Matrices

Factor pairs of 36:

$(1,36), (2,18), (3,12), (4,9), (6,6), (9,4), (12,3), (18,2), (36,1)$

Number of possible orders = 9.

Given: $a_{ij}=\max (i,j)-\min (i,j)$

So,

$$\begin{gathered} a_{11}=\max (1,1)-\min (1,1)=1-1=0, \\ {a}_{12}=\max (1,2)-\min (1,2)=2-1=1, \\ {a}_{21}=\max (2,1)-\min (2,1)=2-1=1, \\ {a}_{22}=\max (2,2)-\min (2,2)=2-2=0 \end{gathered}$$

The matrix $A$ is: $A=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]$

Hence,

$$\begin{gathered} A^2=A\times A \\ =\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\times \left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Given: \(\left[\begin{array}{cc}x+y & 2 \\ 5 & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]\)

Corresponding elements must be equal,

$x+y=6$, $xy=8$
Now, $\frac{24}{x}+\frac{24}{y}=24\left(\frac{1}{x}+\frac{1}{y}\right)$
$$\begin{gathered} =24\left(\frac{x+y}{xy}\right) \\ =24\times \frac{6}{8} \\ =18 \end{gathered}$$

From equality of matrices:

$x+y=6$
$x y=8$
$\frac{24}{x}+\frac{24}{y}=24\left(\frac{1}{x}+\frac{1}{y}\right)$
$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$
$\frac{x+y}{xy}=\frac{6}{8}=\frac{3}{4}$
$\Rightarrow 24\left(\frac{1}{x}+\frac{1}{y}\right)=24·\frac{3}{4}=18$

Let the scalar matrix be \(k I_3 \Rightarrow\) diagonal entries \(k\), others \(0\).
Sum of all elements \(=3 k=9 \Rightarrow k=3\)
Product of all \(9\) elements \(=0\) (six zeros present)

$\left|\mathrm{A}\right|=3,| \mathrm{B}|=2$

$\left|{\mathrm{A}}^{\mathrm{T}}\mathrm{A}(adj(2 \mathrm{A}\left){)}^{-1}\right(adj(4 \mathrm{B})\left)\right(adj\left(\mathrm{AB}\right){)}^{-1}{\mathrm{AA}}^{\mathrm{T}}\right|$

$=3\times 3\times \mid \left(adj(2 \mathrm{A}{)}^{-1}\left|\times |adj(4 \mathrm{B}\left)\right|\times \left|(adj(\mathrm{AB}){)}^{-1}\right|\times 3\times 3\right.\right.$

$\downarrow$ $\downarrow$ $\downarrow$

$\frac{1}{|adj(2 \mathrm{A}\left)\right|}$ $2^{12}\times 2^2$ $\frac{1}{|adj(\mathrm{AB}\left)\right|}$

$=\frac{1}{2^6|adj \mathrm{A}|} =\frac{1}{|adj \mathrm{B}·adj \mathrm{A}|}$

$=\frac{1}{2^6·3^2} =\frac{1}{2^2·3^2}$

$=3^4·\frac{1}{2^6·3^2}·2^{12}·2^2·\frac{1}{2^2·3^2}=64$

$F\left(x\right)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{cc}R\left(x\right)& 0\\ 0& 1\end{array}\right]$,

$R\left(x\right)=\left[\begin{array}{cc}\cos x& -\sin x\\ \sin x& \cos x\end{array}\right]$

${\left[F\left(x\right)\right]}^2=\left[\begin{array}{cc}R(x{)}^2& 0\\ 0& 1\end{array}\right]$

$R{\left(x\right)}^2=R\left(2x\right)=\left[\begin{array}{cc}\cos 2x& -\sin 2x\\ \sin 2x& \cos 2x\end{array}\right]$

$\Rightarrow {\left[F\left(x\right)\right]}^2=\left[\begin{array}{ccc}\cos 2x& -\sin 2x& 0\\ \sin 2x& \cos 2x& 0\\ 0& 0& 1\end{array}\right]=F\left(2x\right)$

