Question

Find the inverse of the matrix $A=\left[\begin{array}{cc}a& b\\ c& \frac{1+bc}{a}\end{array}\right]$ and show that $a{A}^{-1}=\left(a^2+bc+1\right)I-aA.$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ A=\left[\begin{array}{cc}a& b\\ c& \frac{1+bc}{a}\end{array}\right] \\ \mathrm{So},\mathrm{adj}\left(A\right)=\left[\begin{array}{cc}\frac{1+bc}{a}& -b\\ -c& a\end{array}\right] \\ \mathrm{and}\left|A\right|=1 \\ \therefore A^{-1}=\left[\begin{array}{cc}\frac{1+bc}{a}& -b\\ -c& a\end{array}\right] \\ \mathrm{Now},a{A}^{-1}=\left(a^2+bc+1\right)I-aA \\ \mathrm{LHS}=a{A}^{-1}=a\left[\begin{array}{cc}\frac{1+bc}{a}& -b\\ -c& a\end{array}\right]=\left[\begin{array}{cc}1+bc& -ba\\ -ca& a^2\end{array}\right] \\ \mathrm{RHS}=\left(a^2+bc+1\right)I-aA \\ =a^2+bc+1\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}{a}^2& ba\\ ca& 1+bc\end{array}\right] \\ =\left[\begin{array}{cc}{a}^2+bc+1& 0\\ 0& a^2+bc+1\end{array}\right]-\left[\begin{array}{cc}{a}^2& ba\\ ca& 1+bc\end{array}\right] \\ =\left[\begin{array}{cc}1+bc& -ba\\ -ca& a^2\end{array}\right]=\mathrm{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$