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 - True-False Review - Page 206: i

Answer

True

Work Step by Step

Given $\begin{vmatrix} a_1 & a_2 &0 & 0\\ a_3 & a_4 & 0 & 0\\ 0 & 0 & b_1 & b_2\\ 0 & 0 & b_3 & b_4 \end{vmatrix}$ We have $\begin{vmatrix} a_1 & a_2 &0 & 0\\ a_3 & a_4 & 0 & 0\\ 0 & 0 & b_1 & b_2\\ 0 & 0 & b_3 & b_4 \end{vmatrix}\\ =a_1(a_4b_1b_4+0.b_2.0+0.0.b_3-(0.b_1.0+a_4b_2b_3+0.0.b_4))-a_2(a_3b_1b_4+0.b_2.0+0.0.b_3-(0.b_1.0+a_3b_2b_3+0.0.b_4))\\ =a_1(a_4b_1b_4-a_4b_2b_3)-a_2(a_3b_1b_4-a_3b_2b_3)\\ =a_1a_4b_1b_4-a_1a_4b_2b_3-a_2a_3b_1b_4+a_2a_3b_2b_3$ then $\begin{vmatrix} a_1 & a_2 \\ a_3 & a_4 \end{vmatrix}=a_1a_4-a_2a_3$ and $\begin{vmatrix} b_1 & b_2 \\ b_3 & b_4 \end{vmatrix}=b_1b_4-b_2b_3\\ \rightarrow \begin{vmatrix} a_1 & a_2 \\ a_3 & a_4 \end{vmatrix}\begin{vmatrix} b_1 & b_2 \\ b_3 & b_4 \end{vmatrix}\\=(a_1a_4-a_2a_3)(b_1b_2-b_2b_3)\\ =a_1a_4b_1b_4-a_1a_4b_2b_3-a_2a_3b_1b_4+a_2a_3b_2b_3$ Hence, the statement is true.
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