Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 3 - Determinants - 3.2 Determinants and Elementary Operations - 3.2 Exercises - Page 119: 34

Answer

7441

Work Step by Step

To find the value of the determinant $$ \left|\begin{array}{cccc} 0 & -3 & 8 & 2 \\ 8 & 1 & -1 & 6 \\ -4 & 6 & 0 & 9 \\ -7 & 0 & 0 & 14 \end{array}\right| $$ we can replace the fourth row with the first row; then we have $$ \left|\begin{array}{cccc} 0 & -3 & 8 & 2 \\ 8 & 1 & -1 & 6 \\ -4 & 6 & 0 & 9 \\ -7 & 0 & 0 & 14 \end{array}\right|=-\left|\begin{array}{cccc} -7 & 0 & 0 & 14 \\ 8 & 1 & -1 & 6 \\ -4 & 6 & 0 & 9 \\ 0 & -3 & 8 & 2 \end{array}\right|=7\left|\begin{array}{cccc} 1 & 0 & 0 & -2 \\ 8 & 1 & -1 & 6 \\ -4 & 6 & 0 & 9 \\ 0 & -3 & 8 & 2 \end{array}\right| $$ Multiply the first row with (-8) and then add the output to the second row. Furthermore, we multiply the first row with (4) and add the output to the third row: $$ \left|\begin{array}{cccc} 0 & -3 & 8 & 2 \\ 8 & 1 & -1 & 6 \\ -4 & 6 & 0 & 9 \\ -7 & 0 & 0 & 14 \end{array}\right|=7\left|\begin{array}{cccc} 1 & 0 & 0 & -2 \\ 0 & 1 & -1 & 22 \\ 0 & 6 & 0 & 1 \\ 0 & -3 & 8 & 2 \end{array}\right| $$ Multiply the second row with (-6) and add the output to the third row. Furthermore, we multiply the second row with ( 3 ) and then add the output to the fourth row $$ \begin{array}{llll} \left|\begin{array}{ccc} 0 & -3 & 8 & 2 \\ 8 & 1 & -1 & 6 \\ -4 & 6 & 0 & 9 \\ -7 & 0 & 0 & 14 \end{array}\right|=7 \left|\begin{array}{ccccc} 1 & 0 & 0 & -2 \\ 0 & 1 & -1 & 22 \\ 0 & 0 & 6 & -131 \\ 0 & 0 & 5 & 68 \end{array}\right| \\ & \cr =7*1*1*[6 * 68+5 * 131]=7441 \end{array} $$
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