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: 29

Answer

a. $AB=[30]$ b. $BA=\left[\begin{array}{llll} 1 & 2 & 3 & 4\\ 2 & 4 & 6 & 8\\ 3 & 6 & 9 & 12\\ 4 & 8 & 12 & 16 \end{array}\right]$

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. ----------------- a. $A$ is a $1\times\underline{4}$ matrix, B is a $\underline{4}\times 1$ matrix $AB$ exists, and is a $1\times 1$ matrix (single number). $AB=[1(1)+2(2)+3(3)+4(4)]=[1+4+9+16]$ $AB=[30]$ b. $B$ is a $4\times\underline{1}$ matrix, $A$ is a $\underline{1}\times 4$ matrix $BA$ exists, and is a $4\times 4$ matrix. $BA=\left[\begin{array}{llll} 1(1) & 1(2) & 1(3) & 1(4)\\ 2(1) & 2(2) & 2(3) & 2(4)\\ 3(1) & 3(2) & 3(3) & 3(4)\\ 4(1) & 4(2) & 4(3) & 4(4) \end{array}\right]=\left[\begin{array}{llll} 1 & 2 & 3 & 4\\ 2 & 4 & 6 & 8\\ 3 & 6 & 9 & 12\\ 4 & 8 & 12 & 16 \end{array}\right]$
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