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: 40

Answer

$\left[\begin{array}{ll} 24 & 0\\ -33 & -5\\ -3 & -1 \end{array}\right]$

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 and C are both $2\times 2$ matrices, so their sum exists. $A$ is a $3\times\underline{2}$ matrix, $(B+C)$ is a $\underline{2}\times 2$ matrix $A(B+C)$ exists, and is a $3\times 2$ matrix. $B+C=\left[\begin{array}{ll} 5+1 & 1-1\\ -2-1 & -2+1 \end{array}\right]=\left[\begin{array}{ll} 6 & 0\\ -3 & -1 \end{array}\right]$ $A(B+C)=\left[\begin{array}{ll} 4 & 0\\ -3 & 5\\ 0 & 1 \end{array}\right]\cdot\left[\begin{array}{ll} 6 & 0\\ -3 & -1 \end{array}\right]=$ $=\left[\begin{array}{ll} 4(6)+0(-3) & 4(0)+0(-1)\\ -3(6)+5(-3) & -3(0)+5(-1)\\ 0(6)+1(-3) & 0(0)+1(-1) \end{array}\right]$ $=\left[\begin{array}{ll} 24 & 0\\ -33 & -5\\ -3 & -1 \end{array}\right]$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.