It is given that $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$

To show: $P(n) A^{n}=\left[\begin{array}{lll}3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1}\end{array}\right], n \in \mathrm{N}$

We shall prove the result by using the principles of mathematical induction. For $n=1,$ we have :

$P(1)=\left[\begin{array}{lll}3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1}\end{array}\right]$ $=\left[\begin{array}{lll}3^{0} & 3^{0} & 3^{0} \\ 3^{0} & 3^{0} & 3^{0} \\ 3^{0} & 3^{0} & 3^{0}\end{array}\right]$ $=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$ $=A$

Therefore, the result is true for $n=1$

Let the result be true for $n=k$

$P(k): A^{k}=\left[\begin{array}{lll}3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1}\end{array}\right]$

That is Now, we prove that the result is true for $n=k+1$

Now, $A^{k+1}=A \cdot A^{k}$

$=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{lll}3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1}\end{array}\right]$

$=\left[\begin{array}{lll}3.3^{k-1} & 3.3^{k-1} & 3.3^{k-1} \\ 3.3^{k-1} & 3.3^{k-1} & 3.3^{k-1} \\ 3.3^{k-1} & 3.3^{k-1} & 3.3^{k-1}\end{array}\right]$

$=\left[\begin{array}{lll}3.3^{(k+1)-1} & 3.3^{(k+1)-1} & 3.3^{(k+1)-1} \\ 3.3^{(k+1)-1} & 3.3^{(k+1)-1} & 3.3^{(k+1)-1} \\ 3.3^{(k+1)-1} & 3.3^{(k+1)-1} & 3.3^{(k+1)-1}\end{array}\right]$

Therefore, the result is true for $n=k+1$

Thus, by the principal of mathematical induction, we have:

$A^{n}=\left[\begin{array}{lll}3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1}\end{array}\right], n \in \mathrm{N}$