How do I solve for z for this inverse matrix? How do I go about solving this for z?
$$
    \begin{pmatrix}
    a & b & c \\
    1 & 2 & 3 \\
    d & e & f \\
    \end{pmatrix}
^{-1}= \begin{pmatrix}
    1 & 2 & 3 \\
    x & y & z \\
    4 & 5 & 6 \\
    \end{pmatrix}
$$

As :
$$\underbrace{\begin{pmatrix}
    a & b & c \\
    \color{red}{1} & \color{red}{2} & \color{red}{3} \\
    d & e & f \\
    \end{pmatrix}}_A
\underbrace{\begin{pmatrix}
    1 & 2 & \color{red}{3} \\
    x & y & \color{red}{z} \\
    4 & 5 & \color{red}{6} \\
    \end{pmatrix}}_B=
\begin{pmatrix}
    1 & 0 & 0 \\
    0 & 1 & \color{red}{0} \\
    0 & 0 & 1 \\
    \end{pmatrix}$$
multiplication of the second row of $A$ by the third column of $B$ will give $\color{red}{0}$, i.e.,
$$21+2z=0 \implies z=-21/2$$
Besides, it isn't evident a priori that a global solution exists with these constraints. I have checked using a Computer Algebra System that there is indeed a unique solution with this data:
$$a=-3/2, b=-1, c=-1, d=1/6, e=-1, f=-5/3, x=-13/2, y=-8, z=-21/2$$
