Question

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

Answer 1

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