Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 3 - Determinants - 3.1 Exercises - Page 170: 13

Answer

$6$

Work Step by Step

To compute the determinant efficiently, we expand the original matrix down the second column, the resulting matrix across the second row, and the third ($3\times3$) matrix down the first column. Only a single term is non-vanishing in each of the first two expansions. Note the initial negative sign because the sum of the indices of the $(3,2)$ position is odd. $ \begin{align} \det A &=-3\begin{vmatrix}4&-7&3&-5\\0&2&0&0\\5&5&2&-3\\0&9&-1&2\end{vmatrix} =-3\cdot2\begin{vmatrix}4&3&-5\\5&2&-3\\0&-1&2\end{vmatrix}\\ &=-6\cdot(4\begin{vmatrix}2&-3\\-1&2\end{vmatrix}-5\begin{vmatrix}3&-5\\-1&2\end{vmatrix})\\ &=-6\cdot(4-5)=6 \end{align} $
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