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Q. If $A = \begin{bmatrix}3&3&3\\ 3&3&3\\ 3&3&3\end{bmatrix}$, then $A^4 = $

Matrices

Solution:

Given $A = 3 \begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix} $
$ \Rightarrow A^{2} = 3\times3 \begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}^{2} $
$ =9 \begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}$
$ = 9\begin{bmatrix}3&3&3\\ 3&3&3\\ 3&3&3\end{bmatrix} =9A $ ......(1)
Hence $ A^{4} = \left(A^{2}\right)^{2} = \left(9A\right)^{2} = 81A^{2} $
$ =81\left(9A\right) $ [Using (1)]
$=729 A$