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

Published by Pearson

# Chapter 3 - Determinants - 3.1 The Definition of the Determinant - Problems - Page 207: 36

#### Answer

$$\operatorname{det}(A)=\left| \begin{array}{cccc}{4} & {1} & {8} & {6} \\ {0} & {-2} & {13} & {5} \\ {0} & {0} & {-6} & {-3} \\ {0} & {0} & {0} & {-3}\end{array} \right|=-144$$

#### Work Step by Step

Given $$A=\left[ \begin{array}{cccc}{4} & {1} & {8} & {6} \\ {0} & {-2} & {13} & {5} \\ {0} & {0} & {-6} & {-3} \\ {0} & {0} & {0} & {-3}\end{array} \right]$$ since,if we have \begin{aligned}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] \end{aligned} So, 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}{cccc}{4} & {1} & {8} & {6} \\ {0} & {-2} & {13} & {5} \\ {0} & {0} & {-6} & {-3} \\ {0} & {0} & {0} & {-3}\end{array}\right|, \ \\ & \text{from fourth row, we obtain}\\ \operatorname{det}(A)&=-3 \cdot\left|\begin{array}{ccc}{4} & {1} & {8} \\ {0} & {-2} & {13} \\ {0} & {0} & {-6}\end{array}\right|\\ &=-3 \cdot(4 \cdot(-2) \cdot(-6)+1 \cdot 13 \cdot 0+8 \cdot 0 \cdot 0-4 \cdot 13 \cdot 0-1 \cdot 0 \cdot(-6)-8 \cdot(-2) \cdot 0)\\ & =3 \cdot(48+0+0-0+0+0)\\ &=-3 \cdot 48\\ &=-144 \end{aligned}

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