College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Matrices and Determinants - Exercise Set 6.4 - Page 640: 38

Answer

a. $\left[\begin{array}{lll} 1 & 2 & 5\\ 2 & 3 & 8\\ -1 & 1 & 2 \end{array}\right]\left[\begin{array}{l} x\\ y\\ z \end{array}\right]=\left[\begin{array}{l} 2\\ 3\\ 3 \end{array}\right]$ b. Solution set: $\{( -2$ , $-3$ , $2 ) \}$

Work Step by Step

$a.$ $A=\left[\begin{array}{lll} 1 & 2 & 5\\ 2 & 3 & 8\\ -1 & 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{l} 2\\ 3\\ 3 \end{array}\right]$ $AX=B$ $\left[\begin{array}{lll} 1 & 2 & 5\\ 2 & 3 & 8\\ -1 & 1 & 2 \end{array}\right]\left[\begin{array}{l} x\\ y\\ z \end{array}\right]=\left[\begin{array}{l} 2\\ 3\\ 3 \end{array}\right]$ $b.\qquad X=A^{-1}B$ ( given the matrix $A^{-1}$) $X=\left[\begin{array}{lll} 2 & -1 & -1\\ 12 & -7 & -2\\ -5 & 3 & 1 \end{array}\right]\left[\begin{array}{l} 2\\ 3\\ 3 \end{array}\right]=\left[\begin{array}{l} 2(2)-1(3)-1(3)\\ 12(2)-7(3)-2(3)\\ -5(2)+3(3)+1(3) \end{array}\right]=\left[\begin{array}{l} -2\\ -3\\ 2 \end{array}\right]$ Solution set: $\{( -2$ , $-3$ , $2 ) \}$
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