2A){|^2}$$</p>
<p>$$ = {2^6}|2A{|^{2 \times 2}}$$</p>
<p>$$ = {2^6}.\,{2^{12}}.\,{1 \over {{2^{12}}}} = 64$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l57p0n9c | maths | matrices-and-determinants | properties-of-determinants | <p>The positive value of the determinant of the matrix A, whose</p>
<p>Adj(Adj(A)) = $$\left( {\matrix{
{14} & {28} & { - 14} \cr
{ - 14} & {14} & {28} \cr
{28} & { - 14} & {14} \cr
} } \right)$$, is _____________.</p> | [] | null | 14 | <p>$$\left| {adj(adj(A))} \right| = {\left| A \right|^{{2^2}}} = {\left| A \right|^4}$$</p>
<p>$$\therefore$$ $${\left| A \right|^4} = \left| {\matrix{
{14} & {28} & { - 14} \cr
{ - 14} & {14} & {28} \cr
{28} & { - 14} & {14} \cr
} } \right|$$</p>
<p>$$ = {(14)^3}\left| {\matrix{
1 & 2 & { - 1} \cr
{ - 1} & 1 & 2 \cr
2 & { - 1} & 1 \cr
} } \right|$$</p>
<p>$$ = {(14)^3}(3 - 2( - 5) - 1( - 1))$$</p>
<p>$${\left| A \right|^4} = {(14)^4} \Rightarrow \left| A \right| = 14$$</p> | integer | jee-main-2022-online-27th-june-morning-shift |
1l587f1jw | maths | matrices-and-determinants | properties-of-determinants | <p>Let A be a 3 $$\times$$ 3 invertible matrix. If |adj (24A)| = |adj (3 adj (2A))|, then |A|<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "6<sup>6</sup>"}, {"identifier": "B", "content": "2<sup>12</sup>"}, {"identifier": "C", "content": "2<sup>6</sup>"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <p>We know, $$|adj\,A| = |A{|^{n - 1}}$$</p>
<p>Now, $$|adj\,24A| = |adj\,3(adj\,2A)|$$</p>
<p>$$ \Rightarrow |24A{|^{3 - 1}} = |3\,adj\,2A{|^{3 - 1}}$$</p>
<p>$$ \Rightarrow |24A{|^2} = |3\,adj\,2A{|^2}$$</p>
<p>Also, we know, $$|KA| = {K^n}|A|$$</p>
<p>$$ \Rightarrow {\left( {{{(24)}^2}} \right)^2}|A{|^2} = {\left( {{{(3)}^3}} \right)^2}|adj\,2A{|^2}$$</p>
<p>$$ \Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,{\left( {|2A{|^{3 - 1}}} \right)^2}$$</p>
<p>$$ \Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,|2A{|^4}$$</p>
<p>$$ \Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,{\left( {{2^3}} \right)^4}\,.\,|A{|^4}$$</p>
<p>$$ \Rightarrow {3^6}\,.\,{8^6}\,.\,|A{|^2} = {3^6}\,.\,{8^4}\,.\,|A{|^4}$$</p>
<p>$$ \Rightarrow {8^2} = |A{|^2}$$</p>
<p>$$ \Rightarrow |A{|^2} = 64$$ = 2<sup>6</sup> </p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l5c14prc | maths | matrices-and-determinants | properties-of-determinants | <p>Let S = {$$\sqrt{n}$$ : 1 $$\le$$ n $$\le$$ 50 and n is odd}.</p>
<p>Let a $$\in$$ S and $$A = \left[ {\matrix{
1 & 0 & a \cr
{ - 1} & 1 & 0 \cr
{ - a} & 0 & 1 \cr
} } \right]$$.</p>
<p>If $$\sum\limits_{a\, \in \,S}^{} {\det (adj\,A) = 100\lambda } $$, then $$\lambda$$ is equal to :</p> | [{"identifier": "A", "content": "218"}, {"identifier": "B", "content": "221"}, {"identifier": "C", "content": "663"}, {"identifier": "D", "content": "1717"}] | ["B"] | null | <p>Given, $$A = {\left[ {\matrix{
1 & 0 & a \cr
{ - 1} & 1 & 0 \cr
{ - a} & 0 & 1 \cr
} } \right]_{3 \times 3}}$$</p>
<p>S = {$$\sqrt{n}$$ : 1 $$\le$$ n $$\le$$ 50 and n is odd}</p>
<p>$$ \therefore $$ S = $$\left\{ {1,\sqrt 3 ,\sqrt 5 ,\sqrt 7 ,....,\sqrt {49} } \right\}$$</p>
<p>We know,</p>
<p>$$\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}$$</p>
<p>Here, n = order of matrix.</p>
<p>Here, n = 3</p>
<p>$$\therefore$$ $$\left| {adj\,A} \right| = {\left| A \right|^{3 - 1}} = {\left| A \right|^2}$$</p>
<p>Now, $$\left| A \right| = \left| {\matrix{
1 & 0 & a \cr
{ - 1} & 1 & 0 \cr
{ - a} & 0 & 1 \cr
} } \right|$$</p>
<p>$$ = 1(1 - 0) - 0 + a(0 - ( - a))$$</p>
<p>$$ = {a^2} + 1$$</p>
<p>$$\therefore$$ $$\left| {adj\,A} \right| = {\left| A \right|^2} = {({a^2} + 1)^2}$$</p>
<p>Now, $$\sum\limits_{a\, \in \,S}^{} {\det (adj\,A)} $$</p>
<p>$$ = \sum\limits_{a\, \in \,S}^{} {{{({a^2} + 1)}^2}} $$</p>
<p>= $${\left( {{1^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 3 } \right)}^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 5 } \right)}^2} + 1} \right)^2} + .... + {\left( {{{\left( {\sqrt {49} } \right)}^2} + 1} \right)^2}$$</p>
<p>= $${\left( {{1^2} + 1} \right)^2} + {\left( {3 + 1} \right)^2} + {\left( {5 + 1} \right)^2} + .... + {\left( {49 + 1} \right)^2}$$</p>
<p>= $${2^2} + {4^2} + {6^2} + .... + {50^2}$$</p>
<p>= $${2^2}\left( {{1^2} + {2^2} + {3^2} + .... + {{25}^2}} \right)$$</p>
<p>= $$4.{{25.26.51} \over 6} = 100.221$$</p>
<p>$$\therefore$$ $$100K = 100.221$$</p>
<p>$$ \Rightarrow K = 221$$</p> | mcq | jee-main-2022-online-24th-june-morning-shift |
1l5vzdw08 | maths | matrices-and-determinants | properties-of-determinants | <p>Let A and B be two square matrices of order 2. If $$det\,(A) = 2$$, $$det\,(B) = 3$$ and $$\det \left( {(\det \,5(det\,A)B){A^2}} \right) = {2^a}{3^b}{5^c}$$ for some a, b, c, $$\in$$ N, then a + b + c is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "14"}] | ["B"] | null | <p>Given,</p>
<p>$$\det (A) = 2$$,</p>
<p>$$\det (B) = 3$$</p>
<p>and $$\det \left( {\left( {\det \left( {5\left( {\det A} \right)B} \right)} \right){A^2}} \right) = {2^a}{3^b}{5^c}$$</p>
<p>$$ \Rightarrow \left| {\det \left( {5\left( {\det A} \right)B} \right){A^2}} \right| = {2^a}{3^b}{5^c}$$</p>
<p>$$ \Rightarrow \left| {\left| {5\left( {\det A} \right)\left. B \right|{A^2}} \right.} \right| = {2^a}{3^b}{5^c}$$</p>
