## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 932: 13

#### Answer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} \frac{2}{7} & \frac{-3}{7} \\ \frac{1}{7} & \frac{2}{7} \\ \end{matrix} \right]$.

#### Work Step by Step

Consider the given matrix $A=\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]$ 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=3 \\ & 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 2-3\times \left( -1 \right) \right|}\left[ \begin{matrix} 2 & -3 \\ 1 & 2 \\ \end{matrix} \right] \\ & =\frac{1}{4+3}\left[ \begin{matrix} 2 & -3 \\ 1 & 2 \\ \end{matrix} \right] \\ & =\frac{1}{7}\left[ \begin{matrix} 2 & -3 \\ 1 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{2}{7} & \frac{-3}{7} \\ \frac{1}{7} & \frac{2}{7} \\ \end{matrix} \right] \end{align} So, therefore the inverse of the matrix is given by ${{A}^{-1}}=\left[ \begin{matrix} \frac{2}{7} & \frac{-3}{7} \\ \frac{1}{7} & \frac{2}{7} \\ \end{matrix} \right]$ Now, check the result for $A{{A}^{-1}}={{I}_{2}}$ And ${{A}^{-1}}A={{I}_{2}}$ Here, $A=\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} \frac{2}{7} & \frac{-3}{7} \\ \frac{1}{7} & \frac{2}{7} \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 2\times \frac{2}{7}+3\times \frac{1}{7} & 2\times \frac{-3}{7}+3\times \frac{2}{7} \\ \left( -1 \right)\times \frac{2}{7}+2\times \frac{1}{7} & \left( -1 \right)\times \frac{-3}{7}+2\times \frac{2}{7} \\ \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 ${{A}^{-1}}A={{I}_{2}}$.

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