I am struggling to understand the following problem
Consider the $n\times n$ matrix:
$$ M(x,y)=\begin{bmatrix} m_{11} & m_{12} & \cdots & m_{1.n-1} & x\\ m_{21} & m_{22}&&m_{2,n-1}&m_{2,n}\\ \vdots &\vdots &\ddots&\vdots&\vdots\\ m_{n-1,1}&m_{n-1,2} &&m_{n-1,n-1}&m_{n-1,n}\\ y & m_{n,2} & \cdots &m_{n,n-1}& m_{n.n} \end{bmatrix}. $$
Find necessary and sufficient conditions on the entries $m_{ij}$ such that $M(x,y)$ admits LU- and UL-factorization and for all $i,j$ if $|i-j| > 1$ then $(M(x,y)^{-1})_{ij}=0$, i.e. the inverse is a 1-band matrix, for some complex numbers $x$ and $y$. In this case find such $x,y$.
I found out that if $M(x,y)$ is invertible, then it admits an LU factorization if and only if all its leading principal minors are nonzero [1 ], where leading principal minors are defined as
Def: The leading k-th order principal sub-matrix of $M$ is the $k\times k$ matrix obtained by deleting the last $n-k$ rows and columns. The k-th order leading principal minor is its determinant.
So, I observe that $$ \det(m_{11}),\quad \det(\begin{bmatrix} m_{11}&m_{12}\\ m_{21}&m_{22}\end{bmatrix}),\quad\ldots\quad , \det\begin{bmatrix} m_{11} & m_{12} & \cdots & m_{1.n-1} & x\\ m_{21} & m_{22}&&m_{2,n-1}&m_{2,n}\\ \vdots &\vdots &\ddots&\vdots&\vdots\\ m_{n-1,1}&m_{n-1,2} &&m_{n-1,n-1}&m_{n-1,n}\\ y & m_{n,2} & \cdots &m_{n,n-1}& m_{n.n} \end{bmatrix} $$
are the 1st,2nd,...,nth order leading principal minors.
Here I get stuck. So, I thought it would be a good idea to reduce the dimension to get a simpler form.
consider the $3\times 3$ matrix $$ A(x,y)=\begin{bmatrix} a_{11}&a_{12}&x\\ a_{21}&a_{22}&a_{23}\\ y&a_{32}&a_{33} \end{bmatrix} $$
The 1st, 2nd and 3rd order leading principle minors are:
$$ a_{11}\\ a_{11}a_{22}-a_{12}a_{21} \tag{1}\\ -a_{11} a_{23} a_{32}+a_{22} a_{11} a_{33}-a_{12} a_{21} a_{33}+a_{12} a_{23} y+a_{21} a_{32} x-a_{22} x y $$
So, if $\det(A(x,y))\not =0$, then its invertible. And if equations (1) are nonzero, then $A(x,y)$ admits an LU factorization (according to the theorem).
The inverse of $A(x,y)$ will have the following form $$ \frac{1}{det(A(x,y))} \begin{bmatrix}a_{22} a_{33}-a_{23} a_{32} & a_{32} x-a_{12} a_{33} & a_{12} a_{23}-a_{22} x \\ a_{23} y-a_{21} a_{33} & a_{11} a_{33}-x y & a_{21} x-a_{11} a_{23} \\ a_{21} a_{32}-a_{22} y & a_{12} y-a_{11} a_{32} & a_{11} a_{22}-a_{12} a_{21}\end{bmatrix}. $$
Therefore, to get 1-bandness, i.e. if $|i-j| > 1$ then $(A(x,y)^{-1})_{ij}=0$, we let $$ a_{12} a_{23}-a_{22} x =0\\ a_{21} a_{32}-a_{22} y=0 $$
and obtain $x=a_{12} a_{23}/a_{22}$ and $y=a_{21} a_{32}/a_{22}$. However,
I am not able to find (explicitly) necessary and sufficient conditions on the entries and hence I fail to find complex numbers $x$ and $y$. Is there an alternative method to solve this? How can I finish this proof
Any hints are appreciated.