Given: \({[\mathrm{F}(\mathrm{x})]^2=\mathrm{F}(\mathrm{kx})} \)
\( {\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\cos k x & -\sin k x & 0 \\
\sin k x & \cos k x & 0 \\
0 & 0 & 1
\end{array}\right]} \)
\(\Rightarrow {\left[\begin{array}{ccc}
\cos ^2 x-\sin ^2 x & -\cos x \sin x-\sin x \cos x & 0 \\ \sin x \cos x+\cos x \sin x & -\sin ^2 x+\cos ^2 x & 0 \\ 0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\cos k x & -\sin k x & 0 \\
\sin k x & \cos k x & 0 \\
0 & 0 & 1
\end{array}\right]} \)
\(\Rightarrow {\left[\begin{array}{ccc}
\cos 2 x & -2 \sin x \cos x & 0 \\
2 \sin x \cos x & \cos 2 x & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\cos k x & -\sin k x & 0 \\
\sin k x & \cos k x & 0 \\
0 & 0 & 1
\end{array}\right]}\)

\((\because \cos 2 x=\cos ^2 x-\sin ^2 x\))

\(\Rightarrow\left[\begin{array}{ccc}
\cos 2 x & -\sin 2 x & 0 \\
\sin 2 x & \cos 2 x & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\cos k x & -\sin k x & 0 \\
\sin k x & \cos k x & 0 \\
0 & 0 & 1
\end{array}\right]\)

\((\because \sin 2 x=2 \sin x \cos x)\)
The above matrix equation, we get \(\cos 2 x=\cos k x\) and \(\sin 2 x=\sin k x\)

\( \Rightarrow 2 x=k x \)
\( \Rightarrow k=2 \)

Factor pairs of \(36\):

$(1,36), (2,18), (3,12), (4,9), (6,6), (9,4), (12,3), (18,2), (36,1)$

Number of possible orders \(=9\).

Let the scalar matrix be \(k I_3 \Rightarrow\) diagonal entries \(k\), others 0
Sum of all elements \(=3 k=9 \Rightarrow k=3\)
Product of all 9 elements \(=0\) (six zeros present)

According to question,

\(a_{i j}=i-3 j, i, j \in\{1,2,3\}\)

(A) \(a_{11}=1-3=-2<0 \Rightarrow\) A true

(B) \(a_{12}=1-6=-5, a_{21}=2-3=-1\)

\(\Rightarrow a_{12}+a_{21}=-6 \Rightarrow \mathrm{~B}\) true

(C) \(a_{13}=1-9=-8, a_{31}=3-3=0\)

\(\Rightarrow a_{13}=-8<0=a_{31} \Rightarrow\) C false

(D) \(a_{31}=0 \Rightarrow \mathrm{D}\) true

For symmetric matrix :

\(A^{\prime}=A\) and

For skew symmetric matrix :

\(A^{\prime}=-A\)

As per the condition

\(\Rightarrow A=-A\)

\(\Rightarrow 2 A=O \Rightarrow A=O\)

So, \(A\) is a null matrix.

Given: $\left[\begin{array}{lll}x& 2& 0\end{array}\right]\left[\begin{array}{c}5\\ -1\\ x\end{array}\right]=\left[\begin{array}{ll}3& 1\end{array}\right]\left[\begin{array}{c}-2\\ x\end{array}\right]$

$$\begin{gathered} \Rightarrow \left[5x-2+0\right]=\left[-6+x\right] \\ \Rightarrow 5x-2=x-6 \\ \Rightarrow 4x=-4 \\ \Rightarrow x=-1 \end{gathered}$$

$$\begin{gathered} \left[\begin{array}{lll}x& 2& 0\end{array}\right]\left[\begin{array}{c}5\\ -1\\ x\end{array}\right]=\left[\begin{array}{ll}3& 1\end{array}\right]\left[\begin{array}{c}-2\\ x\end{array}\right] \\ [5x-2+0]=[-6+x] \\ 5x-2=x-6 \\ 4x=-4 \\ x=-1 \end{gathered}$$

We know, \((A-B)(A+B)=A^2+A B-B A-B^2\)

Given, $A^2-B^2=(A-B)(A+B)$

\(\Rightarrow A^2-B^2=A^2+A B-B A-B^2\)

\(\Rightarrow A B-B A=0\)

\(\Rightarrow A B=B A\)

\(A^{\prime}=A\) and \(A^{\prime}=-A \Rightarrow A=-A\)

\(\Rightarrow 2 A=O \Rightarrow A=O\)

So, A is a null matrix.

$4AB+3(AB+BA)-4BA$

$=4AB+3AB+3BA-4BA$

$=7AB-BA$

So, Bina's answer is correct.

$4AB+3(AB+BA)-4BA$

$=4AB+3AB+3BA-4BA$

$=7AB-BA$

So, Bina's answer is correct.

