## College Algebra (6th Edition)

Published by Pearson

# Chapter 6 - Matrices and Determinants - Exercise Set 6.4 - Page 639: 3

#### Answer

$B$ is not the multiplicative inverse of $A$.

#### Work Step by Step

(See p.629) Let A be an $n\times n$ square matrix. If there is a square matrix $A^{-1}$ such that $AA^{-1}=I_{n}$ and $A^{-1}A=I_{n},$ then $A^{-1}$ is the multiplicative inverse of $A$. -------------- The products AB and BA should both be $I_{2}=\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right].$ $AB=\left[\begin{array}{ll} -4(-2)+0(0) & -4(4)+0(1)\\ 1(-2)+3(0) & 1(4)+3(1) \end{array}\right]==\left[\begin{array}{ll} 8 & ...\\ ... & ... \end{array}\right]\neq I_{2}$ Since at least one of the products is NOT $I_{2}$, $B$ is not the multiplicative inverse of $A$.

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