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  4. if A = [1&1 0&2], then A4 - 24(A-I) =

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Q. if $A = \begin{bmatrix}1&1\\ 0&2\end{bmatrix}$, then $A^{4} - 2^{4}\left(A-I\right) =$

Matrices

Solution:

We have $A = \begin{bmatrix}1&1\\ 0&2\end{bmatrix} $
$A^2 = \begin{bmatrix}1&1\\ 0&2\end{bmatrix}\begin{bmatrix}1&1\\ 0&2\end{bmatrix} = \begin{bmatrix}1&3\\ 0&4\end{bmatrix}$
$A^3 = \begin{bmatrix}1&7\\ 0&4\end{bmatrix} \begin{bmatrix}1&1\\ 0&2\end{bmatrix}\begin{bmatrix}1&7\\ 0&8\end{bmatrix}$
$A^4 = \begin{bmatrix}1&7\\ 0&8\end{bmatrix} \begin{bmatrix}1&1\\ 0&2\end{bmatrix} = \begin{bmatrix}1&15\\ 0&16\end{bmatrix}$
Now $A^4 - 2^4 ( A - I )=\begin{bmatrix}1&15\\ 0&16\end{bmatrix} -16\begin{bmatrix}0&1\\ 0&1\end{bmatrix}$
$ = \begin{bmatrix}1&15\\ 0&16\end{bmatrix} -\begin{bmatrix}1&16\\ 0&16\end{bmatrix} = \begin{bmatrix}1&-1\\ 0&0\end{bmatrix}$
Also $ 2I - A = \begin{bmatrix}2&0\\ 0&2\end{bmatrix}-\begin{bmatrix}1&1\\ 0&2\end{bmatrix} =\begin{bmatrix}1&-1\\ 0&0\end{bmatrix} $
$\therefore A^4 - 2^4 ( A-I ) = 2I - A$