We shall prove the result by mathematical induction on n.
Step 1 : When n = 1, by the definition or integral powers of a matrix, we have
\({ A }^{ 1 }=\left[ \begin{matrix} 1+2\left( 1 \right) & -4n \\ n & 1-2\left( 1 \right) \end{matrix} \right] =\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \end{matrix} \right] \)
So, the result is true for n = 1.
Step 2 : Let the result be true for n = m. Then,
\({ A }^{ m }=\left[ \begin{matrix} 1+2m & -4m \\ m & 1-2m \end{matrix} \right] \)
Now, we will show that the result is true for n = m + 1, i.e.,
\({ A }^{ m+1 }=\left[ \begin{matrix} 1+2\left( m+1 \right) & -4\left( m+1 \right) \\ \left( m+1 \right) & 1-2\left( m+1 \right) \end{matrix} \right] \)
By the definition of integral powers of a square matrix, we have
\({ A }^{ m+1 }={ A }^{ m }.A\)
\(\Rightarrow { A }^{ m+1 }=\left[ \begin{matrix} 1+2m & -4m \\ m & 1-2m \end{matrix} \right] \left[ \begin{matrix} 3 & -4 \\ 1 & -1 \end{matrix} \right] \)
[by supposition (i)]
\(\Rightarrow { A }^{ m+1 }=\left[ \begin{matrix} 3+6m-4m & -4-8m+4m \\ 3m+1-2m & -4m-4+2m \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+2\left( m+1 \right) & -4\left( m+1 \right) \\ \left( m+1 \right) & 1-2\left( m+1 \right) \end{matrix} \right] \)
This shows that the result is true for n = m + 1, whenever it is true for n = m.
Hence, by the principle of mathematical induction, the result is true for any positive integer n.

We have, \(A=\left[ \begin{matrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix} \right] \)
\({ A }^{ T }=\left[ \begin{matrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix} \right] \)
\(\Rightarrow A+{ A }^{ T }=\left[ \begin{matrix} 2cos\theta & 0 \\ 0 & cos\theta \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\Rightarrow 2cos\theta =1\)
\(\Rightarrow cos\theta =\frac { 1 }{ 2 } \Rightarrow \theta =\cos ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } \)
\(\because \quad \theta =\frac { \pi }{ 3 } \)

We have, \(A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)
\({ A }^{ \prime }=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{matrix} \right] \)
\({ A+A }^{ \prime }=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A+A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] =0\)
and \({ A-A }^{ \prime }=\left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A-A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)

\(A=\left[ \begin{matrix} 0 & x & -4 \\ -2 & 0 & -1 \\ y & -1 & 0 \end{matrix} \right] \)
\({ A }^{ \prime }=\left[ \begin{matrix} 0 & -2 & y \\ x & 3 & -1 \\ -4 & -1 & 0 \end{matrix} \right] \)
For skew symmetric
\(A={ -A }^{ \prime }\)
\(\Rightarrow \left[ \begin{matrix} 0 & x & -4 \\ -2 & 0 & -1 \\ y & -1 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & -2 & y \\ x & 3 & -1 \\ -4 & -1 & 0 \end{matrix} \right] \)
x = 2, y = 4

We have, \({ A }_{ 3\times x }\) and \({ B }_{ 4\times y }\) = \({ AB }_{ 3\times 3 }\)
(i) \(\left( { A }_{ 3\times x }, \right) \left( { B }_{ 4\times y } \right) =\left( { AB }_{ 3\times 3 } \right) \)
\(\therefore\)   x = 4 and y = 3
(ii) \(\left( { A }_{ x\times 2 } \right) \left( { B }_{ y\times 4 } \right) =\left( { AB }_{ 3\times 4 } \right) \)
\(\therefore\)  y = 2, x = 3

We have, \(A=\left[ \begin{matrix} 1 & 2 & 3 \end{matrix} \right] \) and \(B=\left[ \begin{matrix} -2 \\ 3 \\ 1 \end{matrix} \right] \)
\(\therefore \ AB=\left[ \begin{matrix} 1 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} -2 \\ 3 \\ 1 \end{matrix} \right] \)
\(\Rightarrow\)  AB = [ -2 + 6 + 3] = [7]
and \(BA=\left[ \begin{matrix} -2 \\ 3 \\ 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 2 & 3 \end{matrix} \right] \)
\(\Rightarrow BA=\left[ \begin{matrix} -2 & -4 & -6 \\ 3 & 6 & 9 \\ 1 & 2 & 3 \end{matrix} \right] \)

We have, \(\left[ \begin{matrix} { x }^{ 2 } \\ { y }^{ 2 } \end{matrix} \right] -3\left[ \begin{matrix} x \\ 2y \end{matrix} \right] =\left[ \begin{matrix} -2 \\ -9 \end{matrix} \right] \)
\(\Rightarrow\)  x² - 3x = - 2 and y² - 6y = - 9
\(\Rightarrow\)  x² - 3x + 2 = 0 and y² - 6y + 9 = 0
\(\Rightarrow\)  x² - 2x - x + 2 = 0 and y² - 3y - 3y + 9 = 0
\(\Rightarrow\)  x(x - 2) - 1(x - 2) = 0 and y(y - 3) - 3(y - 3) = 0
\(\Rightarrow\)  (x - 2)(x - 1) = 0 and (y - 3)(y - 3) = 0
\(\therefore\)   x = 1, 2 and y = 3, 3

We have A + B - X = 0
By adding X on both the sides,
A + B - X + X = 0 + X
\(\Rightarrow\)  A + B = X
\(\Rightarrow X=\left[ \begin{matrix} 1 & 4 \\ 3 & 2 \\ 2 & 1 \end{matrix} \right] +\left[ \begin{matrix} 5 & 2 \\ -1 & 0 \\ 1 & 1 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 6 & 6 \\ 2 & 2 \\ 3 & 2 \end{matrix} \right] \)

We have, \(\left[ \begin{matrix} 2x+y & x-y \\ x-z & x+y+z \end{matrix} \right] =\left[ \begin{matrix} 10 & -1 \\ 2 & 8 \end{matrix} \right] \)
\(\Rightarrow\) 2x + y = 10, x - y = - 1
x - z = 2 and x + y + z = 8
\(\therefore\) 2(y - 1) + y = 10  \(\Rightarrow\)  2y + y + 2 = 10
\(\Rightarrow\)3y = 12  \(\Rightarrow\)  y = 4
\(\therefore\)  x = 3
3 - z,  \(\Rightarrow\)  z = 1
\(\therefore\)  x = 3, y = 4, z = 1

We shall prove the result by using principle of mathematical induction.
Let \(P\left( n \right) { :A }^{ n }=\left[ \begin{matrix} { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \end{matrix} \right] \)
Now, \(P\left( 1 \right) { :A }^{ 1 }=\left[ \begin{matrix} { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \)
The result is true for n = 1.
Let the result be true for n = k.
So, \({ A }^{ k }=\left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
Now, we prove that P(k + 1) is true.
Now, A^k+1 = A. A^k
\(=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \end{matrix} \right] \)
= A^k+1
Hence, it is true n = k + 1.
Hence, by principle of mathematical induction P(n) is true for all \(n\epsilon N\)