Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 932: 16

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} -1 & 3 \\ -\frac{1}{2} & 1 \\ \end{matrix} \right]$

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