# Chapter 6 - Matrices and Determinants - Exercise Set 6.4 - Page 639: 7

$B$ is the multiplicative inverse of A. $(B=A^{-1})$

#### Work Step by Step

(See p.629) Let A be an $\mathrm{n}\times \mathrm{n}$ 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 $I_{3}=\left[\begin{array}{lll} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]$ $AB=\left[\begin{array}{lll} 0+1+0 & 0+0+0 & 0+0+0\\ 0+0+0 & 0+0+1 & 0+0+0\\ 0+0+0 & 0+0+0 & 1+0+0 \end{array}\right]=I_{3}$ $BA=\left[\begin{array}{lll} 0+0+1 & 0+0+0 & 0+0+0\\ 0+0+0 & 1+0+0 & 0+0+0\\ 0+0+0 & 0+0+0 & 0+1+0 \end{array}\right]=I_{3}.$ $AB=BA=I_{3},$ so $B$ is the multiplicative inverse of A. $(B=A^{-1})$

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