## College Algebra (6th Edition)

Published by Pearson

# Chapter 6 - Matrices and Determinants - Exercise Set 6.3 - Page 625: 39

#### Answer

$\left[\begin{array}{ll} 11 & -1\\ -7 & -3 \end{array}\right]$

#### Work Step by Step

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 element in the ith row and $j\mathrm{t}\mathrm{h}$ column of $AB$ is found by multiplying each element in the ith row of $A$ by the corresponding element in the $j\mathrm{t}\mathrm{h}$ column of $B$ and adding the products. ----------------- $B$ is a $2\times\underline{2}$ matrix, $C$ is a $\underline{2}\times 2$ matrix $BC$ exists, and is a $2\times 2$ matrix. $C$ is a $2\times\underline{2}$ matrix, $B$ is a $\underline{2}\times 2$ matrix $CB$ exists, and is a $2\times 2$ matrix. BC and CB have the same order, $2\times 2$, so their sum exists. $BC=\left[\begin{array}{ll} 5(1)+1(-1) & 5(-1)+1(1)\\ -2(1)-2(-1) & -2(-1)-2(1) \end{array}\right]=\left[\begin{array}{ll} 4 & -4\\ 0 & 0 \end{array}\right]$ $CB=\left[\begin{array}{ll} 1(5)-1(-2) & 1(1)-1(-2)\\ -1(5)+1(-2) & -1(1)+1(-2) \end{array}\right]=\left[\begin{array}{ll} 7 & 3\\ -7 & -3 \end{array}\right]$ $BC+CB=\left[\begin{array}{ll} 4+7 & -4+3\\ 0-7 & 0-3 \end{array}\right]=\left[\begin{array}{ll} 11 & -1\\ -7 & -3 \end{array}\right]$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.