## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 932: 25

#### Answer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} -3 & 2 & -4 \\ -1 & 1 & -1 \\ 8 & -5 & 10 \\ \end{matrix} \right]$.

#### Work Step by Step

Consider the given matrix $A=\left[ \begin{matrix} 5 & 0 & 2 \\ 2 & 2 & 1 \\ -3 & 1 & -1 \\ \end{matrix} \right]$. Compute matrix in the form of: $\left[ \left. A \right|I \right]$ The augment with identity matrix is: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 5 & 0 & 2 & 1 & 0 & 0 \\ 2 & 2 & 1 & 0 & 1 & 0 \\ -3 & 1 & -1 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, by using the row operations we will reduce the matrix into row-echelon form for the inverse as below: \begin{align} & {{R}_{2}}\to {{R}_{2}}-\frac{2}{5}\times {{R}_{1}}, \\ & {{R}_{3}}\to {{R}_{3}}+\frac{3}{5}\times {{R}_{1}}, \\ & {{R}_{3}}\to {{R}_{3}}-\frac{1}{2}\times {{R}_{2}} \\ \end{align} The resulting matrix is: $\left[ \begin{matrix} 5 & 0 & 2 & 1 & 0 & 0 \\ 0 & 2 & \frac{1}{5} & -\frac{2}{5} & 1 & 0 \\ 0 & 0 & \frac{1}{10} & \frac{4}{5} & -\frac{1}{2} & 1 \\ \end{matrix} \right]$ Again, apply the row operations as below: \begin{align} & {{R}_{3}}\to 10\times {{R}_{3}}, \\ & {{R}_{2}}\to {{R}_{2}}-\frac{1}{5}\times {{R}_{3}}, \\ & {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{3}}, \\ & {{R}_{2}}\to \frac{1}{2}\times {{R}_{2}}, \\ & {{R}_{1}}\to \frac{1}{5}\times {{R}_{1}} \\ \end{align} \begin{align} & \left[ \left. A \right|I \right]=\left[ \begin{matrix} 1 & 0 & 0 & -3 & 2 & -4 \\ 0 & 1 & 0 & -1 & 1 & -1 \\ 0 & 0 & 1 & -8 & -5 & 10 \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align} Where ${{A}^{-1}}=\left[ B \right]$ So, The inverse of matrix is ${{A}^{-1}}=\left[ \begin{matrix} -3 & 2 & -4 \\ -1 & 1 & -1 \\ 8 & -5 & 10 \\ \end{matrix} \right]$ Now, check the result for $A{{A}^{-1}}={{I}_{3}}$ And, ${{A}^{-1}}A={{I}_{3}}$ Here, $A=\left[ \begin{matrix} 5 & 0 & 2 \\ 2 & 2 & 1 \\ -3 & 1 & -1 \\ \end{matrix} \right]$ And, ${{A}^{-1}}=\left[ \begin{matrix} -3 & 2 & -4 \\ -1 & 1 & -1 \\ 8 & -5 & 10 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 5 & 0 & 2 \\ 2 & 2 & 1 \\ -3 & 1 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} -3 & 2 & -4 \\ -1 & 1 & -1 \\ -8 & -5 & 10 \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} \left( -15 \right)+0+16 & 10+0+\left( -10 \right) & \left( -20 \right)+0+20 \\ \left( -6 \right)+\left( -2 \right)+8 & 4+2+\left( -5 \right) & \left( -8 \right)+\left( -2 \right)+10 \\ 9+\left( -1 \right)+\left( -8 \right) & \left( -6 \right)+1+5 & 12+\left( -1 \right)+\left( -10 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} And, \begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} -3 & 2 & -4 \\ -1 & 1 & -1 \\ 8 & -5 & 10 \\ \end{matrix} \right]\left[ \begin{matrix} 5 & 0 & 2 \\ 2 & 2 & 1 \\ -3 & 1 & -1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \left( -15 \right)+4+12 & 0+4+\left( -4 \right) & \left( -6 \right)+2+4 \\ \left( -5 \right)+2+3 & 0+2+\left( -1 \right) & \left( -2 \right)+1+1 \\ 40+\left( -10 \right)+\left( -30 \right) & 0+\left( -10 \right)+10 & 16+\left( -5 \right)+\left( -10 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{3}} \end{align} Thus, $A{{A}^{-1}}={{I}_{3}}$ and ${{A}^{-1}}A={{I}_{_{3}}}$.

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