<p>$$ \Rightarrow \left| {\left| {5\left| {A\left| B \right.\left| {{A^2}} \right.} \right.} \right.} \right| = {2^a}{3^b}{5^c}$$</p>
<p>$$ \Rightarrow \left| {\left| {5\,.\,2\,.\,B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>$$\Rightarrow \left| {\left| {10B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>$$ \Rightarrow \left| {{{10}^2}\,.\,\left| B \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>As $$\left| {k\,.\,A} \right| = {k^n}|A|$$</p>
<p>$$ \Rightarrow \left| {100 \times 3{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>$$ \Rightarrow {(300)^2}\,.\,|{A^2}| = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>$$ \Rightarrow {(300)^2}\,.\,|A{|^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>$$ \Rightarrow {(300)^2}\,.\,{2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>$$ \Rightarrow 9 \times 100 \times 100 \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>$$ \Rightarrow {3^2} \times {2^2} \times {5^2} \times {2^2} \times {5^2} \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>$$ \Rightarrow {2^6}\,.\,{3^2}\,.\,{5^4} = {2^a}\,.\,{3^b}\,.\,{5^c}$$</p>
<p>Comparing both sides, we get</p>
<p>$$a = 6$$, $$b = 2$$, $$c = 4$$</p>
<p>$$\therefore$$ $$a + b + c = 6 + 2 + 4 = 12$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6gginlk | maths | matrices-and-determinants | properties-of-determinants | <p>Let A be a 2 $$\times$$ 2 matrix with det (A) = $$-$$ 1 and det ((A + I) (Adj (A) + I)) = 4. Then the sum of the diagonal elements of A can be :</p> | [{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$- \\sqrt2$$"}] | ["B"] | null | <p>$$|(A + I)(adj\,A + I)| = 4$$</p>
<p>$$ \Rightarrow |A\,adj\,A + A + adj\,A + I| = 4$$</p>
<p>$$ \Rightarrow |(A)I + A + adj\,A + I| = 4$$</p>
<p>$$|A| = - 1 \Rightarrow |A + adj\,A| = 4$$</p>
<p>$$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]\,adj\,A = \left[ {\matrix{
a & { - b} \cr
{ - c} & d \cr
} } \right]$$</p>
<p>$$ \Rightarrow \left| {\matrix{
{(a + d)} & 0 \cr
0 & {(a + d)} \cr
} } \right| = 4$$</p>
<p>$$ \Rightarrow a + d = \, \pm \,2$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6kik0ub | maths | matrices-and-determinants | properties-of-determinants | <p>Let $$A=\left(\begin{array}{rr}4 & -2 \\ \alpha & \beta\end{array}\right)$$.</p>
<p>If $$\mathrm{A}^{2}+\gamma \mathrm{A}+18 \mathrm{I}=\mathrm{O}$$, then $$\operatorname{det}(\mathrm{A})$$ is equal to _____________.</p> | [{"identifier": "A", "content": "$$-$$18"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "$$-$$50"}, {"identifier": "D", "content": "50"}] | ["B"] | null | <p>Characteristic equation of A is given by</p>
<p>$$\left| {A - \lambda I} \right| = 0$$</p>
<p>$$\left| {\matrix{
{4 - \lambda } & { - 2} \cr
\alpha & {\beta - \lambda } \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow {\lambda ^2} - (4 + \beta )\lambda + (4\beta + 2\alpha ) = 0$$</p>
<p>So, $${A^2} - (4 + \beta )A + (4\beta + 2\alpha )I = 0$$</p>
<p>$$|A| = 4\beta + 2\alpha = 18$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6klber7 | maths | matrices-and-determinants | properties-of-determinants | <p>Consider a matrix $$A=\left[\begin{array}{ccc}\alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta\end{array}\right]$$, where $$\alpha, \beta, \gamma$$ are three distinct natural numbers.</p>
<p>If $$\frac{\operatorname{det}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))))}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}=2^{32} \times 3^{16}$$, then the number of such 3 - tuples $$(\alpha, \beta, \gamma)$$ is ____________.</p> | [] | null | 42 | <p>$$\det (A) = \left| {\matrix{
\alpha & \beta & \gamma \cr
{{\alpha ^2}} & {{\beta ^2}} & {{\gamma ^2}} \cr
{\beta + \gamma } & {\gamma + \alpha } & {\alpha + \beta } \cr
} } \right|$$</p>
<p>$${R_3} \to {R_3} + {R_1}$$</p>
<p>$$ \Rightarrow (\alpha + \beta + \gamma )\left| {\matrix{
\alpha & \beta & \gamma \cr
{{\alpha ^2}} & {{\beta ^2}} & {{\gamma ^2}} \cr
1 & 1 & 1 \cr
} } \right|$$</p>
<p>$$\therefore$$ $$\det (A) = (\alpha + \beta + \gamma )(\alpha - \beta )(\beta - \gamma )(\gamma - \alpha )$$</p>
<p>Also, $$\det (adj\,(adj\,(adj\,(adj\,(A)))))$$</p>
<p>$$ = {(\det (A))^{{2^4}}} = (\det {(A)^{16}}$$</p>
<p>$$\therefore$$ $${{{{(\alpha + \beta + \gamma )}^{16}}{{(\alpha - \beta )}^{16}}{{(\beta - \gamma )}^{16}}{{(\gamma - \alpha )}^{16}}} \over {{{(\alpha - \beta )}^{16}}{{(\beta - \gamma )}^{16}}{{(\gamma - \alpha )}^{16}}}} = {(4.13)^{16}}$$</p>
<p>$$ \Rightarrow \alpha + \beta + \gamma = 12$$</p>
<p>$$ \Rightarrow (\alpha ,\beta ,\gamma )$$ distinct natural triplets</p>
<p>$$ = {}^{11}{C_2} - 1 - {}^3{C_2}(4) = 55 - 1 - 12$$</p>
<p>$$ = 42$$</p> | integer | jee-main-2022-online-27th-july-evening-shift |
1l6m5y7bn | maths | matrices-and-determinants | properties-of-determinants | <p>Let the matrix $$A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]$$ and the matrix $$B_{0}=A^{49}+2 A^{98}$$. If $$B_{n}=A d j\left(B_{n-1}\right)$$ for all $$n \geq 1$$, then $$\operatorname{det}\left(B_{4}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$3^{28}$$"}, {"identifier": "B", "content": "$$3^{30}$$"}, {"identifier": "C", "content": "$$3^{32}$$"}, {"identifier": "D", "content": "$$3^{36}$$"}] | ["C"] | null | <p>$$A = \left[ {\matrix{
0 & 1 & 0 \cr
0 & 0 & 1 \cr
1 & 0 & 0 \cr
} } \right]$$</p>
