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: 23

Answer

$$ \left[ \begin {array}{cccc} 1&2&1&-1\\ 0&1&0&2\\0&0&-1&-5\\0&0&0&-19\end{array} \right]=19. $$

Work Step by Step

Given $$ \left[ \begin {array}{cccc} 1&2&1&-1\\ 0&1&0&2\\0&3&-1&1\\0&0&4&1\end{array} \right]. $$ Now, multiplying the second row by $-3$ and adding it to the third row, we get $$ \left[ \begin {array}{cccc} 1&2&1&-1\\ 0&1&0&2\\0&0&-1&-5\\0&0&4&1\end{array} \right]. $$ Multiplying the third row by $4$ and adding it to the fourth row, we get $$ \left[ \begin {array}{cccc} 1&2&1&-1\\ 0&1&0&2\\0&0&-1&-5\\0&0&0&-19\end{array} \right]. $$ Hence, it is very easy to note that $$ \left[ \begin {array}{cccc} 1&2&1&-1\\ 0&1&0&2\\0&0&-1&-5\\0&0&0&-19\end{array} \right]=19. $$
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