How do I solve for z for this inverse matrix? How do I go about solving this for z?
$$
    \begin{pmatrix}
    a & b & c \\
    1 & 2 & 3 \\
    d & e & f \\
    \end{pmatrix}
^{-1}= \begin{pmatrix}
    1 & 2 & 3 \\
    x & y & z \\
    4 & 5 & 6 \\
    \end{pmatrix}
$$
 A: 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:
$$a=-3/2, b=-1, c=-1, d=1/6, e=-1, f=-5/3, x=-13/2, y=-8, z=-21/2$$
A: The equation $A^{-1} = B$ implies that $AB = I_3$ where $I$ is the identity matrix, so there are two ways you could go about this:

*

*Take the second equation I wrote and multiply it out, i.e. start with

$\left[\begin{matrix} a & b & c \\ 1 & 2 & 3 \\ d & e & f \end{matrix}\right]
\left[\begin{matrix} 1 & 2 & 3 \\ x & y & z \\ 4 & 5 & 6 \end{matrix}\right] = 
\left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$
and then do the full multiplication to get a set of equations, which you can then use to find an equation (or possibly multiple) for $z$ to solve.


*Use the standard method for manually finding matrix inverses - take the augmented matrix and row-reduce it:

$\left[\begin{array}{c c c | c c c}
a & b & c & 1 & 0 & 0 \\
1 & 2 & 3 & 0 & 1 & 0 \\
d & e & f & 0 & 0 & 1\end{array}\right] \rightarrow \left[\begin{array}{c c c | c c c}
1 & 0 & 0 & ? & ? & ? \\
0 & 1 & 0 & ? & ? & ? \\
0 & 0 & 1 & ? & ? & ? \end{array}\right]$
You should find that the question marks match up with the second matrix in your question, and hence you can find the value of $z$ in there.
