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