question stringlengths 79 9.83k | answer stringlengths 33 9.39k |
|---|---|
If $$A = \left[ {\matrix{
a & b \cr
b & a \cr
} } \right]$$ and $${A^2} = \left[ {\matrix{
\alpha & \beta \cr
\beta & \alpha \cr
} } \right]$$, then
Options:
[{"identifier": "A", "content": "$$\\alpha = 2ab,\\,\\beta = {a^2} + {b^2}$$ "}, {"identifier": "B", "content": "$... | ["C"]
Explanation:
$${A^2} = \left[ {\matrix{
\alpha & \beta \cr
\beta & \alpha \cr
} } \right] = \left[ {\matrix{
a & b \cr
b & a \cr
} } \right]\left[ {\matrix{
a & b \cr
b & a \cr
} } \right]$$
<br><br>$$ = \left[ {\matrix{
{{a^2} + {b^2}} & {2a... |
Let A = $$\left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 0 \cr
1 & 1 & 1 \cr
} } \right]$$ and B = A<sup>20</sup>. Then the sum of the elements of the first column of B is :
Options:
[{"identifier": "A", "content": "210"}, {"identifier": "B", "content": "211"}, {"identifier": "C", "co... | ["C"]
Explanation:
A = $$\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... |
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 _______ .
Options:
[] | 10
Explanation:
$${A^2} = \left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right]\left[ {\matrix{
x & 1 \cr
1 & 0 \cr
} } \right] = \left[ {\matrix{
{{x^2} + 1} & x \cr
x & 1 \cr
} } \right]$$<br><br>$${A^4} = \left[ {\matrix{
{{x^2} + 1} & x \cr
x &... |
If $$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} $$ then which ... | ["D"]
Explanation:
$$ \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 ... |
If the matrix $$A = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right]$$ satisfies the equation<br/><br/> $${A^{20}} + \alpha {A^{19}} + \beta A = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 4 & 0 \cr
0 & 0 & 1 \cr
} } \ri... | 4
Explanation:
$${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 &a... |
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 ___________.
Options:
[] | 2020
Explanation:
$$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right],\,B = \left[ {\matrix{
\alpha \cr
\beta \cr
} } \right]$$<br><br>$$AB = B$$<br><br>$$\left[ {\matrix{
a & b \cr
c & d \cr
} } \right]\left[ {\matrix{
\alpha \cr
\beta \cr
} } \ri... |
Let $$A = \left[ {\matrix{
0 & 1 & 0 \cr
1 & 0 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$. Then the number of 3 $$\times$$ 3 matrices B with entries from the set {1, 2, 3, 4, 5} and satisfying AB = BA is ____________.
Options:
[] | 3125
Explanation:
Let matrix $$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{
... |
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 :
Options:
[{"identifier": "A", "content": "A<sup>6</sup> $$-$$ A"}, {"identifier": "B", "content": "A<sup>5</sup>"}, {"identifier": "C",... | ["A"]
Explanation:
$$A = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right] \Rightarrow {A^2} = \left[ {\matrix{
1 & 0 & 0 \cr
1 & 1 & 1 \cr
1 & 0 & 0 \cr
} } \right]$$<br><br>$${A^3} = \left[ {\matrix{
1 & 0 &... |
<p>Let $$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$$ {1, 2, ......, 100} : A<sup>n</sup> = A } is ____________.</p>
Options:
[] | 25
Explanation:
<p>$$\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} ... |
<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>
Options:
[{... | ["D"]
Explanation:
<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>$$ = \left[ {\matrix{
{{9^2} + {{12}^2} - {{15}^2}}... |
<p>Let $$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$$ is equal to</p>
Options:
[{"identifier": "A", "content": "$$-$$10"}, {"identifier": "B", "content": "$$-$$6"}, {"identifier": "C", "conte... | ["D"]
Explanation:
<p>$${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 } ... |
<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 $$(\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+\left[\begin{array}{l... | 2
Explanation:
<p>$${(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 & \al... |
<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>
Options:
[] | 10
Explanation:
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 ... |
