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 118: 29

Answer

$$ \left|\begin{array}{rrr}{4} & {3} & {-2} \\ {5} & {4} & {1} \\ {-2} & {3} & {4}\end{array}\right|=6\left|\begin{array}{rrr}{2} & {-3} & {-2} \\ {7} & {-10} & {0} \\ {1} & {0} & {0}\end{array}\right|=-60. $$

Work Step by Step

Given $$ \left|\begin{array}{rrr}{4} & {3} & {-2} \\ {5} & {4} & {1} \\ {-2} & {3} & {4}\end{array}\right|. $$ Adding the first row to $2$ times the second row, adding $2$ times the first row to the third row, we get $$ \left|\begin{array}{rrr}{4} & {3} & {-2} \\ {14} & {11} & {0} \\ {6} & {9} & {0}\end{array}\right|. $$ Factor $2$ out of the first column and $3$ out of the third row, we have $$ \left|\begin{array}{rrr}{4} & {3} & {-2} \\ {14} & {11} & {0} \\ {6} & {9} & {0}\end{array}\right|=6\left|\begin{array}{rrr}{2} & {3} & {-2} \\ {7} & {11} & {0} \\ {1} & {3} & {0}\end{array}\right|. $$ Adding $-3$ times the first column to the second column, we have $$ 6\left|\begin{array}{rrr}{2} & {-3} & {-2} \\ {7} & {-10} & {0} \\ {1} & {0} & {0}\end{array}\right|. $$ Finally, we have $$ \left|\begin{array}{rrr}{4} & {3} & {-2} \\ {5} & {4} & {1} \\ {-2} & {3} & {4}\end{array}\right|=6\left|\begin{array}{rrr}{2} & {-3} & {-2} \\ {7} & {-10} & {0} \\ {1} & {0} & {0}\end{array}\right|=-60. $$
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