Question
If $\text{A} = \begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix},$ prove that $\text{A}'' = \begin{bmatrix}3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\end{bmatrix}n \in \text{N}.$

Answer

Given: $\text{A}=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
Let p(n): $\text A^{n}=\begin{bmatrix}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{bmatrix}\ $
p(1): $\Rightarrow \ \ \ \text A=\begin{bmatrix}3^{0}&3^{0}&3^{0}\\3^{0}&3^{0}&3^{0}\\3^{0}&3^{0}&3^{0}\end{bmatrix}\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
$\therefore$ p(1) is true for n = 1.
Now p(k): $\text A^{k}=\begin{bmatrix}3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\end{bmatrix}...\text{(ii)}$
Multiplying eq. (ii) by eq. (i) $\text{A}^{k}\text{A}=\begin{bmatrix}3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\end{bmatrix}\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
$\Rightarrow \text{ A}^ \text{k+1}=\begin{bmatrix}3^{k}&3^{k}&3^{k}\\3^{k}&3^{k}&3^{k}\\3^{k}&3^{k}&3^{k}\end{bmatrix}=\text{p(k+1)}$
Therefore, p(n) is true for all natural numbers by P.M.I.

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