Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 3 - Determinants - 3.2 Properties of Determinants - Problems - Page 219: 12

Answer

$\det A=84$

Work Step by Step

Given $$\begin{array}{c}{A=\left[\begin{array}{ccc} 7 & {-1} & 3&4 \\ 14 & {2} & 4&6\\ 21&1 &3& 4\\ -7&4 &5 &8\end{array}\right] }\end{array} $$ So, we get $$A=\begin{vmatrix} 7 & {-1} & 3&4 \\ 14 & {2} & 4&6\\ 21&1 &3& 4\\ -7&4 &5 &8 \end{vmatrix} \xrightarrow{R_2-2R_1,R_3-3R_{1},R_4+R_1} \begin{vmatrix} 7 & {-1} & 3&4 \\ 0 & {4} & -2&-2\\ 0&4 &-6& -8\\ 0&3 &8 &12 \end{vmatrix} \xrightarrow{R_{3}-R_{2},R_4-\frac{3}{2}R_2,\frac{9}{8}R_4+\frac{19}{8}R_3 }\begin{vmatrix} 7 & {-1} & 3&4 \\ 0 & {4} & -2&-2\\ 0&0 &-4& -6\\ 0&0 &0 &-\frac{3}{4} \end{vmatrix} $$ \begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=7 \cdot 4\cdot(-4)\cdot (-\frac{3}{4})=84}\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.