<p>$$ \Rightarrow {A^2} = \left[ {\matrix{
0 & 1 & 0 \cr
0 & 0 & 1 \cr
1 & 0 & 0 \cr
} } \right] \times \left[ {\matrix{
0 & 1 & 0 \cr
0 & 0 & 1 \cr
1 & 0 & 0 \cr
} } \right] = \left[ {\matrix{
0 & 0 & 1 \cr
1 & 0 & 0 \cr
0 & 1 & 0 \cr
} } \right]$$</p>
<p>$$ \Rightarrow {A^3} = \left[ {\matrix{
0 & 0 & 1 \cr
1 & 0 & 0 \cr
0 & 1 & 0 \cr
} } \right]\left[ {\matrix{
0 & 1 & 0 \cr
0 & 0 & 1 \cr
1 & 0 & 0 \cr
} } \right] = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right] = l$$</p>
<p>Now $${B_0} = {A^{49}} + 2{A^{98}} = {({A^3})^{16}}\,.\,A + 2{({A^3})^{32}}\,.\,{A^2}$$</p>
<p>$${B_0} = A + 2{A^2} = \left[ {\matrix{
0 & 1 & 0 \cr
0 & 0 & 1 \cr
1 & 0 & 0 \cr
} } \right] + \left[ {\matrix{
0 & 0 & 2 \cr
2 & 0 & 0 \cr
0 & 2 & 0 \cr
} } \right] = \left[ {\matrix{
0 & 1 & 2 \cr
2 & 0 & 1 \cr
1 & 2 & 0 \cr
} } \right]$$</p>
<p>$$|{B_0}| = 9$$</p>
<p>Since, $${B_n} = Adj\,|{B_{n - 1}}| \Rightarrow |{B_n}| = |{B_{n - 1}}{|^2}$$</p>
<p>Hence $$|{B_4}| = |{B_3}{|^2} = |{B_2}{|^4} = |{B_1}{|^8} = |{B_0}{|^{16}}$$</p>
<p>$$ = |{3^2}{|^{16}} = {3^{32}}$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
ldoa9wue | maths | matrices-and-determinants | properties-of-determinants | Let A be a $n \times n$ matrix such that $|\mathrm{A}|=2$. If the determinant of the matrix
$\operatorname{Adj}\left(2 \cdot \operatorname{Adj}\left(2 \mathrm{~A}^{-1}\right)\right) \cdot$ is $2^{84}$, then $\mathrm{n}$ is equal to : | [] | null | 5 | $\because\left|\operatorname{adj}\left(2 \cdot \operatorname{adj}\left(2 A^{-1}\right)\right)\right|=2^{84}$
<br/><br/>$\Rightarrow 2^{n \cdot(n-1)}\left|\operatorname{adj}\left(2 A^{-1}\right)\right|^{(n-1)}=2^{84}$
<br/><br/>$\Rightarrow 2^{n(n-1)}\left|2 A^{-1}\right|^{(n-1)^{2}}=2^{84}$
<br/><br/>$\Rightarrow 2^{n(n-1)} \cdot 2^{n(n-1)^{2}} \cdot \frac{1}{|A|^{(n-1)^{2}}}=2^{84}$
<br/><br/>$\Rightarrow 2^{n(n-1)+n(n-1)^{2}-(n-1)^{2}}=2^{84}\{\because|4|=2\}$
<br/><br/>$\therefore n(n-1)+(n-1)^{3}=84$
<br/><br/>$\therefore n=5$ | integer | jee-main-2023-online-31st-january-evening-shift |
ldqwv9fj | maths | matrices-and-determinants | properties-of-determinants | If $P$ is a $3 \times 3$ real matrix such that $P^T=a P+(a-1) I$, where $a>1$, then : | [{"identifier": "A", "content": "$|A d j P|=1$"}, {"identifier": "B", "content": "$|A d j P|>1$"}, {"identifier": "C", "content": "$|A d j P|=\\frac{1}{2}$"}, {"identifier": "D", "content": "$P$ is a singular matrix"}] | ["A"] | null | <p>$$P = \left[ {\matrix{
{{a_1}} & {{b_1}} & {{c_1}} \cr
{{a_2}} & {{b_2}} & {{c_2}} \cr
{{a_3}} & {{b_3}} & {{c_3}} \cr
} } \right]$$</p>
<p>Given : $${P^T} = aP + (a - 1)I$$</p>
<p>$$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{b_1}} & {{b_2}} & {{b_3}} \cr
{{c_1}} & {{c_2}} & {{c_3}} \cr
} } \right] = \left[ {\matrix{
{a{a_1} + a - 1} & {a{b_1}} & {a{c_1}} \cr
{a{a_2}} & {a{b_2} + a - 1} & {a{c_2}} \cr
{a{a_3}} & {a{b_3}} & {a{c_3} + a - 1} \cr
} } \right]$$</p>
<p>$$ \Rightarrow {a_1} = a{a_1} + a - 1 \Rightarrow {a_1}(1 - a) = a - 1 \Rightarrow {a_1} = - 1$$</p>
<p>Similarly, $${a_1} = {b_2} = {c_3} = - 1$$</p>
<p>Now, $$\left. \matrix{
{a_2} = a{b_1} \hfill \cr
{b_1} = a{a_2} \hfill \cr} \right] \to {a_2} = {a^2}{a_2} \Rightarrow {a_2} = 0 \Rightarrow {b_1} = 0$$</p>
<p>$${c_1} = a{a_3}$$</p>
<p>Similarly, all other elements will also be 0</p>
<p>$${a_2} = {a_3} = {b_1} = {b_3} = {c_1} = {c_2} = 0$$</p>
<p>$$\therefore$$ $$P = \left[ {\matrix{
{ - 1} & 0 & 0 \cr
0 & { - 1} & 0 \cr
0 & 0 & { - 1} \cr
} } \right]$$</p>
<p>$$|P| = - 1$$</p>
<p>$$|Adj(P){|_{n \times n}} = |A{|^{(n - 1)}}$$</p>
<p>$$ \Rightarrow |Adj(P)| = {( - 1)^2} = 1$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldr7bd4v | maths | matrices-and-determinants | properties-of-determinants | <p>Let $$A=\left(\begin{array}{cc}\mathrm{m} & \mathrm{n} \\ \mathrm{p} & \mathrm{q}\end{array}\right), \mathrm{d}=|\mathrm{A}| \neq 0$$ and $$\mathrm{|A-d(A d j A)|=0}$$. Then </p> | [{"identifier": "A", "content": "$$1+\\mathrm{d}^{2}=\\mathrm{m}^{2}+\\mathrm{q}^{2}$$"}, {"identifier": "B", "content": "$$1+d^{2}=(m+q)^{2}$$"}, {"identifier": "C", "content": "$$(1+d)^{2}=m^{2}+q^{2}$$"}, {"identifier": "D", "content": "$$(1+d)^{2}=(m+q)^{2}$$"}] | ["D"] | null | <p>$$\left| {A - d\left( {\matrix{
q & { - n} \cr
{ - p} & m \cr
} } \right)} \right| = 0$$</p>
<p>$$\left| {\matrix{
{m - qd} & {n(1 + d)} \cr
{p(1 + d)} & {q - md} \cr
} } \right| = 0$$</p>
<p>$$(m - qd)(q - md) = np{(1 + d)^2}$$</p>
<p>$$mq - ({q^2} + {m^2})d + qm{d^2} = np(1 + {d^2}) + 2npd$$</p>
<p>$${d^2}(mq - np) + 1(mq - np) = (2np + {m^2} + {q^2})d$$</p>
<p>$$({d^2} + 1)(mq - np) = (2np + m + a)d$$</p>
<p>$${d^2} + 1 = 2np + {m^2} + {q^2}$$</p>
<p>$$2d = 2mq - 2np$$</p>
<p>$$ \Rightarrow {(1 + d)^2} = {(m + q)^2}$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldsfnv8m | maths | matrices-and-determinants | properties-of-determinants | <p>The set of all values of $$\mathrm{t\in \mathbb{R}}$$, for which the matrix <br/><br/>$$\left[ {\matrix{
{{e^t}} & {{e^{ - t}}(\sin t - 2\cos t)} & {{e^{ - t}}( - 2\sin t - \cos t)} \cr
{{e^t}} & {{e^{ - t}}(2\sin t + \cos t)} & {{e^{ - t}}(\sin t - 2\cos t)} \cr
{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr
