Given $A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]$, compute $A^{-1}$ and show that $2 A^{-1}=9 I-A$
We have, $A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]$
Now, $\operatorname{adj}(A)=\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]$
and $|A|=2$
$\therefore A^{-1}=\frac{1}{2}\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]$
Now, $2 A^{-1}=9 I-A$
LHS $=2 A^{-1}=\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]$
$\mathrm{RHS}=9 I-A=9\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]=\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]=\mathrm{LHS}$
Hence proved.
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