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: 7

Answer

-72

Work Step by Step

Given $$\begin{array}{c}{A=\left[\begin{array}{ccc} {1} & {-1} & 2&4 \\ {3} & {1} & 2 &4 \\ -1& 1 &1& 3\\ 2 &1 & 4 &2\end{array}\right] }\end{array} $$ So, we get $$A=\begin{vmatrix} {1} & {-1} & 2&4 \\ 3& 1 & 2 &4 \\ -1& 1 &1& 3\\ 2 &1 & 4 &2 \end{vmatrix} \xrightarrow{R_{2}-3R_{1} ,R_3+R_{1},R_4-2R_1} \begin{vmatrix} 1 & -1 & 2&4 \\ 0& 4 & -4 &-8 \\ 0& 0&5& 6\\ 0 &3 & 0&-6 \end{vmatrix} \xrightarrow{R_{4}-\frac{3}{4}R_{2}} \begin{vmatrix} 1 & -1 & 2&4 \\ 0& 4 & -4 &-8 \\ 0& 0&5& 6\\ 0 &0 & 3&0 \end{vmatrix} \xrightarrow{R_{4}-\frac{3}{5}R_{3}} \begin{vmatrix} 1 & -1 & 2&4 \\ 0& 4 & -4 &-8 \\ 0& 0&5& 6\\ 0 &0 & 0&-\frac{18}{5} \end{vmatrix}$$ \begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=1 \cdot 4 \cdot 5\cdot-\frac{18}{5}=-72}\end{array}
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