## Differential Equations and Linear Algebra (4th Edition)

Published by Pearson

# Chapter 3 - Determinants - 3.1 The Definition of the Determinant - Problems - Page 207: 28

#### Answer

\begin{aligned} \operatorname{det}(A) &=\left|\begin{array}{ccc}{0} & {0} & {-3} \\ {0} & {4} & {3} \\ {-2} & {1} & {5}\end{array}\right| =-24 \end{aligned}

#### Work Step by Step

Given $$A=\left[ \begin{array}{ccc}{0} & {0} & {-3} \\ {0} & {4} & {3} \\ {-2} & {1} & {5}\end{array} \right]$$ Since, if we have$$A=\left[\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right]$$ we get \begin{aligned}\operatorname{det}(A)&=\left|\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right|\\ &=a_{11}\left|\begin{array}{ll} {a_{22}} & {a_{23}} \\ {a_{32}} & {a_{33}}\end{array}\right|-a_{12}\left|\begin{array}{ll} {a_{21}} & {a_{23}} \\ {a_{31}} & {a_{33}}\end{array}\right|+a_{13}\left|\begin{array}{ll} {a_{21}} & {a_{22}} \\ {a_{31}} & {a_{32}}\end{array}\right|\\ &=a_{11}( a_{22} a_{33}-a_{23} a_{32})-a_{12}( a_{21} a_{33}-a_{23} a_{31})+a_{13}( a_{21} a_{32}-a_{22} a_{31})\\ &=a_{11} a_{22} a_{33}+a_{12} a_{23} a_{31}+a_{13} a_{21} a_{32}-a_{11} a_{23} a_{32}-a_{12} a_{21} a_{33}-a_{13} a_{22} a_{31}\\ \end{aligned} Therefore, we get \begin{aligned} \operatorname{det}(A) &=\left|\begin{array}{ccc}{0} & {0} & {-3} \\ {0} & {4} & {3} \\ {-2} & {1} & {5}\end{array}\right| \\ &=0 \cdot 4 \cdot 5+0 \cdot 3 \cdot(-2)+(-3) \cdot 0 \cdot 1-0 \cdot 3 \cdot(-1)-0 \cdot 0 \cdot 5-(-3) \cdot 4 \cdot(-2) \\ &=0-0-0+0-0-24\\ &=-24 \end{aligned}

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