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

Question

If $A=\left[\begin{array}{ll}3 & 2 \ 0 & 1\end{array}\right]$, then $\left(A^{-1}\right)^3$ is equal to, $\frac{1}{27}\left[\begin{array}{cc}1 & -26 \ 0 & 27\end{array}\right]$, $\frac{1}{27}\left[\begin{array}{cc}-1 & -26 \ 0 & 27\end{array}\right]$, $\frac{1}{27}\left[\begin{array}{ll}1 & -26 \ 0 & -27\end{array}\right]$, $\frac{1}{27}\left[\begin{array}{cc}-1 & -26 \ 0 & -27\end{array}\right]$

MARKS App · Accepted Answer

$|\mathrm{A}|=\left|\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right|=3-0=3 \neq 0 \quad \therefore \mathrm{A}^{-1} \text { exists }$ Adj. $A=\left[\begin{array}{cc}1 & 0 \ -2 & 3\end{array}\right]^{\prime}=\left[\begin{array}{cc}1 & -2 \ 0 & 3\end{array}\right]$ $\therefore \quad \mathrm{A}^{-1}=\frac{1}{3}\left[\begin{array}{cc} 1 & -2 \ 0 & 3 \end{array}\right]$ $\begin{aligned} & \therefore\left(A^{-1}\right)^3=\frac{1}{27}\left[\begin{array}{cc} 1 & -2 \ 1 & 3 \end{array}\right]\left[\begin{array}{cc} 1 & -2 \ 0 & 3 \end{array}\right]\left[\begin{array}{cc} 1 & -2 \ 0 & 3 \end{array}\right] \ & =\frac{1}{27}\left[\begin{array}{cc} 1 & -8 \ 0 & 9 \end{array}\right]\left[\begin{array}{cc} 1 & -2 \ 0 & 3 \end{array}\right]=\frac{1}{27}\left[\begin{array}{cc} 1 & -26 \ 0 & 27 \end{array}\right] \end{aligned}$

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