<p>If $$A = {1 \over 2}\left[ {\matrix{
1 & {\sqrt 3 } \cr
{ - \sqrt 3 } & 1 \cr
} } \right]$$, then :</p>
Options:
[{"identifier": "A", "content": "$$\\mathrm{A^{30}-A^{25}=2I}$$"}, {"identifier": "B", "content": "$$\\mathrm{A^{30}+A^{25}-A=I}$$"}, {"identifier": "C", "content": "$$\\mathrm{A^{30}... | ["B"]
Explanation:
$$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}
\c... |
<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 $$\sum_\limits{n=1}^{50} B^{n}$$... | ["C"]
Explanation:
$$
\begin{aligned}
& \text { Let } C=\left[\begin{array}{cc}
1 & 2 \\
-1 & -1
\end{array}\right], \mathrm{D}=\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\right] \\\\
& \mathrm{DC}=\left[\begin{array}{cc}
1 & 2 \\
-1 & -1
\end{array}\right]\left[\begin{array}{cc}
-1 & -2 \\
1 & 1
\end{array}\... |
Let $A=I_2-2 M M^T$, where $M$ is a real matrix of order $2 \times 1$ such that the relation $M^T M=I_1$ holds. If $\lambda$ is a real number such that the relation $A X=\lambda X$ holds for some non-zero real matrix $X$ of order $2 \times 1$, then the sum of squares of all possible values of $\lambda$ is equal to ____... | 2
Explanation:
$\begin{aligned} & A=I_2-2 M^T \\\\ & A^2=\left(I_2-2 M M^T\right)\left(I_2-2 M^T\right) \\\\ & =I_2-2 M^T-2 M M^T+4 M^T M^T \\\\ & =I_2-4 M M^T+4 M M^T \\\\ & =I_2\end{aligned}$
<br/><br/>$\begin{aligned} & \mathrm{AX}=\lambda \mathrm{X} \\\\ & \mathrm{A}^2 \mathrm{X}=\lambda \mathrm{AX} \\\\ & \mathrm... |
Let $A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right], B=\left[B_1, B_2, B_3\right]$, where $B_1, B_2, B_3$ are column matrics, and
<br/><br/>$$
\mathrm{AB}_1=\left[\begin{array}{l}
1 \\
0 \\
0
\end{array}\right], \mathrm{AB}_2=\left[\begin{array}{l}
2 \\
3 \\
0
\e... | 28
Explanation:
<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>$$\mathrm{B}_1=\left[\begin{array}{l}
\mathrm{x}_1 \\
\mathrm{y}_1 \\
\mathrm{z}_1
\end{array}\right], \quad \mathrm{B}_2=\le... |
Let $$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.$$ Then
Options:
[{"identifier": "A", "content": "there cannot exist any $$B$$ such that $$AB=BA$$ "}, {"identifier": "B", "content": "there exist m... | ["D"]
Explanation:
$$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 \c... |
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?
Options:
[{"identifier": "A", "content": "$$A=B$$ "}, {"identifier": "B", "content": "$$AB=BA$$ "}, {"identifier": "C", "c... | ["B"]
Explanation:
$${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$$
<br><br>$${A^2} - {B^2} = {A^2} + AB - BA - {B^2}$$
<br><br>$$ \Rightarrow AB = BA$$ |
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 :
Options:
[{"i... | ["D"]
Explanation:
$$P = \left[ {\matrix{
1 & 0 & 0 \cr
3 & 1 & 0 \cr
9 & 3 & 1 \cr
} } \right]$$
<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>$${P^3} = \left[ {... |
Let $$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 $$ is
Options:
[{"identifie... | ["D"]
Explanation:
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265941/exam_images/vlkqlnh7isvrysclmlsm.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263666/exam_images/rf4qtzmkrumvqo5bw5x9.webp"><so... |
Let $$\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]$$<br/><br/> then the matrix
A<sup>31</sup> is equal... | ["D"]
Explanation:
x<sup>2</sup> + x + 1 = 0
<br><br>$$ \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 &... |
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 :
Options:
[{"identifier": "A", "content": "6I \u2013 A"}, {"identifier": "B", "content": "4I \u2013 A"}, {"identifier": "C"... | ["C"]
Explanation:
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$$... |
Let $$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 \c... | 1
Explanation:
$$XB = A$$
<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>$$ \Rightarrow $$ $${1 \over ... |
Let $$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$$ 3. If B = [b<sub>ij</sub>], then b<sub>13</sub>is equal to _____________.
