We shall prove the result by using principle of mathematical induction
\(\mathrm{P}(n): \text { If } \mathrm{A}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, then } \mathrm{A}^n=\left[\begin{array}{cc} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right], n \in \mathbf{N}\)
\(P(1): A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, so } A^1=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\)
Therefore, the result is true for n = 1.
Let the result be true for n = k. So
\(\mathrm{P}(k): \mathrm{A}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, then } \mathrm{A}^k=\left[\begin{array}{cc} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right]\)
Now, we prove that the result holds for n = k +1
\( \mathrm{A}^{k+1} =\mathrm{A} \cdot \mathrm{A}^k=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right] \)
\(=\left[\begin{array}{cc} \cos \theta \cos k \theta-\sin \theta \sin k \theta & \cos \theta \sin k \theta+\sin \theta \cos k \theta \\ -\sin \theta \cos k \theta+\cos \theta \sin k \theta & -\sin \theta \sin k \theta+\cos \theta \cos k \theta \end{array}\right]\)
\(=\left[\begin{array}{cc} \cos (\theta+k \theta) & \sin (\theta+k \theta) \\ -\sin (\theta+k \theta) & \cos (\theta+k \theta) \end{array}\right]=\left[\begin{array}{cc} \cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta \end{array}\right]\)
Therefore, the result is true for n = k + 1. Thus by principle of mathematical induction, we have \(A^n=\left[\begin{array}{rr} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right]\), holds for all natural numbers.

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\(i.e.\ \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\( \Rightarrow \begin{bmatrix} -2 & 1 \\ 6 & -3 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}A\quad [Applying\quad { R }_{ 1 }\leftrightarrow { R }_{ 2 }]\)
\( \Rightarrow \begin{bmatrix} -2 & 1 \\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 3 \end{bmatrix}A\quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }+3{ R }_{ 1 }]\)
Since second row of LHS matrix has all zero,
\(\therefore \ { A }^{ -1 }\) does not exist.

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\(i.e.\ \begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\(\Rightarrow \begin{bmatrix} 1 & -4 \\ 1 & -2 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}A \quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }]\)
\(\Rightarrow \begin{bmatrix} 1 & -4 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}A\quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }]\)
\(\Rightarrow \begin{bmatrix} 1 & -4 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -\frac { 1 }{ 2 } & 1 \end{bmatrix}A\quad [Applying\quad { R }_{ 2 }\rightarrow \left( \frac { 1 }{ 2 } \right) { R }_{ 2 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} -1 & 3 \\ -\frac { 1 }{ 2 } & 1 \end{bmatrix}A\quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }+4{ R }_{ 2 }]\)
\(Hence,\ { A }^{ -1 }=\begin{bmatrix} -1 & 3 \\ -\frac { 1 }{ 2 } & 1 \end{bmatrix}.\)

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\(i.e.\ \begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\(\Rightarrow \begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}A \quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -3 & 4 \end{bmatrix}A \quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-{ 3R }_{ 1 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 4 & -5 \\ -3 & 4 \end{bmatrix}A.\quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }]\)
\(Hence,\ { A }^{ -1 }=\begin{bmatrix} 4 & -5 \\ -3 & 4 \end{bmatrix}.\)

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\( i.e.\ \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\(\Rightarrow \begin{bmatrix} 6 & 2 \\ 5 & 2 \end{bmatrix}=\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}A \quad [Applying\quad { R }_{ 1 }\leftrightarrow 2{ R }_{ 1 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 0 \\ 5 & 2 \end{bmatrix}=\begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}A\quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 2 & -1 \\ -10 & 6 \end{bmatrix}A \quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-{ 5R }_{ 1 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}A\quad [Applying\quad { R }_{ 2 }\rightarrow \frac { 1 }{ 2 } { R }_{ 2 }]\)
\(Hence,{ A }^{ -1 }=\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}\)

\(\left[ \left( \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right) +\left( \begin{matrix} \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \\ \omega & { \omega }^{ 2 } & 1 \end{matrix} \right) \right] \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\( \left[ \left( \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right) +\left( \begin{matrix} \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \\ \omega & { \omega }^{ 2 } & 1 \end{matrix} \right) \right] \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] =\left( \begin{matrix} 1+\omega & \omega +{ \omega }^{ 2 } & { \omega }^{ 2 }+1 \\ \omega +{ \omega }^{ 2 } & { \omega }^{ 2 }+1 & 1+\omega \\ { \omega }^{ 2 }+\omega & 1+{ \omega }^{ 2 } & \omega +1 \end{matrix} \right) \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] \)
\(=\left( \begin{matrix} -{ \omega }^{ 2 } & -1 & -\omega \\ -1 & -\omega & -{ \omega }^{ 2 } \\ -1 & -\omega & -{ \omega }^{ 2 } \end{matrix} \right) \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] \quad \quad \left[ \because \quad 1+\omega +{ \omega }^{ 2 }=0 \right] \)
\(=\left( \begin{matrix} -{ \omega }^{ 2 } & -\omega & -{ \omega }^{ 3 } \\ -1 & -{ \omega }^{ 2 } & -{ \omega }^{ 4 } \\ -1 & -{ \omega }^{ 2 } & -{ \omega }^{ 4 } \end{matrix} \right) =\left[ \begin{matrix} -{ \omega }^{ 2 } & -\omega & -1 \\ -1 & -{ \omega }^{ 2 } & -{ \omega } \\ -1 & -{ \omega }^{ 2 } & -{ \omega } \end{matrix} \right] \quad \left[ \because { \omega }^{ 3 }=1\quad \& \quad { \omega }^{ 4 }=\omega \quad \right] \)
 \(=\left[ \begin{matrix} -0 \\ -0 \\ -0 \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \) 
which is true.