} } \right]$$ is invertible, is :</p> | [{"identifier": "A", "content": "$$\\left\\{ {k\\pi ,k \\in \\mathbb{Z}} \\right\\}$$"}, {"identifier": "B", "content": "$$\\mathbb{R}$$"}, {"identifier": "C", "content": "$$\\left\\{ {(2k + 1){\\pi \\over 2},k \\in \\mathbb{Z}} \\right\\}$$"}, {"identifier": "D", "content": "$$\\left\\{ {k\\pi + {\\pi \\over 4},k \\in \\mathbb{Z}} \\right\\}$$"}] | ["B"] | null | If the matrix is invertible then its determinant
should not be zero.
<br/><br/>So, $$
\left|\begin{array}{ccc}
e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\
e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\
e^t & e^{-t} \cos t & e^{-t} \sin t
\end{array}\right| \neq 0
$$
<br/><br/>$$
\Rightarrow e^t \times e^{-t} \times e^{-t}\left|\begin{array}{ccc}
1 & \sin t-2 \cos t & -2 \sin t-\cos t \\
1 & 2 \sin t+\cos t & \sin t-2 \cos t \\
1 & \cos t & \sin t
\end{array}\right| \neq 0
$$
<br/><br/>Applying, $R_1 \rightarrow R_1-R_2$ then $R_2 \rightarrow R_2-R_3$,
<br/><br/>$$
\begin{aligned}
& e^{-t}\left|\begin{array}{ccc}
0 & -\sin t-3 \cos t & -3 \sin t+\cos t \\
0 & 2 \sin t & -2 \cos t \\
1 & \cos t & \sin t
\end{array}\right| \neq 0 \\\\
& \Rightarrow e^{-t}\left(2 \sin t \cos t+6 \cos ^2 t+6 \sin ^2 t-2 \sin t \cos t\right) \neq 0 \\\\
& \Rightarrow 6e^{-t} \neq 0, \text { for } \forall t \in R .
\end{aligned}
$$ | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldv1q57x | maths | matrices-and-determinants | properties-of-determinants | <p>Let $$x,y,z > 1$$ and $$A = \left[ {\matrix{
1 & {{{\log }_x}y} & {{{\log }_x}z} \cr
{{{\log }_y}x} & 2 & {{{\log }_y}z} \cr
{{{\log }_z}x} & {{{\log }_z}y} & 3 \cr
} } \right]$$. Then $$\mathrm{|adj~(adj~A^2)|}$$ is equal to</p> | [{"identifier": "A", "content": "$$6^4$$"}, {"identifier": "B", "content": "$$2^8$$"}, {"identifier": "C", "content": "$$4^8$$"}, {"identifier": "D", "content": "$$2^4$$"}] | ["B"] | null | $$
\begin{aligned}
& |A|=\frac{1}{\log x \log y \log z}\left|\begin{array}{ccc}
\log x & \log y & \log z \\
\log x & 2 \log y & \log z \\
\log x & \log y & 3 \log z
\end{array}\right|=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 3
\end{array}\right|=2 \\\\
& \Rightarrow\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right|=\left|\operatorname{adj}\left(A^2\right)\right|^2=\left(\left|A^2\right|^2\right)^2=|A|^8=2^8
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldww8dwi | maths | matrices-and-determinants | properties-of-determinants | <p>Let A be a 3 $$\times$$ 3 matrix such that $$\mathrm{|adj(adj(adj~A))|=12^4}$$. Then $$\mathrm{|A^{-1}~adj~A|}$$ is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "2$$\\sqrt3$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$\\sqrt6$$"}] | ["B"] | null | $|A|^{(n-1)^{3}}=12^{4}$
<br/><br/>
$$
\begin{aligned}
&|A|^{8}=12^{4} \\\\
&|A|=\sqrt{12} \\\\
&\left|A^{-1} \operatorname{adj} A\right|=\left|A^{-1}\right| \cdot|A|^{2} \\\\
&=|A| = 2\sqrt3
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldybt7ax | maths | matrices-and-determinants | properties-of-determinants | <p>Let $$\alpha$$ be a root of the equation $$(a - c){x^2} + (b - a)x + (c - b) = 0$$ where a, b, c are distinct real numbers such that the matrix $$\left[ {\matrix{
{{\alpha ^2}} & \alpha & 1 \cr
1 & 1 & 1 \cr
a & b & c \cr
} } \right]$$ is singular. Then, the value of $${{{{(a - c)}^2}} \over {(b - a)(c - b)}} + {{{{(b - a)}^2}} \over {(a - c)(c - b)}} + {{{{(c - b)}^2}} \over {(a - c)(b - a)}}$$ is</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "9"}] | ["A"] | null | $$
\begin{aligned}
& \Delta=0=\left|\begin{array}{ccc}
\alpha^2 & \alpha & 1 \\\\
1 & 1 & 1 \\\\
\mathrm{a} & \mathrm{b} & \mathrm{c}
\end{array}\right| \\\\
& \Rightarrow \alpha^2(\mathrm{c}-\mathrm{b})-\alpha(\mathrm{c}-\mathrm{a})+(\mathrm{b}-\mathrm{a})=0
\end{aligned}
$$<br/><br/>
It is singular when $\alpha=1$<br/><br/>
$$
\begin{aligned}
& \frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)} \\\\
& \frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} \\\\
& =3 \frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnxwdvt | maths | matrices-and-determinants | properties-of-determinants | Let the determinant of a square matrix A of order $m$ be $m-n$, where $m$ and $n$<br/><br/> satisfy $4 m+n=22$ and $17 m+4 n=93$.<br/><br/> If $\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A)))=3^{a} 5^{b} 6^{c}$ then $a+b+c$ is equal to : | [{"identifier": "A", "content": "96"}, {"identifier": "B", "content": "84"}, {"identifier": "C", "content": "109"}, {"identifier": "D", "content": "101"}] | ["A"] | null | Given that $|A|=m-n$, and let's solve the system of linear equations to find the values of $m$ and $n$ :
<br/><br/>$4m + n = 22$ ...... (1)
<br/><br/>$17m + 4n = 93$ ....... (2)
<br/><br/>We can multiply equation (1) by 4 to make the coefficients of $n$ in both equations equal:
<br/><br/>$16m + 4n = 88$ ......... (3)
<br/><br/>Now, subtract equation (3) from equation (2):
<br/><br/>$(17m + 4n) - (16m + 4n) = 93 - 88$
<br/><br/>$m = 5$
<br/><br/>Now, we can find the value of $n$ by substituting $m$ into equation (1):
<br/><br/>$4(5) + n = 22$
<br/><br/>$20 + n = 22$
<br/><br/>$n = 2$
<br/><br/>Since the order of matrix A is $m$, the order is 5. Now, let's find the determinant of $n \operatorname{adj}(\operatorname{adj}(mA))$:
<br/><br/>We know that $\operatorname{det}(A) = m-n = 3$.
