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)

$\frac{1}{27}\left[\begin{array}{cc}1& -26\\ 0& 27\end{array}\right]$

$A=\left[\begin{array}{cc}3& 2\\ 0& 1\end{array}\right]$

$|A|=3$ and $adjA=\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]$

$A^{-1}=\frac{adjA}{|A|}=\frac{1}{3}\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]$

$A^{-3}=(A^{-1}{)}^3=\frac{1}{27}\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]\left[\begin{array}{cc}1& -2\\ 0& 3\end{array}\right]$

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

Hence, option C.