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 6.3: 44

Answer

$\left[\begin{array}{ll} 28 & 12\\ -56 & -24\\ -7 & -3 \end{array}\right]$

Work Step by Step

A product of two matrices exists if the first has as many columns as the second matrix has rows. The product of an $m\times\underline{n}$ matrix $A$ and an $\underline{n}\times p$ matrix $B$ is an $m\times p$ matrix $AB$. The product $CB$ exists because $C$ is a 2$\times\underline{2}$ matrix, and $B$ is a $\underline{2}\times$2 matrix. ($CB$) is 2$\times$2. The product of A ( a $3\times\underline{2}$ matrix) and $CB$ ( a $\underline{2}\times$2 matrix) exists and is a $3\times 2$ matrix. $CB=\left[\begin{array}{ll} 1(5)-1(-2) & 1(1)-1(-2)\\ -1(5)+1(-2) & -1(1)+1(-2) \end{array}\right] =\left[\begin{array}{ll} 7 & 3\\ -7 & -3 \end{array}\right]$ $A(CB)=\left[\begin{array}{ll} 4 & 0\\ -3 & 5\\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 7 & 3\\ -7 & -3 \end{array}\right]$ $=\left[\begin{array}{ll} 28+0 & 12+0\\ -21-35 & -9-15\\ 0-7 & 0-3 \end{array}\right]$ $=\left[\begin{array}{ll} 28 & 12\\ -56 & -24\\ -7 & -3 \end{array}\right]$
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