Question

Given $A=\left[\begin{array}{cc}2& -3\\ -4& 7\end{array}\right]$, compute A⁻¹ and show that $2A^{-1}=9I-A.$

Answer 1

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