Let \(P(n):(A B)^{n}=A^{n} B^{n}\)
\(\therefore \ P(1):(A B)^{1}=A^{1} B^{1}\)
\(\Rightarrow A B=A B\)
So, P(1) is true.
Let P(n) is true for some \(k \in N\)
So, \(P(\mathrm{k}):(A B)^{k}=A^{k} B^{k}, k \in N \ \ldots(\mathrm{i})\)
Now
\( (A B)^{k+1} =(A B)^{k}(A B) \quad(\text { using }(i)) \)
\(=A^{k} B^{k}(A B) \)
\(=A^{k} B^{k-1}(B A) B \)
\(=A^{k} B^{k-1}(A B) B \quad(\text { as given } A B=B A) \)
\(=A^{k} B^{k-1} A B^{2} \)
\(=A^{k} B^{k-2}(B A) B^{2}\)
\(=A^{k} B^{k-2} A B B^{2} \)
\(=A^{k} B^{k-2} A B^{3} \)
\( \ldots \)
\(\ldots\)
\(=A^{k+1} B^{k+1} \)
Thus P(1) is true and whenever P(k) is true P(k+1) is true. So, P(n) is true for all n \( \in\) N

\(\left[\begin{array}{rrr} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right]+\left[\begin{array}{rrr} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right]\)

We have,
\(A=\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]\) 
\(\begin{array}{l} \Rightarrow A^{\prime}=\left[\begin{array}{rrr} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right] \end{array}\) 
\(\text { Let } P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left\{\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]+\left[\begin{array}{ccc} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right]\right\}\) 
\(=\frac{1}{2}\left[\begin{array}{ccc} 4 & 11 & -5 \\ 11 & 6 & 3 \\ -5 & 3 & 8 \end{array}\right]=\left[\begin{array}{ccc} 2 & \frac{11}{2} & -\frac{5}{2} \\ \frac{11}{2} & 3 & \frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 4 \end{array}\right]\) 
which is symmetric matrix and
\(Q =\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left\{\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]-\left[\begin{array}{rrr} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{array}\right]\right\} \)
\(=\frac{1}{2}\left[\begin{array}{rrr} 0 & -3 & -7 \\ 3 & 0 & 7 \\ 7 & -7 & 0 \end{array}\right]=\left[\begin{array}{rrr} 0 & -\frac{3}{2} & -\frac{7}{2} \\ \frac{3}{2} & 0 & \frac{7}{2} \\ \frac{7}{2} & -\frac{7}{2} & 0 \end{array}\right] \) 
which is skew-symmetric matrix
Now,\(P+Q=\frac{1}{2}\left(A+A^{\prime}\right)+\frac{1}{2}\left(A-A^{\prime}\right)\) 
 \(=\left[\begin{array}{rrr} 2 & \frac{11}{2} & -\frac{5}{2} \\ \frac{11}{2} & 3 & \frac{3}{2} \\ -\frac{5}{2} & \frac{3}{2} & 4 \end{array}\right]+\left[\begin{array}{rrr} 0 & -\frac{3}{2} & -\frac{7}{2} \\ \frac{3}{2} & 0 & \frac{7}{2} \\ \frac{7}{2} & -\frac{7}{2} & 0 \end{array}\right]=\left[\begin{array}{rrr} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right]=A\) 
Hence, A is represented as sum of symmetric and skew-symmetric matrix.

Let X be of order \(m\times n\) 
Since  \(\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] \) is order \(2\times 3\) ,
\(\therefore \ n=2.\)
Now LHS is order \(m\times 2\).
But RHS is order \(2\times 3\).
\(\therefore \ m=2.\)
Thus X is of order  \(2\times 2\)
Let    \(X=\begin{bmatrix} a & b \\ c & d \end{bmatrix}.\)
\(\therefore \ X\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right] \)
\(\Rightarrow \begin{bmatrix} a & b \\ c & d \end{bmatrix}\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{matrix} \right] =\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{matrix} \right] \)
Equating corresponding elements,
a+ 4b = -7....(1)
2a + 5b = -8.....(2)    
3a + 6b = -9 ..(3)
c + 4d = 2 ...(4)
2c + 5d = 4...(5)
3c + 6d = 6...(6)
Solving (1) and (2), a = 1, b = -2.
These also satisfy (3).
Solving (4) and (5), c = 2, d = 0.
These also satisfy (6).
Hence, \(X=\begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}\)