We have: \(A=\left[ \begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{matrix} \right] .\)
\(\therefore \ A\prime =\left[ \begin{matrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{matrix} \right]\)
\(\therefore \ AA\prime =\left[ \begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{matrix} \right] \)
\(=\begin{bmatrix} 9+1+1 & 0+1-2 \\ 0+1-2 & 0+1+4 \end{bmatrix}=\begin{bmatrix} 11 & -1 \\ -1 & 5 \end{bmatrix}.\)
Which is a symmetric matrix.

We have: \(A=\begin{bmatrix} 2 & 4 \\ 5 & 6 \end{bmatrix}.\)
\(\therefore\ A\prime =\begin{bmatrix} 2 & 5 \\ 4 & 6 \end{bmatrix}.\)
\(\therefore \ A+A\prime =\begin{bmatrix} 2 & 4 \\ 5 & 6 \end{bmatrix}+\begin{bmatrix} 2 & 5 \\ 4 & 6 \end{bmatrix}\)
\( =\begin{bmatrix} 2+2 & 4+5 \\ 5+4 & 6+6 \end{bmatrix}=\begin{bmatrix} 4 & 9 \\ 9 & 12 \end{bmatrix}...(1)\)
Now \((A+A\prime )\prime =\begin{bmatrix} 4 & 9 \\ 9 & 12 \end{bmatrix}=A+A\prime \)   [Using (1)]
Hence,   \(A+A\prime \)   is a symmetric matrix.

We have:   \(A=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{matrix} \right] \) and\(B=\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix} \right] .\)
\(\therefore \ 2A+3B=2\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{matrix} \right] +3\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 14 \end{matrix} \right] +\left[ \begin{matrix} -3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 12 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 6-3 & 0+0 & 0+0 \\ 0+0 & -10+6 & 0+0 \\ 0+0 & 0+0 & 14+12 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 3 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 26 \end{matrix} \right] \) =diag.\(\left[ \begin{matrix} 3, & -4, & 26 \end{matrix} \right] .\)

Here,
\(f(x)f(y)=\left[ \begin{matrix} cosx & -sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} cosy & -siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(=\left[ \begin{matrix} cosxcosy-sinxsiny & -sinycosx-sinxcosy & 0 \\ sinxcosy+cosxsiny & -sinxsiny+cosxcosy & 0 \\ 0 & 0 & 1 \end{matrix} \right]\)
\( =\left[ \begin{matrix} cos(x+y) & -sin(x+y) & 0 \\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(=f(x+y).\)

\(We\quad have:\quad 3\begin{bmatrix} x & y \\ z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix},\)
\(\Rightarrow \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}=\begin{bmatrix} x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix}.\)
Equating corresponding elements:
3x = x + 4 \(\Rightarrow \) 2x = 4 \(\Rightarrow \) x = 2
3y = x + y + 6 \(\Rightarrow \) 2y = 2 + 6 \(\Rightarrow \) 2y = 8  \(\Rightarrow \) y = 4
3w = 2w + 3  \(\Rightarrow \) w = 3 
and 3z = z + w-1 \(\Rightarrow \) 2z = 3-1 = 2  \(\Rightarrow \) z = 1.
Hence, x = 2, y = 4, z = 1 and w = 3.

We have: \(3\begin{bmatrix} x & y \\ z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix},\)
\(\Rightarrow \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}=\begin{bmatrix} x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{bmatrix},\)
Equating corresponding elements:
3x = x + y  \(\Rightarrow \) 2x = 4 \(\Rightarrow \) x = 2
3y = x + y + 6 \(\Rightarrow \) 2y = 2 + 6 \(\Rightarrow \) 2y = 8 \(\Rightarrow \) y = 4
3w = 2w + 3 \(\Rightarrow \) w = 3   and  
3z = z + w - 1\(\Rightarrow \) 2z = 3 - 1 = 2 \(\Rightarrow \) z = 1.
Hence, x = 2, y = 4, z = 1 and w = 3.

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6), and (6, 3)
Hence, the possible orders of a matrix having 18 elements are:
1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3
(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.
Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.
The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and  (6, 4)
Hence, the possible orders of a matrix having 24 elements are:
1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4
(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.