Options... | 910
Explanation:
Let $$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 &a... |
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 :
Options:
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier... | ["C"]
Explanation:
$$A = \left[ {\matrix{
{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right]$$<br><br>Let $$x = \left[ {\matrix{
1 \cr
1 \cr
1 \cr
} } \right]$$<br><br>$$AX = \l... |
If $$A = \left[ {\matrix{
1 & 1 & 1 \cr
0 & 1 & 1 \cr
0 & 0 & 1 \cr
} } \right]$$ and M = A + A<sup>2</sup> + A<sup>3</sup> + ....... + A<sup>20</sup>, then the sum of all the elements of the matrix M is equal to _____________.
Options:
[] | 2020
Explanation:
$${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... |
If $$P = \left[ {\matrix{
1 & 0 \cr
{{1 \over 2}} & 1 \cr
} } \right]$$, then P<sup>50</sup> is :
Options:
[{"identifier": "A", "content": "$$\\left[ {\\matrix{\n 1 & 0 \\cr \n {25} & 1 \\cr \n\n } } \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {\\matrix{\n 1 & {50} ... | ["A"]
Explanation:
$$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 ... |
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 ... | ["B"]
Explanation:
$$A{A^T} = \left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr
{{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr
} } \right)\left( {\matrix{
{{1 \over {\sqrt 5 }}} & {{{ - 2} \over {\sqrt 5 }}} \cr
{{2 \over {\sqrt 5 }}} & {{1 \over {\sqrt ... |
If the matrix $$A = \left( {\matrix{
0 & 2 \cr
K & { - 1} \cr
} } \right)$$ satisfies $$A({A^3} + 3I) = 2I$$, then the value of K is :
Options:
[{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$$-$$$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$-$$1"}, {"i... | ["A"]
Explanation:
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>$$ \Rightarrow \left| {\matrix{
{0 - \lambda } & 2 \... |
The number of elements in the set $$\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$$ 2 identity matrix, is :
Options:
[] | 8
Explanation:
$${(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]$$<br><br>$$... |
<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>
Options:
[{"identifier": "A", "content": "$$\\left( {{{{3^{10}} - 3} \\over 2}} \\right)A$$"}, {"identifier": ... | ["A"]
Explanation:
<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>$$ = \left[ {\matrix{
1 & 2 & 4 \cr
{{1 \over 2}} & 1 & 2 \cr
{{1 ... |
<p>Let $$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$$ is ____________.</p>
Options:
[] | 1
Explanation:
$M=\left[\begin{array}{cc}0 & -\alpha \\ \alpha & 0\end{array}\right], M^{2}=\left[\begin{array}{cc}-\alpha^{2} & 0 \\ 0 & -\alpha^{2}\end{array}\right]=-\alpha^{2}$ I
<br/><br/>
$N=M^{2}+M^{4}+\ldots+M^{98}$
<br/><br/>
$=\left[-\alpha^{2}+\alpha^{4}-\alpha^{6}+\ldots\right] I$
<br/><br/>
$=\frac{-\alph... |
<p>Let $$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$$ {1, 2, .........., 10} and nA<sup>n</sup> + mB<sup>m</sup> = I} is _____... | 1
Explanation:
<p>$${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[ {\m... |
<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>
Options:
[] | 100
Explanation:
$$
\begin{aligned}
&\mathrm{A}=\left[\begin{array}{cc}
-1 & \mathrm{a} \\\\
0 & \mathrm{~b}
\end{array}\right] \\\\
&\mathrm{A}^2=\left[\begin{array}{cc}
-1 & \mathrm{a} \\\\
0 & \mathrm{~b}
\end{array}\right]\left[\begin{array}{cc}
-1 & \mathrm{a} \\\\
0 & \mathrm{~b}
\end{array}\right] \\\\
&=\left[... |
<p>Let $$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\}$$ is equal to ____________... | 17
Explanation:
Here $A=\left(\begin{array}{ccc}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{array}\right)$
