If $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&1&1\\1&1&1\end{array}} \right]$, prove that ${A^n} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\end{array}} \right]$, $n \in N.$

.:

We shall prove it by mathematical induction.

To prove that n = 1 is true.

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

We have, A = $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&1&1\\1&1&1\end{array}} \right]$ and ${A^n} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\end{array}} \right]$ ….(i)

Thus, it is true for n = 1.
Let us assume that (i) is true for n = k, i.e.,

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

Then, ${A^{k + 1}} = {A^k} \cdot A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\1&1&1\\1&1&1\end{array}} \right]$

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

$\Rightarrow$ (i) is true for n = k + 1

So, by mathematical induction
${A^n} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\\{{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}\end{array}} \right],n \in N$is true.