$a_{ij}=\max (i,j)-\min (i,j)$

$$\begin{gathered} a_{11}=\max (1,1)-\min (1,1)=1-1=0 \\ {a}_{12}=\max (1,2)-\min (1,2)=2-1=1 \\ {a}_{21}=\max (2,1)-\min (2,1)=2-1=1 \\ {a}_{22}=\max (2,2)-\min (2,2)=2-2=0 \end{gathered}$$

The matrix $A$ is:

$A=\left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]$

$\begin{array}{rcl}{A}^2& =& A\times A\\ & =& \left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\times \left[\begin{array}{ll}0& 1\\ 1& 0\end{array}\right]\\ & =& \left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\end{array}$

\((A-B)(A+B)=A^2+A B-B A-B^2\)

Given $A^2-B^2=(A-B)(A+B)$

\(\Rightarrow A^2-B^2=A^2+A B-B A-B^2\)

\(\Rightarrow A B-B A=0\)

\(\Rightarrow A B=B A\)

For a skew-symmetric matrix, we have:

$A^T=-A$

$$\begin{gathered} {\left[\begin{array}{ccc}a& c& -1\\ b& 0& 5\\ 1& -5& 0\end{array}\right]}^T=-\left[\begin{array}{ccc}a& c& -1\\ b& 0& 5\\ 1& -5& 0\end{array}\right] \\ \left[\begin{array}{ccc}a& b& 1\\ c& 0& -5\\ -1& 5& 0\end{array}\right]=\left[\begin{array}{ccc}-a& -c& 1\\ -b& 0& -5\\ -1& 5& 0\end{array}\right] \end{gathered}$$

Comparing both sides, we get:

$a=-a\Rightarrow 2a=0\Rightarrow a=0$

$b=-c$ and $c=-b$

Now,

$\begin{array}{rcl}2 a+(b+c)& =& 2\times 0+(b-b)\\ & =& 0\end{array}$

Given, $A=\left[\begin{array}{ccc}1& 12& 4y\\ 6x& 5& 2x\\ 8x& 4& 6\end{array}\right]$

for symmetric matrix : $A^T=A$

$A^T=\left[\begin{array}{ccc}1& 6x& 8x\\ 12& 5& 4\\ 4y& 2x& 6\end{array}\right]$

Now, equating the corresponding elements :

$6x=12\Rightarrow x=2$

$4y=8x\Rightarrow y=2x=4$

$2x=4\Rightarrow x=2$ (consistent)

Now,

$2x+y=2\left(2\right)+4=8$

\((A+B)^2=A^2+A B+B A+B^2\)

Given \((A+B)^2=A^2+B^2\)

\(\Rightarrow A B+B A=0\)

\(\Rightarrow A B=-B A\)

\(a_{i j}=i-3 j, i, j \in\{1,2,3\}\)

\(a_{11}=1-3=-2<0 \Rightarrow\) A true

\(a_{12}=1-6=-5, a_{21}=2-3=-1\)

\(\Rightarrow a_{12}+a_{21}=-6 \Rightarrow \mathrm{~B}\) true

\(a_{13}=1-9=-8, a_{31}=3-3=0\)

\(\Rightarrow a_{13}=-8<0=a_{31} \Rightarrow\) C false

\(a_{31}=0 \Rightarrow \mathrm{D}\) true

Compare each off-diagonal pair \((i, j)\) with \((j, i)\).

$6x=12\Rightarrow x=2$

$4y=8x\Rightarrow y=2x=4$

$2x=4\Rightarrow x=2$ (consistent)

$2x+y=2\left(2\right)+4=8$

For a skew-symmetric matrix, we have:

$A^T=-A$

$$\begin{gathered} \Rightarrow {\left[\begin{array}{ccc}a& c& -1\\ b& 0& 5\\ 1& -5& 0\end{array}\right]}^T=-\left[\begin{array}{ccc}a& c& -1\\ b& 0& 5\\ 1& -5& 0\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}a& b& 1\\ c& 0& -5\\ -1& 5& 0\end{array}\right]=\left[\begin{array}{ccc}-a& -c& 1\\ -b& 0& -5\\ -1& 5& 0\end{array}\right] \end{gathered}$$

Comparing both sides, we get:

$a=-a\Rightarrow 2a=0\Rightarrow a=0$

$b=-c$ and $c=-b$

Now,

$\begin{array}{rcl}2a-\left(b+c\right)& =& 2\times 0-\left(b-b\right)=0\end{array}$

Given \((A+B)^2=A^2+B^2\)

Since, \((A+B)^2=A^2+A B+B A+B^2\)

\(\Rightarrow A B+B A=0\)

\(\Rightarrow A B=-B A\)