<br/><br/>
We get $A^{2}=A$ and similarly for
<br/><br/>
$$
B=A-I=\left[\begin{array}{lll}
1 & -1 & -1 \\
1 & -1 & -1 \\
1 & -1 & -1
\end{array}\right]
$$
<br/><br/>
We get $B^{2}=-B \Rightarrow B^{3}=B$
<br/><br... |
<p>Which of the following matrices can NOT be obtained from the matrix $$\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$$ by a single elementary row operation ?</p>
Options:
[{"identifier": "A", "content": "$$\\left[\\begin{array}{cc}0 & 1 \\\\ 1 & -1\\end{array}\\right]$$"}, {"identifier": "B", "co... | ["C"]
Explanation:
<p>Given matrix $$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$$ Option A can be obtained.</p>
<p>For option B... |
<p>Let $$\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$$, then</p>
Options:
[{"identifier": "A", "content": "$$\\alpha=1$$"}, {"identifier": "B", "content": "$$\\alpha=4$$"}, {"identifier": "C", "content": "$$\\beta=8$$"}, {"identifier": ... | ["D"]
Explanation:
$\mathrm{A}^{2}=3 \mathrm{~A}+\alpha \mathrm{I}$
<br/><br/>
$A^{3}=3 A^{2}+\alpha A$
<br/><br/>
$\mathrm{A}^{3}=3(3 \mathrm{~A}+\alpha \mathrm{I})+\alpha \mathrm{A}$
<br/><br/>
$\mathrm{A}^{3}=9 \mathrm{~A}+\alpha \mathrm{A}+3 \alpha \mathrm{I}$
<br/><br/>
$\mathrm{A}^{4}=(9+\alpha) \mathrm{A}^{2}+3... |
<p>If A and B are two non-zero n $$\times$$ n matrices such that $$\mathrm{A^2+B=A^2B}$$, then :</p>
Options:
[{"identifier": "A", "content": "$$\\mathrm{A^2B=I}$$"}, {"identifier": "B", "content": "$$\\mathrm{A^2=I}$$ or $$\\mathrm{B=I}$$"}, {"identifier": "C", "content": "$$\\mathrm{A^2B=BA^2}$$"}, {"identifier": "D... | ["C"]
Explanation:
Given : $A^{2}+B=A^{2} B\quad...(i)$
<br/><br/>
$\Rightarrow A^{2}+B-I=A^{2} B-I$
<br/><br/>
$\Rightarrow A^{2} B-A^{2}-B+I=I$
<br/><br/>
$\Rightarrow A^{2}(B-I)-I(B-I)=I$
<br/><br/>
$\Rightarrow\left(A^{2}-I\right)(B-I)=I$
<br/><br/>
$\therefore A^{2}-I$ is the inverse matrix of $B-I$ and vice vers... |
<p>Let $$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$$ is equal to ___________.</p>
Options:
[] | 2
Explanation:
$$
\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 &... |
<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... | ["A"]
Explanation:
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/>$$
\begin{aligned}
\therefore \quad A^2 & =\left[\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right]\left[\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array... |
<p>Let $$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$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "18"}, {"identifier": "B", ... | ["C"]
Explanation:
We have, $P^2=I-P$
<br/><br/>$$
\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}
$$
<br/><br/>an... |
<p>Let $$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|$$ is equal to</p>
Options:
[... | ["D"]
Explanation:
<p>$$\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}| \\
& =|\... |
<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>
Options:
[{"identifier": "A", "content": "$$-10$$\n"}, {"identifier": "B", "content":... | ["C"]
Explanation:
<p>$$\begin{aligned}
& |A-\lambda I|=0 \\
& \left|\begin{array}{ccc}
2-\lambda & a & 0 \\
1 & 3-\lambda & 1 \\
0 & 5 & b-\lambda
\end{array}\right|=0 \\
& (2-\lambda)[(3-\lambda)(b-\lambda)-5]-a[b-\lambda-0]+0=0 \\
& (2-\lambda)\left[3 b-3 \lambda-b \lambda+\lambda^2-5\right]-a b+a \lambda=0 \\
& \l... |
Let $$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|$$ equals
Options:
[{"identifier": "A", "content": "$$1/5$$ "}, {"identifier": "B", "content... | ["A"]
Explanation:
$$\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}$$ |
Let $$A$$ be a square matrix all of whose entries are integers.
