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 - 3.1 The Definition of the Determinant - Problems - Page 207: 29

Answer

\begin{aligned} \operatorname{det}(A) &=\left|\begin{array}{ccc}{9} & {1} & {-7} \\ {6} & {2} & {1} \\ {-4} & {0} & {-2}\end{array}\right| =-84 \end{aligned}

Work Step by Step

Given $$ A=\left[ \begin{array}{ccc}{9} & {1} & {-7} \\ {6} & {2} & {1} \\ {-4} & {0} & {-2}\end{array} \right]$$ 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}( 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_{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}{9} & {1} & {-7} \\ {6} & {2} & {1} \\ {-4} & {0} & {-2}\end{array}\right| \\ &=9 \cdot 2 \cdot(-2)+1 \cdot 1 \cdot(-4)+(-7) \cdot 6 \cdot 0-9 \cdot 1 \cdot 0-1 \cdot 6 \cdot(-2)-(-7) \cdot 2 \cdot(-4) \\ &=-36-4-0-0+12-56=-84 \end{aligned}
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