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