Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 932: 28

Answer

The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{2} & 0 & 0 & -\frac{1}{4} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & \frac{1}{2} \\ \end{matrix} \right]$.

Work Step by Step

Consider the given matrix $ A=\left[ \begin{matrix} 2 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 2 \\ \end{matrix} \right]$. Compute the matrix of the form: $\left[ \left. A \right|I \right]$ The augmented matrix with identity is: $\left[ \left. A \right|I \right]=\left[ \begin{matrix} 2 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]$ Now, by using the row operations we will reduce the matrix into row-echelon form for the inverse as below: $\begin{align} & {{R}_{4}}\to \frac{1}{2}\times {{R}_{4}}, \\ & {{R}_{1}}\to {{R}_{1}}-1\times {{R}_{4}}, \\ & {{R}_{3}}\to 1\times {{R}_{3}}, \\ & {{R}_{1}}\to \frac{1}{2}\times {{R}_{1}} \\ \end{align}$ The resulting matrix is: $\begin{align} & \left[ \left. A \right|I \right]=\left[ \begin{matrix} 1 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & -\frac{1}{4} \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & \frac{1}{2} \\ \end{matrix} \right] \\ & =\left[ \left. I \right|B \right] \end{align}$ Where ${{A}^{-1}}=\left[ B \right]$ So, the inverse of the matrix is: ${{A}^{-1}}=\left[ \begin{matrix} \frac{1}{2} & 0 & 0 & -\frac{1}{4} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & \frac{1}{2} \\ \end{matrix} \right]$ Now, check the result for $ A{{A}^{-1}}={{I}_{4}}$ And ${{A}^{-1}}A={{I}_{4}}$ Here, $ A=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 1 & 0 & 0 & 1 \\ \end{matrix} \right]$ and ${{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ -1 & 0 & 0 & 1 \\ \end{matrix} \right]$ So, $\begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 1 & 0 & 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ -1 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\times 1+0+0+0 & 0+0+0+0 & 0+0+0+0 & 0+0+0+0 \\ 0+0+0+0 & 0+\left( -1 \right)+0+0 & 0+0+0+0 & 0+0+0+0 \\ 0+0+0+0 & 0+0+0+0 & 0+0+3\times \frac{1}{3}+0 & 0+0+0+0 \\ 1+0+0+1\times \left( -1 \right) & 0+0+0+0 & 0+0+0+0 & 0+0+0+1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{4}} \end{align}$ And, $\begin{align} & {{A}^{-1}}A=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ -1 & 0 & 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 1 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1+0+0+0 & 0+0+0+0 & 0+0+0+0 & 0+0+0+0 \\ 0+0+0+0 & 0+1+0+0 & 0+0+3+0 & 0+0+0+0 \\ 0+0+0+0 & 0+0+0+0 & 0+0+1+0 & 0+0+0+0 \\ \left( -1 \right)+0+0+1 & 0+0+0+0 & 0+0+0+0 & 0+0+0+1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{4}} \end{align}$ Hence, $ A{{A}^{-1}}={{I}_{4}}$ and ${{A}^{-1}}A={{I}_{4}}$.
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