<br/>Then which one of the following is true?
Options:
[{"identifier": "A", "content": "If det $$A = \\pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are not necessarily integers"}, {"identifier": "B", "content": "If det $$A \\ne \\pm 1,$$ the... | ["C"]
Explanation:
As all entries of square matrix $$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}}$$ are integers. |
Let $$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|$$
Options:
[{"identifier": "A", "content": "statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statemen... | ["D"]
Explanation:
We know that $$\left| {adj\left( {adj\,\,A} \right)} \right| = {\left| {Adj\,\,A} \right|^{2 - 1}}$$b
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left| A \right|^{2 - 1}} = \left| A \right|... |
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 :
Options:
[{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$0$$ "}, {"... | ["C"]
Explanation:
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><... |
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 $$A{u_2} = \left( {\matrix{
0 \cr
1 \cr
... | ["D"]
Explanation:
Let $$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 ... |
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 :
Options:
[{"identifier": "A", "content": "$$4$$ "}, {"identifier": "B", "co... | ["B"]
Explanation:
$$\left| P \right| = 1\left( {12 - 12} \right) - \alpha \left( {4 - 6} \right) + $$
<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>$$ \Rightarrow {\left| A ... |
Let A and B be two invertible matrices of order 3 $$ \times $$ 3. If det(ABA<sup>T</sup>) = 8 and det(AB<sup>–1</sup>) = 8,
<br/>then det (BA<sup>–1</sup> B<sup>T</sup>) is equal to :
Options:
[{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "$... | ["C"]
Explanation:
$${\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 ... |
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:
Options:
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$${1 \\over 3}$... | ["C"]
Explanation:
|B| = $$\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 ... |
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 :
Options:
[{"identifier": "A", "content": "8"}, {"identifier": "B", "cont... | ["A"]
Explanation:
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>= $${{\left| {adj\left( {adjA} \right)} \right|} \over {\left| {3A} \r... |
Let A be a 3 $$ \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 $$ , then the ordered pair,
<br... | ["C"]
Explanation:
$$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|... |
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 ____________.
Options:
[] | 7
Explanation:
$$A = \left[ {\matrix{
x & y & z \cr
y & z & x \cr
z & x & y \cr
} } \right]$$
<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>$$ \therefore $$ $${({x^3} + {y^3} + {z^... |
Let $$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 + ... | 36
Explanation:
$$|{P^{ - 1}}AP - I{|^2}$$<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>$$ =... |
If $$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$$ is :
Options:
[{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6}... | ["A"]
Explanation:
$${A^2} = \left[ {\matrix{
0 & {\sin \alpha } \cr
{\sin \alpha } & 0 \cr
} } \right]\left[ {\matrix{
0 & {\sin \alpha } \cr
{\sin \alpha } & 0 \cr
} } \right] = \left[ {\matrix{
{{{\sin }^2}\alpha } & 0 \cr
0 & {{{\sin }^2}\alpha } \cr
} } ... |
If $$A = \left[ {\matrix{
2 & 3 \cr
0 & { - 1} \cr
} } \right]$$, then the value of det(A<sup>4</sup>) + det(A<sup>10</sup> $$-$$ (Adj(2A))<sup>10</sup>) is equal to _____________.
Options:
[] | 16
Explanation:
$$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
} ... |
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 e... | 108
Explanation:
$$A = \left[ {\matrix{
2 & { - 1} & 1 \cr
{ - 1} & 2 & { - 1} \cr
1 & { - 1} & 2 \cr
} } \right]$$<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... |
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 _____________.