\(A^{2}-4 A \left.+7 I=\left[\begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array}\right]\left[\begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array}\right]-4\left[\begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array}\right]+7 \mid \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\(=\left[\begin{array}{rr} 4-3 & 6+6 \\ -2-2 & -3+4 \end{array}\right]-\left[\begin{array}{rr} 8 & 12 \\ -4 & 8 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \)
\(=\left[\begin{array}{rr} 1 & 12 \\ -4 & 1 \end{array}\right]-\left[\begin{array}{rr} 8 & 12 \\ -4 & 8 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \)
\(=\left[\begin{array}{rr} 1-8+7 & 12-12+0 \\ -4+4+0 & 1-8+7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\)
\(\Rightarrow A^{2} =4 A-7 I \Rightarrow A^{5}=4 A^{4}-7 I A^{3} \)
\(A^{5} =4(4 A-7 I)(4 A-7 I)-7(4 A-7 I) A \)
\(=4\left(16 A^{2}-28 \mathrm{~A}-28 \mathrm{~A}+49 I^{2}\right)-28 A^{2}+49 A \)
\(=64 A^{2}-224 A+1961-28 A^{2}+49 A \)
\(=36 A^{2}-175 A+196 I \)
\(=36(4 A-7 I)-175 A+196 I \)
\(=144 A-252 I-175 A+196 I=-31 A-56 I \)
\(=-31\left[\begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array}\right]-56\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\(=\left[\begin{array}{rr} -62 & -93 \\ 31 & -62 \end{array}\right]-\left[\begin{array}{rr} 56 & 0 \\ 0 & 56 \end{array}\right] \)
\(=\left[\begin{array}{cr} -62-56 & -93-0 \\ 31-0 & -62-56 \end{array}\right]=\left[\begin{array}{rr} -118 & -93 \\ 31 & -118 \end{array}\right]\)

\((I+A)^{3}=(I+A)(I+A)(I+A)\\ =\left(I^{2}+I A+A I+A^{2}\right)(I+A)=(I+A+A+A)(I+A)\\ =(I+3 A)(I+A) \quad\left[\because A^{2}=A\right]\)
\(=I^{2}+I A+3 A I+3 A^{2}\\=I+A+3 A+3 A\\=I+7 A\)

Consider 3(2X + 3 Y) - 2(3X + 2 Y)
\(=3\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]-2\left[\begin{array}{rr} -2 & 2 \\ 1 & -5 \end{array}\right]\)
\(\Rightarrow 6 X+9 Y-6 X-4 Y=\left[\begin{array}{cc} 6 & 9 \\ 12 & 0 \end{array}\right]-\left[\begin{array}{rr} -4 & 4 \\ 2 & -10 \end{array}\right]\)
\(=\left[\begin{array}{cc} 6+4 & 9-4 \\ 12-2 & 0+10 \end{array}\right]=\left[\begin{array}{cc} 10 & 5 \\ 10 & 10 \end{array}\right]\)
\(\Rightarrow 5 Y=\left[\begin{array}{rr} 10 & 5 \\ 10 & 10 \end{array}\right] \)
\(\Rightarrow \ Y=\frac{1}{5}\left[\begin{array}{rr} 10 & 5 \\ 10 & 10 \end{array}\right]=\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right]\)
\(\text { Now } \ 2 X =\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]-3 Y=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]-3\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right] \)
\(=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]-\left[\begin{array}{ll} 6 & 3 \\ 6 & 6 \end{array}\right]=\left[\begin{array}{lr} 2-6 & 3-3 \\ 4-6 & 0-6 \end{array}\right] \)
\(=\left[\begin{array}{ll} -4 & 0 \\ -2 & -6 \end{array}\right] \)
\(\Rightarrow \ X=\frac{1}{2}\left[\begin{array}{rr} -4 & 0 \\ -2 & -6 \end{array}\right]=\left[\begin{array}{lr} -2 & 0 \\ -1 & -3 \end{array}\right]\)
\(\text {Hence, } \quad X=\left[\begin{array}{rr} -2 & 0 \\ -1 & -3 \end{array}\right] \text { ; }Y=\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right]\)