Question

If $A=\left[\begin{array}{cc}3& 2\\ 0& 1\end{array}\right]$, then $A^{-3}$ is

• A. $\frac{1}{27}\left[\begin{array}{cc}1& -26\\ 0& -27\end{array}\right]$
• B. $\frac{1}{27}\left[\begin{array}{cc}-1& -26\\ 0& -27\end{array}\right]$
• C. $\frac{1}{27}\left[\begin{array}{cc}1& -26\\ 0& 27\end{array}\right]$
• D. $\frac{1}{27}\left[\begin{array}{cc}-1& 26\\ 0& -27\end{array}\right]$

Answer 1

Answer: (C)

We have $A^{-1}=\frac{1}{3}\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]$
$\Rightarrow A^{-3}=\frac{1}{27}\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]$
$=\frac{1}{27}\left[\begin{array}{cc}1& -8\\ 0& 9\end{array}\right]\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]=\frac{1}{27}\left[\begin{array}{cc}1& -26\\ 0& 27\end{array}\right]$