Let $P(n): A^n=\left[\begin{array}{lll}3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1}\end{array}\right]$
where $\mathrm{A}=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$
For $\mathrm{n}=1 \quad$ L.H.S. $=\mathrm{A}^{\mathrm{n}}=\mathrm{A}=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$
R.H.S. $=\mathbf{A}^{\mathrm{n}}=\left[\begin{array}{ccc}3^{\circ} & 3^{\circ} & 3^{\circ} \\ 3^{\circ} & 3^{\circ} & 3^{\circ} \\ 3^{\circ} & 3^{\circ} & 3^{\circ}\end{array}\right]=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$
$\therefore \mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$
Let it be true for $\mathrm{n}=\mathrm{k}$
$\therefore A^k=\left[\begin{array}{lll}3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1}\end{array}\right]$
Multiplying both sides by A
L. H.S. $=\mathrm{A}^{\mathrm{k}} \mathrm{A}=\mathrm{A}^{\mathrm{k}+1}$
R.H.S. $=\left[\begin{array}{lll}3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1}\end{array}\right]$ A
$\mathrm{A}^{\mathrm{k}+1}=\left[\begin{array}{lll}3^{\mathrm{k}-1} & 3^{\mathrm{k}-1} & 3^{\mathrm{k}-1} \\ 3^{\mathrm{k}-1} & 3^{\mathrm{k}-1} & 3^{\mathrm{k}-1} \\ 3^{\mathrm{k}-1} & 3^{\mathrm{k}-1} & 3^{\mathrm{k}-1}\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$
$=\left[\begin{array}{lll}3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k\end{array}\right]=\left[\begin{array}{lll}3^{(k+1)-1} & 3^{(k+1)-1} & 3^{(k+1)-1} \\ 3^{(k+1)-1} & 3^{(k+1)-1} & 3^{(k+1)-1} \\ 3^{(k+1)-1} & 3^{(k+1)-1} & 3^{(k+1)-1}\end{array}\right]$
$\therefore P(n)$ is true for $\mathrm{n}=\mathrm{k}+1$
By principle of mathematical induction that $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n}, \mathrm{n} \in \mathrm{N}$.