Solution
= Solution
The <determinant> is $k-6$, so for $k\ne6$ the <inverse matrix> is
$$
\boxed{A^{-1}=\frac1{k-6}\begin{pmatrix}k&-3&0\\-2&1&0\\-4k&12&k-6\end{pmatrix}.}
$$
The specified equations use $k=4$, with right-hand side $(2,3,1)^T$. Multiplying by $A^{-1}|_{k=4}$ gives
$$
\boxed{(x,y,z)=\left(\frac12,\frac12,-1\right).}
$$
In particular the parameter in the inverse formula must be set to four, not left arbitrary for this numerical system.