Opti... | 16
Explanation:
| 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 $$\times$$... |
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 :
Options:
[{"ident... | ["D"]
Explanation:
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<... |
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 __________.
Options:
[] | 4
Explanation:
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<s... |
<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 :
Options:
[{"identifier": "A", "content": "$${{ - 2b} \\over {b + 1}}$$"}, {"identifier": "B", "content": "$${{2b} \\over {b + 1}}$$"}, ... | ["D"]
Explanation:
$$\left| {{1 \over 2}\left| {\matrix{
a & 0 & 1 \cr
b & {2b + 1} & 1 \cr
0 & b & 1 \cr
} } \right|} \right| = 1$$<br><br>$$ \Rightarrow \left| {\matrix{
a & 0 & 1 \cr
b & {2b + 1} & 1 \cr
0 & b & 1 \cr
} } \right| = \... |
<p>Let $$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}$$, then the sum of all elements of the matrix B is</p>
Options:
[{"identifier": "A", "content": "$$-$$5"}, {"identifier": "B", "conten... | ["C"]
Explanation:
<p>Given $$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}$$</p>
<p>Cofactor... |
<p>Let A be a matrix of order 3 $$\times$$ 3 and det (A) = 2. Then det (det (A) adj (5 adj (A<sup>3</sup>))) is equal to _____________.</p>
Options:
[{"identifier": "A", "content": "512 $$\\times$$ 10<sup>6</sup>"}, {"identifier": "B", "content": "256 $$\\times$$ 10<sup>6</sup>"}, {"identifier": "C", "content": "1024 ... | ["A"]
Explanation:
<p>$$|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$$</p> |
<p>Let $$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$$, is</p>
Options:
[{"identifier": "A", "content": "117"}, {"iden... | ["C"]
Explanation:
<p>$$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$$</p>
<p>$$ \Rig... |
<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>
Options:
[{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "64"}, {"identifier": "D", "content": "128"}] | ["C"]
Explanation:
<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>$$ = {... |
<p>The positive value of the determinant of the matrix A, whose</p>
<p>Adj(Adj(A)) = $$\left( {\matrix{
{14} & {28} & { - 14} \cr
{ - 14} & {14} & {28} \cr
{28} & { - 14} & {14} \cr
} } \right)$$, is _____________.</p>
Options:
[] | 14
Explanation:
<p>$$\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 ... |
<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>
Options:
[{"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"]
Explanation:
<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>$$ \Rightarrow {\left( {{{(24)}^2}} \right)^2... |
<p>Let S = {$$\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$$ is eq... | ["B"]
Explanation:
<p>Given, $$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\}$$</p>
<p>W... |
<p>Let A and B be two square matrices of order 2. If $$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$$ N, then a + b + c is equal to :</p>
Options:
[{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "12"}, {"identifi... | ["B"]
Explanation:
<p>Given,</p>
<p>$$\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}$$</p>
... |
<p>Let A be a 2 $$\times$$ 2 matrix with det (A) = $$-$$ 1 and det ((A + I) (Adj (A) + I)) = 4. Then the sum of the diagonal elements of A can be :</p>
Options:
[{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$- \\sqr... | ["B"]
Explanation:
<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>$$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]\,adj\,A = \left[ {\matrix{
a &... |
<p>Let $$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})$$ is equal to _____________.</p>
Options:
[{"identifier": "A", "content": "$$-$$18"}, {"identifier": "B", "content": "18... | ["B"]
Explanation:
<p>Characteristic equation of A is given by</p>
<p>$$\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$$</p>
<... |
<p>Consider a matrix $$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}(\operatorna... | 42
Explanation:
<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>$$ \Rightarrow (\alpha + \beta + \gamma )\left| {... |
<p>Let the matrix $$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)$$ is equal to :</p>
Options:
[{"identifier": "... | ["C"]
Explanation:
<p>$$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
... |
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 :
Options:
[] | 5
Explanation:
$\because\left|\operatorname{adj}\left(2 \cdot \operatorname{adj}\left(2 A^{-1}\right)\right)\right|=2^{84}$
<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/><b... |
If $P$ is a $3 \times 3$ real matrix such that $P^T=a P+(a-1) I$, where $a>1$, then :
Options:
[{"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"]
Explanation:
<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>$$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{b_1}} & {{b_2}} & {{b_3}} \cr
{{c... |
<p>Let $$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}$$. Then </p>
Options:
[{"identifier": "A", "content": "$$1+\\mathrm{d}^{2}=\\mathrm{m}^{2}+\\mathrm{q}^{2}$$"}, {"identifier": "B", "content":... | ["D"]
Explanation:
<p>$$\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} = n... |
<p>The set of all values of $$\mathrm{t\in \mathbb{R}}$$, for which the matrix <br/><br/>$$\left[ {\matrix{
{{e^t}} & {{e^{ - t}}(\sin t - 2\cos t)} & {{e^{ - t}}( - 2\sin t - \cos t)} \cr
{{e^t}} & {{e^{ - t}}(2\sin t + \cos t)} & {{e^{ - t}}(\sin t - 2\cos t)} \cr
{{e^t}} & {{e^{ - t}... | ["B"]
Explanation:
If the matrix is invertible then its determinant
should not be zero.