<br/><br/>
$$
\begin{aligned}
& \therefore \operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A))) \\\\
& =|2 \operatorname{adj}(\operatorname{adj}(5 A))| \\\\
& =2^5|5 A|^{16} \\\\
& =2^5 5^{80}|A|^{16}=2^5 \cdot 3^{16} \cdot 5^{80} \\\\
& =3^{11} 5^{80} 6^5
\end{aligned}
$$
<br/><br/>So, $a+b+c=96$ | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgowmdng | maths | matrices-and-determinants | properties-of-determinants | <p>Let for $$A = \left[ {\matrix{
1 & 2 & 3 \cr
\alpha & 3 & 1 \cr
1 & 1 & 2 \cr
} } \right],|A| = 2$$. If $$\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}$$, then $$3n + \alpha $$ is equal to</p> | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "10"}] | ["A"] | null | $$
\begin{aligned}
& A=\left[\begin{array}{lll}
1 & 2 & 3 \\
\alpha & 3 & 1 \\
1 & 1 & 2
\end{array}\right] \\\\
& |A|=2
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\Rightarrow&1(6-1)-2(2 \alpha-1)+3(\alpha-3)=2\\\\
\Rightarrow&5-4 \alpha+2+3 \alpha-9=2\\\\
\Rightarrow&-\alpha-4=0\\\\
\Rightarrow&\alpha=-4
\end{aligned}
$$
<br/><br/>Now, $$\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}$$
<br/><br/>$$ \Rightarrow $$ $$2^3|\operatorname{adj}(2 \operatorname{adj}(2 A))|$$ = ${32^n}$
<br/><br/>$$
\begin{aligned}
\Rightarrow & 8|\operatorname{Adj}(2 \operatorname{Adj}(2 \mathrm{~A}))| = {32^n}\\\\
\Rightarrow & 8\left|\operatorname{Adj}\left(2 \times 2^2 \operatorname{Adj}(\mathrm{A})\right)\right| = {32^n} \\\\
\Rightarrow & 8\left|\operatorname{Adj}\left(2^3 \operatorname{AdjA}\right)\right| = {32^n} \\\\
\Rightarrow & 8\left|2^6 \operatorname{Adj}(\operatorname{AdjA})\right| = {32^n} \\\\
\Rightarrow & 2^3\left(2^6\right)^3|\operatorname{Adj}(\operatorname{Adj})| = {32^n} \\\\
\Rightarrow & 2^3 \cdot 2^{18}|\mathrm{~A}|^4 = {32^n} \\\\
\Rightarrow& 2^{21} \cdot 2^4 = {32^n}\\\\
\Rightarrow& 2^{25} = {32^n}\\\\
\Rightarrow& \left(2^5\right)^5 = {32^n}\\\\
\Rightarrow& (32)^5 = {32^n}
\end{aligned}
$$
<br/><br/>$$
\begin{array}{ll}
\therefore n=5 \\\\
\Rightarrow 3 n+\alpha=15-4=11
\end{array}
$$ | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgvqep21 | maths | matrices-and-determinants | properties-of-determinants | <p>If $$\mathrm{A}=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{ccc}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right]$$, then $$|\operatorname{adj}(\operatorname{adj}(2 \mathrm{~A}))|$$ is equal to :</p> | [{"identifier": "A", "content": "$$2^{12}$$"}, {"identifier": "B", "content": "$$2^{20}$$"}, {"identifier": "C", "content": "$$2^{8}$$"}, {"identifier": "D", "content": "$$2^{16}$$"}] | ["D"] | null | Given that
<br/><br/>$$
\begin{aligned}
& A=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll}
5 ! & 6 ! & 7 ! \\
6 ! & 7 ! & 8 ! \\
7 ! & 8 ! & 9 !
\end{array}\right] \\\\
& \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !}\left|\begin{array}{lll}
5 ! & 6 ! & 7 ! \\
6 ! & 7 ! & 8 ! \\
7 ! & 8 ! & 9 !
