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: 14

Answer

See below

Work Step by Step

Given $$\begin{array}{c}{A=\left[\begin{array}{ccc} {1} & {2} & 3& 4 & 5\\ {5} & {4} & 3 & 2 & 1 \\ 2 & 3 & 4 & 5 & 6\\ 6& 5 & 4 & 3 & 2\\ 0 & 2 & 4 & 6 & 8 \end{array}\right] }\end{array} $$ So, we get $$A=\begin{vmatrix} {1} & {2} & 3& 4 & 5\\ {5} & {4} & 3 & 2 & 1 \\ 2 & 3 & 4 & 5 & 6\\ 6& 5 & 4 & 3 & 2\\ 0 & 2 & 4 & 6 & 8 \end{vmatrix} \rightarrow \begin{vmatrix} {1} & {2} & 3& 4 & 5\\ {0} & {-6} & -12 & -18 & -24 \\ 0 & -1 & -2 & -3 & -4\\ 0 & -7 & -14 & -21 & -28\\ 0 & 2 & 4 & 6 & 8 \end{vmatrix} \rightarrow \begin{vmatrix} {1} & {2} & 3& 4 & 5\\ {0} & {1} & 2 & 3 & 4 \\ 0 & -1 & -2 & -3 & -4\\ 0 & -7 & -14 & -21 & -28\\ 0 & 2 & 4 & 6 & 8 \end{vmatrix} \rightarrow -6\begin{vmatrix} {1} & {2} & 3& 4 & 5\\ {0} & {1} & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{vmatrix}$$ \begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=-6 \cdot 1 \cdot 1 \cdot 0 \cdot 0 \cdot 0=0}\end{array}
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