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: 33

Answer

136

Work Step by Step

To find the value of the determinant $\left|\begin{array}{cccc}1 & -2 & 7 & 9 \\ 3 & -4 & 5 & 5 \\ 3 & 6 & 1 & -1 \\ 4 & 5 & 3 & 2\end{array}\right|$ we multiply the first row with (-3) and add it to the second row: $$ \left|\begin{array}{cccc} 1 & -2 & 7 & 9 \\ 3 & -4 & 5 & 5 \\ 3 & 6 & 1 & -1 \\ 4 & 5 & 3 & 2 \end{array}\right|=\left|\begin{array}{cccc} 1 & -2 & 7 & 9 \\ 0 & 2 & -16 & -22 \\ 3 & 6 & 1 & -1 \\ 4 & 5 & 3 & 2 \end{array}\right| $$ Now, we multiply the first row with ( -3 ) and add it to the third row. Furthermore, we multiply the first row with (-4) and add it to the fourth row: $$ \left|\begin{array}{cccc} 1 & -2 & 7 & 9 \\ 3 & -4 & 5 & 5 \\ 3 & 6 & 1 & -1 \\ 4 & 5 & 3 & 2 \end{array}\right|=\left|\begin{array}{cccc} 1 & -2 & 7 & 9 \\ 0 & 2 & -16 & -22 \\ 0 & 12 & -20 & -28 \\ 0 & 13 & -25 & -34 \end{array}\right| $$ Taking as the common factor the number $\ 2$ from the second row and the common factor $\ 4$ from the third row, we get $$ \left|\begin{array}{cccc} 1 & -2 & 7 & 9 \\ 3 & -4 & 5 & 5 \\ 3 & 6 & 1 & -1 \\ 4 & 5 & 3 & 2 \end{array}\right|=8\left|\begin{array}{cccc} 1 & -2 & 7 & 9 \\ 0 & 1 & -8 & -11 \\ 0 & 3 & -5 & -7 \\ 0 & 13 & -25 & -34 \end{array}\right| $$ Multiply the second row with (-3) and add it to the third row. Next, we multiply the second row with (-13) and add it to the fourth row to obtain $$ \begin{array}{l} \left|\begin{array}{ccc} 1 & -2 & 7 & 9 \\ 3 & -4 & 5 & 5 \\ 3 & 6 & 1 & -1 \\ 4 & 5 & 3 & 2 \end{array}\right|=8\left|\begin{array}{cccc} 1 & -2 & 7 & 9 \\ 0 & 1 & -8 & -11 \\ 0 & 0 & 19 & 26 \\ 0 & 0 & 79 & 109 \end{array}\right| \\ =8 *(19 * 109-26 * 79)=8 * 17=136 \end{array} $$
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