<br/><br/>So, $$
\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
$$
<br/>... |
<p>Let $$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)|}$$ is equal to</p>
Options:
[{"identifier": "A", "content": "... | ["B"]
Explanation:
$$
\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|\operatorn... |
<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>
Options:
[{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "2$$\\sqrt3$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$\\sqrt6$$"}] | ["B"]
Explanation:
$|A|^{(n-1)^{3}}=12^{4}$
<br/><br/>
$$
\begin{aligned}
&|A|^{8}=12^{4} \\\\
&|A|=\sqrt{12} \\\\
&\left|A^{-1} \operatorname{adj} A\right|=\left|A^{-1}\right| \cdot|A|^{2} \\\\
&=|A| = 2\sqrt3
\end{aligned}
$$ |
<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 $${{{{(... | ["A"]
Explanation:
$$
\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}
$$<br/><br/>
It is singula... |
Let the determinant of a square matrix A of order $m$ be $m-n$, where $m$ and $n$<br/><br/> satisfy $4 m+n=22$ and $17 m+4 n=93$.<br/><br/> If $\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A)))=3^{a} 5^{b} 6^{c}$ then $a+b+c$ is equal to :
Options:
[{"identifier": "A", "content": "96"}, {"identifier": ... | ["A"]
Explanation:
Given that $|A|=m-n$, and let's solve the system of linear equations to find the values of $m$ and $n$ :
<br/><br/>$4m + n = 22$ ...... (1)
<br/><br/>$17m + 4n = 93$ ....... (2)
<br/><br/>We can multiply equation (1) by 4 to make the coefficients of $n$ in both equations equal:
<br/><br/>$16m + 4... |
<p>Let for $$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 $$ is equal to</p>
Options:
[{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "9"},... | ["A"]
Explanation:
$$
\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&... |
<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>
Options:
[{"identifier": "A", "content": "$$2^{12}$$"}, {"identifier": "B... | ["D"]
Explanation:
Given that
<br/><br/>$$
\begin{aligned}
& A=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll}
5 ! & 6 ! & 7 ! \\
6 ! & 7 ! & 8 ! \\
7 ! & 8 ! & 9 !
\end{array}\right] \\\\
& \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !}\left|\begin{array}{lll}
5 ! & 6 ! & 7 ! \\
6 ! & 7 ! & 8 ! \\
7 ! & 8 ! & 9 !