\end{array}\right| \\\\
& \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !} \times 5 ! 6 ! 7 !\left|\begin{array}{lll}
1 & 6 & 7 \times 6 \\
1 & 7 & 8 \times 7 \\
1 & 8 & 9 \times 8
\end{array}\right|
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& R_3 \rightarrow R_3-R_2 \text { and } R_2 \rightarrow R_2-R_1 \\\\
& |A|=\left|\begin{array}{lll}
1 & 6 & 42 \\
0 & 1 & 14 \\
0 & 1 & 16
\end{array}\right|=2 \\\\
& |\operatorname{adj}(\operatorname{adj}(2 A))|=|2 A|^{(n-1)^2} \\\\
& =|2 A|^4=\left(2^3|A|\right)^4=2^{12}|A|^4=2^{16}
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxt8e8l | maths | matrices-and-determinants | properties-of-determinants | <p>If A is a 3 $$\times$$ 3 matrix and $$|A| = 2$$, then $$|3\,adj\,(|3A|{A^2})|$$ is equal to :</p> | [{"identifier": "A", "content": "$${3^{12}}\\,.\\,{6^{10}}$$"}, {"identifier": "B", "content": "$${3^{11}}\\,.\\,{6^{10}}$$"}, {"identifier": "C", "content": "$${3^{12}}\\,.\\,{6^{11}}$$"}, {"identifier": "D", "content": "$${3^{10}}\\,.\\,{6^{11}}$$"}] | ["B"] | null | Given that $A$ is $3 \times 3$ matrix and $|A|=2$
<br/><br/>$$
\begin{aligned}
& \text { Now, | 3adj }\left(|3 A| A^2\right) \text { | } \\\\
& =3^3\left|\operatorname{adj}\left(|3 A| A^2\right)\right| \\\\
& =3^3\left|\operatorname{adj}\left(54 A^2\right)\right| \\\\
& =3^3\left|54 A^2\right|^2 \\\\
& =3^3 \times\left(54^3\right)^2 \times|A|^4 \\\\
& =3^3 \times(54)^6 \times 2^4 ~~~~~ {[|A|=2 \text { given }]} \\\\
& =3^3 \times\left(3^3 \times 2\right)^6 \times 2^4 ~~~~~ {\left[\left(a^m\right)^n=a^{m n}\right]} \\\\
& =3^{11} \times 3^{10} \times 2^{10} ~~~~~~~ {\left[(a b)^m=a^m b^m\right]} \\\\
& =(3)^{11} \times(6)^{10}
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-morning-shift |
lv0vxdmu | maths | matrices-and-determinants | properties-of-determinants | <p>Let $$A$$ be a $$3 \times 3$$ matrix of non-negative real elements such that $$A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=3\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$$. Then the maximum value of $$\operatorname{det}(\mathrm{A})$$ is _________.</p> | [] | null | 27 | <p>Let $$A = \left[ {\matrix{
{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right]$$</p>
<p>Now</p>
<p>$$A\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]=3\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]$$</p>
<p>$$\left[ {\matrix{
{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right]\left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right] = \left[ {\matrix{
3 \cr
3 \cr
3 \cr
} } \right]$$</p>
<p>$$\begin{aligned}
& a_{11}+a_{12}+a_{13}=3 \\
& a_{21}+a_{22}+a_{23}=3 \\
& a_{31}+a_{32}+a_{33}=3
\end{aligned}$$</p>
<p>Now for maximum value of $$\operatorname{det}(A)=a_{i j}\left\{\begin{array}{ll}0 & i \neq j \\ 3 & i=j\end{array}\right\}$$</p>
<p>$$\therefore|A|=27$$</p> | integer | jee-main-2024-online-4th-april-morning-shift |
lv3ve470 | maths | matrices-and-determinants | properties-of-determinants | <p>If $$\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$$ and $$\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$$, then $$\frac{\mathrm{a}}{\alpha-\mathrm{a}}+\frac{\mathrm{b}}{\beta-\mathrm{b}}+\frac{\gamma}{\gamma-\mathrm{c}}$$ is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "0"}] | ["D"] | null | <p>$$\left|\begin{array}{lll}
\alpha & b & c \\
a & \beta & c \\
a & b & \gamma
\end{array}\right|=0$$</p>
<p>$$\begin{aligned}
& R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\
& \Rightarrow\left|\begin{array}{ccc}
\alpha-a & b-\beta & 0 \\
0 & \beta-b & c-\gamma \\
a & b & \gamma
\end{array}\right|=0
\end{aligned}$$</p>
<p>Take $$\alpha$$-a, $$\beta$$-b, $$\gamma$$-c common from column-1, 2 and 3 respectively</p>
<p>$$\begin{aligned}
& (\alpha-a)(\beta-b)(\gamma-c)\left|\begin{array}{ccc}
1 & -1 & 0 \\
0 & 1 & -1 \\
\frac{a}{\alpha-a} & \frac{b}{\beta-b} & \frac{\gamma}{\gamma-c}
\end{array}\right|=0 \\
& \Rightarrow \frac{\gamma}{\gamma-c}+\frac{b}{\beta-b}+\frac{a}{\alpha-a}=0
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv7v3k3w | maths | matrices-and-determinants | properties-of-determinants | <p>Let A and B be two square matrices of order 3 such that $$\mathrm{|A|=3}$$ and $$\mathrm{|B|=2}$$. Then $$|\mathrm{A}^{\mathrm{T}} \mathrm{A}(\operatorname{adj}(2 \mathrm{~A}))^{-1}(\operatorname{adj}(4 \mathrm{~B}))(\operatorname{adj}(\mathrm{AB}))^{-1} \mathrm{AA}^{\mathrm{T}}|$$ is equal to :</p> | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "81"}, {"identifier": "C", "content": "64"}, {"identifier": "D", "content": "108"}] | ["C"] | null | <p>$$\begin{aligned}
& |A|=3 \\
& |B|=2 \\
& \left.\left|A^T\right||A| \mid(\operatorname{adj}(2 A))^{-1}\|\operatorname{adj}(4 B)\|(\operatorname{adj}(A B))^{-1}\right)|A|\left|A^T\right| \\
& 3 \cdot 3 \frac{1}{64 \cdot 9}(64)^2 \cdot 4 \cdot \frac{1}{9 \cdot 4} 3 \cdot 3 \\
& =64
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
jlh0quVNcdwOiEkm | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | If the system of linear equations
<br/>$$x + 2ay + az = 0;$$ $$x + 3by + bz = 0;\,\,x + 4cy + cz = 0;$$
<br/>has a non - zero solution, then $$a, b, c$$. | [{"identifier": "A", "content": "satisfy $$a+2b+3c=0$$"}, {"identifier": "B", "content": "are in A.P"}, {"identifier": "C", "content": "are in G.P"}, {"identifier": "D", "content": "are in H.P."}] | ["D"] | null | For homogeneous system of equations to have non zero solution, $$\Delta = 0$$
<br><br>$$\left| {\matrix{
1 & {2a} & a \cr
1 & {3b} & b \cr
1 & {4c} & c \cr
} } \right| = 0\,{C_2} \to {C_2} - 2{C_3}$$
<br><br>$$\left| {\matrix{
1 & 0 & a \cr
1 & b & b \cr
1 & {2c} & c \cr
} } \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}$$
<br><br>$$\left| {\matrix{
1 & 0 & a \cr
0 & b & {b - a} \cr
0 & {2c - b} & {c - b} \cr
} } \right| = 0$$
<br><br>$$b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0$$
<br><br>On simplification, $${2 \over b} = {1 \over a} + {1 \over c}$$
<br><br>$$\therefore$$ $$a,b,c$$ are in Harmonic Progression. | mcq | aieee-2003 |
58rNudkmSRYVl7JA | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | The system of equations
