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Q. If $A=\begin{bmatrix}3 & 2 \\ 0 & 1\end{bmatrix}$, then $A^{-3}$ is

Matrices

Solution:

We have $A^{-1}=\frac{1}{3}\begin{bmatrix}1 & -2 \\ 0 & 3\end{bmatrix}$
$\Rightarrow A^{-3} =\frac{1}{27}\begin{bmatrix}1 & -2 \\0 & 3\end{bmatrix}\begin{bmatrix}1 & -2 \\0 & 3\end{bmatrix}\begin{bmatrix}1 & -2 \\0 & 3\end{bmatrix}$
$ =\frac{1}{27}\begin{bmatrix}1 & -8 \\0 & 9\end{bmatrix}\begin{bmatrix}1 & -2 \\0 & 3\end{bmatrix}=\frac{1}{27}\begin{bmatrix}1 & -26 \\0 & 27\end{bmatrix}$