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 \(n\epsilon N\)

a = -2, b = 0 and c = -3

Given matrix equation is
\( \left[\begin{array}{ll} 2 x & 3 \end{array}\right]\left[\begin{array}{rr} 1 & 2 \\ -3 & 0 \end{array}\right]\left[\begin{array}{l} x \\ 8 \end{array}\right]=O \)
\(\Rightarrow \left[\begin{array}{ll} 2 x & 3 \end{array}\right]\left[\left[\begin{array}{rr} 1 & 2 \\ -3 & 0 \end{array}\right]\left[\begin{array}{l} x \\ 8 \end{array}\right]\right)=O \) 
[by associative law of multiplication]
\(\Rightarrow \left[\begin{array}{ll}2 x & 3\end{array}\right]\left[\begin{array}{c}x+16 \\ -3 x\end{array}\right]=0\)
\(\Rightarrow [2 x(x+16)-9 x]=[O]\)
\(\Rightarrow\) 2 x^{2} + 32 x - 9x = 0
\(\Rightarrow\)\(2 x^{2}+23 x=0\)
\(\Rightarrow x(2 x+23)=0\)
\(\Rightarrow\) x = 0 and x = -23 / 2

(i) 3 x 3
(ii) 9
(iii)a₂₃ = x² - y, a31 = 0, a₁₂ = 1]