Answer
$B$ 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 $I_{3}=\left[\begin{array}{lll}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]$
$AB=\left[\begin{array}{lll}
-2+2+1 & 0+1-1 & -2+3-1\\
-5+4+1 & 0+2-1 & -5+6-1\\
3-2-1 & 0-1+1 & 3-3+1
\end{array}\right]=I_{3}$
$BA=\left[\begin{array}{lll}
-2+0+3 & 1+0-1 & -1+0+1\\
-4-5+9 & 2+2-3 & -2-1+3\\
2-5+3 & -1+2-1 & 1-1+1
\end{array}\right]=I_{3}$
$AB=BA=I_{3},$
so $B$ is the multiplicative inverse of A.
$(B=A^{-1})$
