Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 934: 90

The inverse of the inverse of $A$ is $A$ itself.

Consider the given matrix, $A=\left[ \begin{matrix} 3 & 5 \\ 2 & 4 \\ \end{matrix} \right]$ The inverse of matrix $A$ is equal to the following, ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Compare the matrix to the original matrix. So, $\begin{align} & a=3 \\ & b=5 \\ & c=2 \\ & d=4 \end{align}$ The inverse is, ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Substitute the values to get, $\begin{align} & {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right] \\ & =\frac{1}{\left( 3 \right)\left( 4 \right)-\left( 5 \right)\left( 2 \right)}\left[ \begin{matrix} 4 & -5 \\ -2 & 3 \\ \end{matrix} \right] \\ & =\frac{1}{12-10}\left[ \begin{matrix} 4 & -5 \\ -2 & 3 \\ \end{matrix} \right] \\ & =\frac{1}{2}\left[ \begin{matrix} 4 & -5 \\ -2 & 3 \\ \end{matrix} \right] \\ & {{A}^{-1}}=\left[ \begin{matrix} 2 & -\frac{5}{2} \\ -1 & \frac{3}{2} \\ \end{matrix} \right] \end{align}$ Therefore, the inverse of the matrix ${{A}^{-1}}$ is $\left[ \begin{matrix} 2 & -\frac{5}{2} \\ -1 & \frac{3}{2} \\ \end{matrix} \right]$. Now, the inverse of ${{A}^{-1}}$ is: $\begin{align} & a=2 \\ & b=-\frac{5}{2} \\ & c=-1 \\ & d=\frac{3}{2} \\ \end{align}$ Simplifying it for the inverse, $\begin{align} & {{\left[ {{A}^{-1}} \right]}^{-1}}=\frac{1}{\left( 2 \right)\left( \frac{3}{2} \right)-\left( -\frac{5}{2} \right)\left( -1 \right)}\left[ \begin{matrix} \frac{3}{2} & \frac{5}{2} \\ 1 & 2 \\ \end{matrix} \right] \\ & =\frac{1}{\frac{1}{2}}\left[ \begin{matrix} \frac{3}{2} & \frac{5}{2} \\ 1 & 2 \\ \end{matrix} \right] \\ & {{\left[ {{A}^{-1}} \right]}^{-1}}=\frac{2}{1}\left[ \begin{matrix} \frac{3}{2} & \frac{5}{2} \\ 1 & 2 \\ \end{matrix} \right] \end{align}$ Which equals the original matrix $A$.