Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 932: 14

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{6} & \frac{1}{4} \\ \frac{1}{3} & 0 \\ \end{matrix} \right]$

Consider the given matrix $ A=\left[ \begin{matrix} 0 & 3 \\ 4 & -2 \\ \end{matrix} \right]$ Now, by using the inverse formula, we get: ${{A}^{-1}}=\frac{1}{\left| ad-bc \right|}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Let, $\begin{align} & a=0 \\ & b=3 \\ & c=4 \\ & d=-2 \end{align}$ Substitute the values to get $\begin{align} & {{A}^{-1}}=\frac{1}{\left| ad-bc \right|}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right] \\ & {{A}^{-1}}=\frac{1}{\left| 0\times \left( -2 \right)-3\times 4 \right|}\left[ \begin{matrix} -2 & -3 \\ -4 & 0 \\ \end{matrix} \right] \\ & =\frac{1}{-12}\left[ \begin{matrix} -2 & -3 \\ -4 & 0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{-2}{-12} & \frac{-3}{-12} \\ \frac{-4}{-12} & \frac{0}{-12} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{1}{6} & \frac{1}{4} \\ \frac{1}{3} & 0 \\ \end{matrix} \right] \end{align}$ So, therefore the inverse of the matrix is ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{6} & \frac{1}{4} \\ \frac{1}{3} & 0 \\ \end{matrix} \right]$ Now, check the result for $ A{{A}^{-1}}={{I}_{2}}$ And ${{A}^{-1}}A={{I}_{2}}$ Here, $ A=\left[ \begin{matrix} 0 & 3 \\ 4 & -2 \\ \end{matrix} \right]$ So, $\begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 0 & 3 \\ 4 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} \frac{1}{6} & \frac{1}{4} \\ \frac{1}{3} & 0 \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 0\times \frac{1}{6}+3\times \frac{1}{3} & 0\times \frac{1}{4}+3\times 0 \\ 4\times \frac{1}{6}+\left( -2 \right)\times \frac{1}{3} & 4\times \frac{1}{4}+\left( -2 \right)\times 0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align}$ This implies that ${{A}^{-1}}A={{I}_{2}}$ Now, evaluate the product ${{A}^{-1}}A={{I}_{2}}$. $\begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} \frac{1}{6} & \frac{1}{4} \\ \frac{1}{3} & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 3 \\ 4 & -2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{1}{6}\times 0+\frac{1}{4}\times 4 & \frac{1}{6}\times 3+\frac{1}{4}\times \left( -2 \right) \\ \frac{1}{3}\times 0+0\times 4 & \frac{1}{3}\times 3+0\times \left( -2 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align}$