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 218: 1

Answer

$$\text {det }A =-36$$

Work Step by Step

Given $$\begin{array}{c}{A=\left[\begin{array}{ccc}{1} & {2} & {3} \\ {2} & {6} & {4} \\ {3} & {-5} & {2}\end{array}\right] }\end{array} $$ So, we get $$A=\begin{array}{c}{\left[\begin{array}{ccc}{1} & {2} & {3} \\ {2} & {6} & {4} \\ {3} & {-5} & {2}\end{array}\right] \xrightarrow[ R_{3}-3 R_{1}]{R_{2}-2 R_{1 }} \left[\begin{array}{ccc}{1} & {2} & {3} \\ {0} & {2} & {-2} \\ {0} & {0} & {-18} \end{array} \right]} \xrightarrow[ ]{R_{3}+\frac{11}{2} R_{2}} \left[\begin{array}{ccc}{1} & {2} & {3} \\ {0} & {2} & {-2} \\ {0} & {0} & {-18}\end{array} \right] \end{array}$$ \begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=1 \cdot 2 \cdot(-18)=-36}\end{array}
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