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 208: 52

Answer

See below

Work Step by Step

Given $B=\begin{bmatrix} 2 & 1 & 5 & 5 & 0\\0 & 2 & 0 & -1 & 5\\0 & 0 & 6 & 1 & 2\\0 & 1 & -1 & -1 & -2 & 3\\0 & 0 & 2 & 0 & -4 \end{bmatrix}$ then $\det (B)=\begin{bmatrix} 2 & 1 & 5 & 5 & 0\\0 & 2 & 0 & -1 & 5\\0 & 0 & 6 & 1 & 2\\0 & 1 & -1 & -1 & -2 & 3\\0 & 0 & 2 & 0 & -4 \end{bmatrix}\\ =2.\det(A)\\ =2.[(2.6.(-2).(-4))-1.(2.1.(-1).(-4))-1.(2.6.3.0)-1.(0.0.(-2).(-4))+(0.1.1.(-4))-1((-1).0.(-1).(-4))-1.((-1).6.1.(-4)-1.(2.2.(-2).2)-1.(5.6.(-2).0)+(0.0.3.0)+(-1.2.1.2)+(5.1.(-1).0)-1(0.1.(-2).0)+(2.1.3.2)-1.(5.0.(-1).0)-(2.2.(-1).0)+(0.2.(-2).0)-1((-1).0.3.2)+(5.0.(-2).2)-(0.2.1.0)+((-1).6.3.0)-1((-1).2.(-1).0)+(5.6.1.0)-1(5.1.1.2)\\ =2.78\\ =156$
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