<br/><p>$$\matrix{
{\alpha \,x + y + z = \alpha - 1} \cr
{x + \alpha y + z = \alpha - 1} \cr
{x + y + \alpha \,z = \alpha - 1} \cr
} $$</p>
<p>has no solutions, if $$\alpha $$ is :</p> | [{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "either $$-2$$ or $$1$$ "}, {"identifier": "C", "content": "not $$-2$$ "}, {"identifier": "D", "content": "$$1$$"}] | ["A"] | null | $$ax + y + z = \alpha - 1$$
<br><br>$$x + \alpha \,y + z = \alpha - 1;$$
<br><br>$$x + y + z\alpha = \alpha - 1$$
<br><br>$$\Delta = \left| {\matrix{
\alpha & 1 & 1 \cr
1 & \alpha & 1 \cr
1 & 1 & \alpha \cr
} } \right|$$
<br><br>$$ = \alpha \left( {{\alpha ^2} - 1} \right) - 1\left( {\alpha - 1} \right) + 1\left( {1 - \alpha } \right)$$
<br><br>$$ = \alpha \left( {\alpha - 1} \right)\left( {\alpha + 1} \right) - 1\left( {\alpha - 1} \right) - 1\left( {\alpha - 1} \right)$$
<br><br>For infinite solutions, $$\Delta = 0$$
<br><br>$$ \Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 1 - 1} \right] = 0$$
<br><br>$$ \Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 2} \right] = 0$$
<br><br>$$ \Rightarrow \alpha = - 2,1;$$
<br><br>But $$\alpha \ne 1.\,\,\,$$ $$\therefore$$ $$\,\,\alpha = - 2$$ | mcq | aieee-2005 |
KZ9FgIlN2I2zF7L0 | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | Let $$a, b, c$$ be any real numbers. Suppose that there are real numbers $$x, y, z$$ not all zero such that $$x=cy+bz,$$ $$y=az+cx,$$ and $$z=bx+ay.$$ Then $${a^2} + {b^2} + {c^2} + 2abc$$ is equal to : | [{"identifier": "A", "content": "$$2$$ "}, {"identifier": "B", "content": "$$-1$$ "}, {"identifier": "C", "content": "$$0$$ "}, {"identifier": "D", "content": "$$1$$ "}] | ["D"] | null | The given equations are
<br><br>$$\matrix{
{ - x + cy + bz = 0} \cr
{cx - y + az = 0} \cr
{bx + ay - z = 0} \cr
} $$
<br><br>As $$x,y,z$$ are not all zero
<br><br>$$\therefore$$ The above system should not have unique (zero) solution
<br><br>$$ \Rightarrow \Delta = 0 \Rightarrow \left| {\matrix{
{ - 1} & c & b \cr
c & { - 1} & a \cr
b & a & { - 1} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$$
<br><br>$$ \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$$
<br><br>$$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$$
| mcq | aieee-2008 |
xbDxrjEriuFOr9Rq | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | Consider the system of linear equations;
$$$\matrix{
{{x_1} + 2{x_2} + {x_3} = 3} \cr
{2{x_1} + 3{x_2} + {x_3} = 3} \cr
{3{x_1} + 5{x_2} + 2{x_3} = 1} \cr
} $$$
<br/>The system has : | [{"identifier": "A", "content": "exactly $$3$$ solutions "}, {"identifier": "B", "content": "a unique solution "}, {"identifier": "C", "content": "no solution "}, {"identifier": "D", "content": "infinitenumber of solutions "}] | ["C"] | null | $$D = \left| {\matrix{
1 & 2 & 1 \cr
2 & 3 & 1 \cr
3 & 5 & 2 \cr
} } \right| = 0$$
<br><br>$${D_1}\left| {\matrix{
3 & 2 & 1 \cr
3 & 3 & 1 \cr
1 & 5 & 2 \cr
} } \right| \ne 0$$
<br><br>$$ \Rightarrow $$ Given system, does not have any solution.
<br><br>$$ \Rightarrow $$ No solution | mcq | aieee-2010 |
XGQveL8rrGXCZ8eT | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | The number of values of $$k$$ for which the linear equations
<br/>$$4x + ky + 2z = 0,kx + 4y + z = 0$$ and $$2x+2y+z=0$$ possess a non-zero solution is : | [{"identifier": "A", "content": "$$2$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "zero"}, {"identifier": "D", "content": "$$3$$ "}] | ["A"] | null | $$\Delta = 0 \Rightarrow \left| {\matrix{
4 & k & 2 \cr
k & 4 & 1 \cr
2 & 2 & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) + $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0$$
<br><br>$$ \Rightarrow 8 - {k^2} + 2k + 4k - 16 = 0$$
<br><br>$$ \Rightarrow {k^2} - 6k + 8 = 0$$
<br><br>$$ \Rightarrow \left( {k - 4} \right)\left( {k - 2} \right) = 0,k = 4,2$$ | mcq | aieee-2011 |
DNXywk8s38SOeEaA | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | The number of values of $$k$$, for which the system of equations : $$$\matrix{
{\left( {k + 1} \right)x + 8y = 4k} \cr
{kx + \left( {k + 3} \right)y = 3k - 1} \cr
} $$$
<br/>has no solution, is <br/> | [{"identifier": "A", "content": "infinite "}, {"identifier": "B", "content": "1 "}, {"identifier": "C", "content": "2 "}, {"identifier": "D", "content": "3"}] | ["B"] | null | From the given system, we have
<br><br>$${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$$
<br><br>( as System has no solution)
<br><br>$$ \Rightarrow {k^2} + 4k + 3 = 8k$$
<br><br>$$ \Rightarrow k = 1,3$$
<br><br>If $$k = 1$$ then $${8 \over {1 + 3}} \ne {{4.1} \over 2}$$ which is false
<br><br>And if $$k = 3$$
<br><br>Then $${8 \over 6} \ne {{4.3} \over {9 - 1}}$$ which is true, therefore $$k=3$$
<br><br>Hence for only one value of $$k.$$ System has no solution. | mcq | jee-main-2013-offline |
brxKL055e9VwbMXT | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | The set of all values of $$\lambda $$ for which the system of linear equations:<br/><br/>
$$\matrix{
{2{x_1} - 2{x_2} + {x_3} = \lambda {x_1}} \cr
{2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr
{ - {x_1} + 2{x_2} = \lambda {x_3}} \cr
} $$<br/><br/>
has a non-trivial solution | [{"identifier": "A", "content": "contains two elements "}, {"identifier": "B", "content": "contains more than two elements "}, {"identifier": "C", "content": "in an empty set "}, {"identifier": "D", "content": "is a singleton"}] | ["A"] | null | $$\left. {\matrix{
{2{x_1} - 2{x_2} + {x^3} = \lambda {x_1}} \cr
{2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr
{\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} = \lambda {x_3}} \cr
} } \right\}$$
<br><br>$$\eqalign{
& \Rightarrow \,\,\,\,\,\,\,\left( {2 - \lambda } \right){x_1} - 2{x_2} + {x_3} = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,2{x_1} - \left( {3 + \lambda } \right){x_2} + 2{x_3} = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} - \lambda {x_3} = 0 \cr} $$
<br><br>For non-trivial solution, $$\Delta = 0$$
<br><br>i.e. $$\,\,\,\left| {\matrix{
{2 - \lambda } & { - 2} & 1 \cr
2 & { - \left( {3 + \lambda } \right)} & 2 \cr
