If $A=\left[\begin{array}{l}1\text{ 1 1}\\ \text{1 1 1}\\ \text{1 1 1}\end{array}\right]$ then $A^n$for every positive integer n is

the result by the principle of mathematical induction on n.

Step1    When $n=1$, by the definition of integral powers of a matrix, we have  

             $A^1=A=\left[\begin{array}{l}1\text{ 1 1}\\ \text{1 1 1}\\ \text{1 1 1}\end{array}\right]=\left[\begin{array}{l}{3}^{1-1}{\text{ 3}}^{1-1}{\text{ 3}}^{1-1}\\ 3^{1-1}{\text{ 3}}^{1-1}{\text{ 3}}^{1-1}\\ 3^{1-1}{\text{ 3}}^{1-1}{\text{ 3}}^{1-1}\end{array}\right]$

So, the result is true for$n=1$.

Step2   Let the result be true for$n=m$ .Then 

$A^m=\left[\begin{array}{l}{3}^{m-1}{\text{ 3}}^{m-1}{\text{ 3}}^{m-1}\\ 3^{m-1}{\text{ 3}}^{m-1}{\text{ 3}}^{m-1}\\ 3^{m-1}{\text{ 3}}^{m-1}{\text{ 3}}^{m-1}\end{array}\right]\mathrm{...}\text{(i)}$

Now we shall show that the result is true for $n=m+1$, i.e.$n=m+1$

$A^{m+1}=\left[\begin{array}{l}{3}^m{\text{ 3}}^m{\text{ 3}}^m\\ 3^m{\text{ 3}}^m{\text{ 3}}^m\\ 3^m{\text{ 3}}^m{\text{ 3}}^m\end{array}\right]$

By the definition of integral powers of a matrix, we have 

$\begin{array}{l}{A}^{m+1}=A^m·A\\ =\left[\begin{array}{ccc}{3}^{m-1}& 3^{m-1}& 3^{m-1}\\ 3^{m-1}& 3^{m-1}& 3^{m-1}\\ 3^{m-1}& 3^{m-1}& 3^{m-1}\end{array}\right]\left[\begin{array}{lll}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]\\ =\left[\begin{array}{lll}{3}^{m-1}+3^{m-1}+3^{m-1}& 3^{m-1}+3^{m-1}+3^{m-1}& 3^{m-1}+3^{m-1}+3^{m-1}\\ 3^{m-1}+3^{m-1}+3^{m-1}& 3^{m-1}+3^{m-1}+3^{m-1}& 3^{m-1}+3^{m-1}+3^{m-1}\\ 3^{m-1}+3^{m-1}+3^{m-1}& 3^{m-1}+3^{m-1}+3^{m-1}& 3^{m-1}+3^{m-1}+3^{m-1}\end{array}\right]\\ =\left[\begin{array}{ccc}3.3^{m-1}& 3.3^{m-1}& 3.3^{m-1}\\ 3.3^{m-1}& 3.3^{m-1}& 3.3^{m-1}\\ 3.3^{m-1}& 3.3^{m-1}& 3.3^{m-1}\end{array}\right]\\ =\left[\begin{array}{ccc}{3}^m& 3^m& 3^m\\ 3^m& 3^m& 3^m\\ 3^m& 3^m& 3^m\end{array}\right]\end{array}$

This shows that the result is true for, whenever it is true for$n=m$

 Hence, by the principle of mathematical induction the result is valid for any positive  integer n.