Question:
If $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$, then $A^{5}=$
(a) $5 \mathrm{~A}$
(b) $10 \mathrm{~A}$
(c) $16 \mathrm{~A}$
(d) $32 \mathrm{~A}$
Solution:
(c) $16 \mathrm{~A}$
$A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$
$\Rightarrow A=2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow A=2 I$
$\Rightarrow A^{5}=(2 I)^{5}$
$\Rightarrow A^{5}=16 \times 2 I$
$\Rightarrow A^{5}=16\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$
$\Rightarrow A^{5}=16 A$