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 $A$.
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The products AB and BA should both be $I_{2}=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right].$
$AB=\left[\begin{array}{ll}
4(\frac{3}{2})+5(-1) & 4(-\frac{5}{2})+5(2)\\
2(\frac{3}{2})+3(-1) & 2(-\frac{5}{2})+3(2)
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=I_{2}$
$BA=\left[\begin{array}{ll}
\frac{3}{2}(4)+(-\frac{5}{2})2 & \frac{3}{2}(5)+(-\frac{5}{2})3\\
-1(4)+2(2) & -1(5)+2(3)
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=I_{2}$
$AB=BA=I_{2},$
so $B$ is the multiplicative inverse of A.
$(B=A^{-1})$
