Consider the following matrix:
$$ M=\begin{bmatrix} 3 & 1 & x \\ 2 & 0 & 1 \\ y & 4 & 1 \end{bmatrix}. $$
I want to find complex numbers $x$ and $y$ such that $M^{-1}$ exist and is a 2-band matrix.
I know that when the determinant for a square matrix is not equal to zero, the inverse for that matrix does exist. We observe that
$$ \det(M)\not=0 \iff \begin{vmatrix} 3 & 1 & x \\ 2 & 0 & 1 \\ y & 4 & 1 \end{vmatrix}\not=0 \iff 8x+y\not=14. \tag{1} $$
However, the inverse of matrix $M$ is (assuming (1)) equal to
$$ \dfrac{1}{8 x+y-14}\begin{bmatrix} -4 & 4 x-1 & 1\\ y-2 & 3-x y & 2 x-3\\ 8 & y-12 & -2 \end{bmatrix} $$
So, $M^{-1}$ is a 2-band matrix iff $$ y\not=2,12 \\ x\not=1/4,3/2 \tag{2}\\ xy\not=3 $$ and therefore we can conclude that $M^{-1}$ exist and is a 2-band matrix iff (1) and (2) holds.
My question is: Is this method correct? If so, what can we say about the uniqueness of the solution?
Edit:
a matrix $A$ is called a $k$-band matrix if $a_{ij}=0$ for $|i-j|>k$.