\end{array}\rig... |
<p>If A is a 3 $$\times$$ 3 matrix and $$|A| = 2$$, then $$|3\,adj\,(|3A|{A^2})|$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "$${3^{12}}\\,.\\,{6^{10}}$$"}, {"identifier": "B", "content": "$${3^{11}}\\,.\\,{6^{10}}$$"}, {"identifier": "C", "content": "$${3^{12}}\\,.\\,{6^{11}}$$"}, {"identifier": "D",... | ["B"]
Explanation:
Given that $A$ is $3 \times 3$ matrix and $|A|=2$
<br/><br/>$$
\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 \\\... |
<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>
Options:
[] | 27
Explanation:
<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>$$A\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]=3\left[\begin{array}{l}
1 \\
1 \\
1
\end{arr... |
<p>If $$\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... | ["D"]
Explanation:
<p>$$\left|\begin{array}{lll}
\alpha & b & c \\
a & \beta & c \\
a & b & \gamma
\end{array}\right|=0$$</p>
<p>$$\begin{aligned}
& R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\
& \Rightarrow\left|\begin{array}{ccc}
\alpha-a & b-\beta & 0 \\
0 & \beta-b & c-\gamma \\
a & b & \gamma
\end{array}\r... |
<p>Let A and B be two square matrices of order 3 such that $$\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}}|$$ is equal to :</p>
Options:
[{"ide... | ["C"]
Explanation:
<p>$$\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}$$<... |
If the system of linear equations
<br/>$$x + 2ay + az = 0;$$ $$x + 3by + bz = 0;\,\,x + 4cy + cz = 0;$$
<br/>has a non - zero solution, then $$a, b, c$$.
Options:
[{"identifier": "A", "content": "satisfy $$a+2b+3c=0$$"}, {"identifier": "B", "content": "are in A.P"}, {"identifier": "C", "content": "are in G.P"}, {"id... | ["D"]
Explanation:
For homogeneous system of equations to have non zero solution, $$\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... |
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>
Options:
[{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "eith... | ["A"]
Explanation:
$$ax + y + z = \alpha - 1$$
<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>$$ = \alpha \left( {{\... |
Let $$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$$ is equal to :
Options:
[{"identifier": "A", "content": "$$2$$ "}, {"identifier": "B", "content": "$$-1$$ "}, {"identifier": "C", "... | ["D"]
Explanation:
The given equations are
<br><br>$$\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... |
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 :
Options:
[{"identifier": "A", "content": "exactly $$3$$ solutions "}, {"identifier": "B", "content": "a unique solutio... | ["C"]
Explanation:
$$D = \left| {\matrix{
1 & 2 & 1 \cr
2 & 3 & 1 \cr
3 & 5 & 2 \cr
} } \right| = 0$$
<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, d... |
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 :
Options:
[{"identifier": "A", "content": "$$2$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "zero"}, {"identifier": "D", "content":... | ["A"]
Explanation:
$$\Delta = 0 \Rightarrow \left| {\matrix{
4 & k & 2 \cr
k & 4 & 1 \cr
2 & 2 & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) + $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0... |
The number of values of $$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
} $$$
<br/>has no solution, is <br/>
Options:
[{"identifier": "A", "content": "infinite "}, {"identifier": "B", "content": "1 "}, {"identifier"... | ["B"]
Explanation:
From the given system, we have
<br><br>$${{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}$$ which is ... |
The set of all values of $$\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
} $$<br/><br/>
has a non-trivial solution
Options:
[{"identifier": ... | ["A"]
Explanation:
$$\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_... |
<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 :
Options:
[{"identifier": "A", "content": "infinitely many values of $$\\lambda .$$ "}, {"identifier": "B", "content": "exa... | ["D"]
Explanation:
<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>$$ \Rightarrow \la... |
If S is the set of distinct values of 'b' for which the following system of linear equations
<br/><br/>x + y + z = 1
<br/>x + ay + z = 1
<br/>ax + by + z = 0
<br/><br/>has no solution, then S is :
Options:
[{"identifier": "A", "content": "an empty set"}, {"identifier": "B", "content": "an infinite set"}, {"identifier... | ["D"]
Explanation:
$$\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 $$ a = 1
<br><br>For a = 1, the... |
The number of real values of $$\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 $$x + 2y + 2z = 0
<br/><br/>has infinitely many solutions, is :
Options:
[{"identifier": "A", "content": "0"}, {"identifier": "B", "con... | ["B"]
Explanation:
<p>The system of equations can be written in the matrix form as</p>
<p>$$\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 \... |
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
Options:
[{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "-10"}, {"identifier": "C", "content": "10"}, {"i... | ["C"]
Explanation:
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) $$... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.