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: 35

Answer

$-1100$

Work Step by Step

To find the value of the determinant, we can multiply the first row with (-2) and then add the output to the second row. Furthermore, we multiply the first row with (-2) and then we add the output to the third row. After that, we replace the second row with the fifth row $$ \left|\begin{array}{ccccc} 1 & -1 & 8 & 4 & 2 \\ 2 & 6 & 0 & -4 & 3 \\ 2 & 0 & 2 & 6 & 2 \\ 0 & 2 & 8 & 0&0 \\ 0 & 1 & 1 & 2 & 2 \end{array}\right|=\left|\begin{array}{ccccc} -1 & -1 & 8 & 4 & 2 \\ 0 & 1 & 1 & 2 & 2 \\ 0 & 2 &-14 & -2 & -2 \\ 0 & 2 & 8 & 0 & 0 \\ 0 & 8 & -16 & -12 & -1 \end{array}\right| $$ Multiply the second row with (-2) and then add the output to the third row. Next, we multiply the second row with (-2) and add the output to the fourth row. Then, we multiply the second row with (-8) and then add the output to the third row. Finally, we replace the third row with the fourth row $$ \left|\begin{array}{ccccc} 1 & -1 & 8 & 4 & 2 \\ 2 & 6 & 0 & -4 & 3 \\ 0 & 0 & 2 & 6 & 2 \\ 0 & 2 & 8 & 0 & 0 \\ 0 & 1 & 1 & 2 & 2 \end{array}\right|=\left|\begin{array}{cccc} 1 & -1 & 8 & 4 & 2 \\ 0 & 1 & 1 & 2 & 2 \\ 0 & 0 & 6 & -4 & -4 \\ 0 & 0 & -16 & -6 & -6 \\ 0 & 0 & -24 & -28 & -17 \\ \end{array}\right| $$ We can multiply the third row with (4) and then add the output to the fifth row. Next, we multiply the third row with $(16 / 6)$ and then we add the output to the fourth row $$ \left|\begin{array}{cccc} 1 & -1 & 8 & 4 & 2 \\ 2 & 6 & 0 & -4 & 3 \\ 2 & 0 & 2 & 6 & 2 \\ 0 & 2 & 8 & 0&0 \\ 0 & 1 & 1 & 2 & 2 \\ \end{array}\right|=\left|\begin{array}{cccc} 1 & -1 & 8 & 4 & 2 \\ 0 & 1 & 1 & 2 & 2 \\ 0 & 0 & 6 & -4 & -4 \\ 0 & 0 & 0 & -50/3 & -50/3 \\ 0 & 0 & 0 & -44 & -33 \\ \end{array}\right|\\=6 * (33 * \frac {50}{3}-44 * \frac {50}{3})=-1100 $$
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