## Precalculus (6th Edition) Blitzer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \\ \end{matrix} \right]$
Consider the given matrix $A=\left[ \begin{matrix} 3 & -1 \\ -4 & 2 \\ \end{matrix} \right]$ 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=3 \\ & b=-1 \\ & 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| 3\times 2-\left( -1 \right)\times \left( -4 \right) \right|}\left[ \begin{matrix} 2 & 1 \\ 4 & 3 \\ \end{matrix} \right] \\ & =\frac{1}{6+\left( -4 \right)}\left[ \begin{matrix} 2 & 1 \\ 4 & 3 \\ \end{matrix} \right] \\ & =\frac{1}{2}\left[ \begin{matrix} 2 & 1 \\ 4 & 3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \\ \end{matrix} \right] \end{align} So, therefore the inverse of the matrix is given by: ${{A}^{-1}}=\left[ \begin{matrix} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \\ \end{matrix} \right]$ Now, check the result for $A{{A}^{-1}}={{I}_{2}}$ And ${{A}^{-1}}A={{I}_{2}}$ Here, $A=\left[ \begin{matrix} 3 & -1 \\ -4 & 2 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 3 & -1 \\ -4 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & \frac{1}{2} \\ 2 & \frac{3}{2} \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 3\times 1+\left( -1 \right)\times 2 & 3\times \frac{1}{2}+\left( -1 \right)\times \frac{3}{2} \\ \left( -4 \right)\times 1+2\times 2 & \left( -4 \right)\times \frac{1}{2}+2\times \frac{3}{2} \\ \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 & \frac{1}{2} \\ 2 & \frac{3}{2} \\ \end{matrix} \right]\left[ \begin{matrix} 3 & -1 \\ -4 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\times 3+\frac{1}{2}\left( -4 \right) & 1\times \left( -1 \right)+\frac{1}{2}\times 2 \\ 2\times 3+\frac{3}{2}\times \left( -4 \right) & 2\times \left( -1 \right)+\frac{3}{2}\times 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align}