{ - 1} & 2 & { - \lambda } \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \left( {2 - \lambda } \right)\left[ {\lambda \left( {3 + \lambda } \right) - 4} \right] + $$
<br><br>$$\,\,\,\,\,\,\,\,\,2\left[ { - 2\lambda + 2} \right] + 1\left[ {4 - \left( {3 + \lambda } \right)} \right] = 0$$
<br><br>$$ \Rightarrow {\lambda ^3} + {\lambda ^2} - 5\lambda + 3 = 0$$
<br><br>$$ \Rightarrow \lambda = 1,1,3$$
<br><br>Hence, $$\lambda $$ has $$2$$ values. | mcq | jee-main-2015-offline |
cZjTe4dWg3qrDPLQ | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | <p>The system of linear equations </p>
<p>$$\matrix{
{x + \lambda y - z = 0} \cr
{\lambda x - y - z = 0} \cr
{x + y - \lambda z = 0} \cr
} $$ </p>
has a non-trivial solution for : | [{"identifier": "A", "content": "infinitely many values of $$\\lambda .$$ "}, {"identifier": "B", "content": "exactly one value of $$\\lambda .$$ "}, {"identifier": "C", "content": "exactly two values of $$\\lambda .$$ "}, {"identifier": "D", "content": "exactly three values of $$\\lambda .$$ "}] | ["D"] | null | <p>For non-trivial solution, we have</p>
<p>$$\left| {\matrix{
1 & \lambda & { - 1} \cr
\lambda & { - 1} & { - 1} \cr
1 & 1 & { - \lambda } \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow 1(\lambda + 1) - \lambda ( - {\lambda ^2} + 1) - 1(\lambda + 1) = 0$$</p>
<p>$$ \Rightarrow \lambda ({\lambda ^2} - 1) = 0$$</p>
<p>$$ \Rightarrow \lambda = - 1,0,1$$</p> | mcq | jee-main-2016-offline |
1JH2i525goy0oScG | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | If S is the set of distinct values of 'b' for which the following system of linear equations
<br/><br/>x + y + z = 1
<br/>x + ay + z = 1
<br/>ax + by + z = 0
<br/><br/>has no solution, then S is : | [{"identifier": "A", "content": "an empty set"}, {"identifier": "B", "content": "an infinite set"}, {"identifier": "C", "content": "a finite set containing two or more elements"}, {"identifier": "D", "content": "a singleton "}] | ["D"] | null | $$\left| {\matrix{
1 & 1 & 1 \cr
1 & a & 1 \cr
a & b & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow $$ 1 [a – b] – 1 [1 – a] + 1 [b – a<sup>2</sup>] = 0
<br><br>$$ \Rightarrow $$ (a - 1)<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ a = 1
<br><br>For a = 1, the equations become
<br><br>x + y + z = 1
<br><br>x + y + z = 1
<br><br>x + by + z = 0
<br><br>These equations give no solution for b = 1
<br><br>$$ \Rightarrow $$ S is singleton set. | mcq | jee-main-2017-offline |
xcycrqfea8aJoM29XdHeA | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | The number of real values of $$\lambda $$ for which the system of linear equations
<br/><br/>2x + 4y $$-$$ $$\lambda $$z = 0
<br/><br/>4x + $$\lambda $$y + 2z = 0
<br/><br/>$$\lambda $$x + 2y + 2z = 0
<br/><br/>has infinitely many solutions, is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <p>The system of equations can be written in the matrix form as</p>
<p>$$\left[ {\matrix{
2 & 4 & { - \lambda } \cr
4 & \lambda & 2 \cr
\lambda & 2 & 2 \cr
} } \right]\left[ {\matrix{
x \cr
y \cr
z \cr
} } \right] = \left[ {\matrix{
0 \cr
0 \cr
0 \cr
} } \right]$$</p>
<p>The system has infinite solutions; thus, we get</p>
<p>$$\left| {\matrix{
2 & 4 & { - \lambda } \cr
4 & \lambda & 2 \cr
\lambda & 2 & 2 \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow 0 = 2(2\lambda - 4) - 4(8 - 2\lambda ) - \lambda (8 - {\lambda ^2})$$</p>
<p>$$ \Rightarrow 4\lambda - 8 - 32 + 8\lambda - 8\lambda + {\lambda ^3} = 0$$</p>
<p>$$ \Rightarrow {\lambda ^3} + 4\lambda - 40 = 0$$</p>
<p>We can solve this by graphical method:</p>
<p>$$y = {x^3} + 4x - 40$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3bip55r/c48ed0bf-90ff-469a-a2d8-184046732c50/02c6fef0-d6a0-11ec-a354-cb09522b689a/file-1l3bip55s.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3bip55r/c48ed0bf-90ff-469a-a2d8-184046732c50/02c6fef0-d6a0-11ec-a354-cb09522b689a/file-1l3bip55s.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2017 (Online) 8th April Morning Slot Mathematics - Matrices and Determinants Question 263 English Explanation"></p>
<p>For x = 0, y = $$-$$40: If we take y = $$-$$40, then we have</p>
<p>$$ - 40 = {x^3} + 4x - 40$$</p>
<p>$$ \Rightarrow {x^3} + 4x = 0$$</p>
<p>$$ \Rightarrow x({x^2} + 4) = 0$$</p>
<p>$$ \Rightarrow x = 0,{x^2} + 4 = 0$$</p>
<p>$$ \Rightarrow x = \pm \,2i$$</p>
<p>The given equation of line intersects x only at one point; therefore, the real value of $$\lambda$$ is only one.</p> | mcq | jee-main-2017-online-8th-april-morning-slot |
jz7BmhJjLuRJkhaJ | maths | matrices-and-determinants | solutions-of-system-of-linear-equations-in-two-or-three-variables-using-determinants-and-matrices | If the system of linear equations
<br/><br/>x + ky + 3z = 0
<br/>3x + ky - 2z = 0
<br/>2x + 4y - 3z = 0
<br/><br/>has a non-zero solution (x, y, z), then $${{xz} \over {{y^2}}}$$ is equal to | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "-10"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "-30"}] | ["C"] | null | System of equations has non-zero solution when determinant of coefficient = 0.
<br><br>So, in this questions,
<br><br>$$\left| {\matrix{
1 & K & 3 \cr
3 & K & { - 2} \cr
2 & 4 & { - 3} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow \,\,\,\,$$ ($$-$$ 3K + 8) $$-$$ K ($$-$$9 + 4) + 3(12 $$-$$ 2K) = 0
<br><br>$$ \Rightarrow \,\,\,\,$$ $$-$$ 3K + 8 + 9K $$-$$ 4K + 36 $$-$$ 6K = 0
<br><br>$$ \Rightarrow \,\,\,\,$$ $$-$$ 4K + 44 = 0
<br><br>$$ \Rightarrow \,\,\,\,$$ K = 11
<br><br>Now the equations become
<br><br>x + 11y + 3z = 0 . . . (1)
<br><br>3x + 11y $$-$$ 2z = 0 . . . (2)
<br><br>2x + 4y $$-$$ 3z = 0 . . . (3)
<br><br>By adding equation (1) and (3) we get,
<br><br>3x + 15y = 0
<br><br>$$ \Rightarrow \,\,\,\,$$ x = $$-$$ 5y
<br><br>Putting x = $$-$$ 5y in equation (1) we get
<br><br>$$-$$ 5y + 11y + 3z = 0
<br><br>$$ \Rightarrow \,\,\,\,$$ 6y + 3z = 0
<br><br>$$ \Rightarrow \,\,\,\,$$ z = $$-$$ 2y
<br><br>$$\therefore\,\,\,\,$$ $${{xz} \over {{y^2}}}$$
<br><br>$$ = {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}$$
<br><br>$$ = {{10{y^2}} \over {{y^2}}}$$
<br><br>$$ = 10$$ | mcq | jee-main-2018-offline |
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