## College Algebra (6th Edition)

Published by Pearson

# Chapter 6 - Matrices and Determinants - Exercise Set 6.3: 30

#### Answer

a. $AB$=$\left[\begin{array}{lll} -1 & -2 & -3\\ -2 & -4 & -6\\ -3 & -6 & -9 \end{array}\right]$ b. $BA=[-14]$

#### 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. ----------------- a. $A$ is a $3\times\underline{1}$ matrix, B is a $\underline{1}\times 3$ matrix $AB$ exists, and is a $3\times 3$ matrix. $AB=\left[\begin{array}{lll} -1(1) & -1(2) & -1(3)\\ -2(1) & -2(2) & -2(3)\\ -3(1) & -3(2) & -3(1) \end{array}\right]$=$\left[\begin{array}{lll} -1 & -2 & -3\\ -2 & -4 & -6\\ -3 & -6 & -9 \end{array}\right]$ b. $B$ is a $1\times\underline{3}$ matrix, $A$ is a $\underline{3}\times 1$ matrix $BA$ exists, and is a $1\times 1$ matrix (single number). $BA=[1(-1)+2(-2)+3(-3)]=[-1-4-9]$ $BA=[-14]$

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