Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 4 - Review - Review Exercises - Page 296: 14

Answer

- The inverse of the given matrix is: \[ \left( \begin{array}{ccc} \frac 23 & \frac 13 & -\frac {10}3 & \frac {10}3\\ \frac 13 & -\frac 13 & \frac 43 & -\frac 73 \\ -\frac 23 & \frac 23 & -\frac 23 & \frac 53 \\ \frac 1 3 & -\frac 13 & \frac 13 & -\frac 13 \end{array} \right)\]

Work Step by Step

1. Put an identity $I$ matrix on the right of the given matrix, to get a $4 \times 8$ one. \[ \left( \begin{array}{ccc} 1 & 2 & 3 & 4 & 1 & 0 & 0 & 0\\ 1 & 3 & 4 & 2 & 0 & 1 & 0 & 0\\ 0 & 1 & 2 & 3 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 & 0 & 1 \end{array} \right)\] 2. Row-reduce the whole matrix: $R_2 = R_2 - R_1$: \[ \left( \begin{array}{ccc} 1 & 2 & 3 & 4 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & -2 & -1 & 1 & 0 & 0\\ 0 & 1 & 2 & 3 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 & 0 & 1 \end{array} \right)\] $R_1 = R_1 - 2R_2$ $R_3 = R_3 - R_2$ \[ \left( \begin{array}{ccc} 1 & 0 & 1 & 8 & 3 & -2 & 0 & 0\\ 0 & 1 & 1 & -2 & -1 & 1 & 0 & 0\\ 0 & 0 & 1 & 5 & 1 & -1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 & 0 & 1 \end{array} \right)\] $R_1 = R_1 - R_3$ $R_2 = R_2 - R_3$ $R_4 = R_4 - R_3$ \[ \left( \begin{array}{ccc} 1 & 0 & 0 & 10 & 4 & -3 & 0 & 0\\ 0 & 1 & 0 & -7 & -2 & 2 & -1 & 0\\ 0 & 0 & 1 & 5 & 1 & -1 & 1 & 0 \\ 0 & 0 & 0 & -3 & -1 & 1 & -1 & 1 \end{array} \right)\] $R_1 = R_1 + \frac{10} 3R_4$ $R_2 = R_2 - \frac 73R_4$ $R_3 = R_3 + \frac 53 R_4$ \[ \left( \begin{array}{ccc} 1 & 0 & 0 & 0 & \frac 23 & \frac 13 & -\frac {10}3 & \frac {10}3\\ 0 & 1 & 0 & 0 & \frac 13 & -\frac 13 & \frac 43 & -\frac 73 \\ 0 & 0 & 1 & 0 & -\frac 23 & \frac 23 & -\frac 23 & \frac 53 \\ 0 & 0 & 0 & -3 & -1 & 1 & -1 & 1 \end{array} \right)\] $R_4 = -\frac 13R_4$ \[ \left( \begin{array}{ccc} 1 & 0 & 0 & 0 & \frac 23 & \frac 13 & -\frac {10}3 & \frac {10}3\\ 0 & 1 & 0 & 0 & \frac 13 & -\frac 13 & \frac 43 & -\frac 73 \\ 0 & 0 & 1 & 0 & -\frac 23 & \frac 23 & -\frac 23 & \frac 53 \\ 0 & 0 & 0 & 1 & \frac 1 3 & -\frac 13 & \frac 13 & -\frac 13 \end{array} \right)\] - Thus, the inverse of the given matrix is: \[ \left( \begin{array}{ccc} \frac 23 & \frac 13 & -\frac {10}3 & \frac {10}3\\ \frac 13 & -\frac 13 & \frac 43 & -\frac 73 \\ -\frac 23 & \frac 23 & -\frac 23 & \frac 53 \\ \frac 1 3 & -\frac 13 & \frac 13 & -\frac 13 \end{array} \right)\]
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