Let $M\in M(n\times n,\mathbb{R})$ such that $$M=\begin{pmatrix}a & c & 0 & \ldots & \ldots & 0 & d \\ e & a & c & \ddots & \ddots & \vdots & d \\ 0 & e & \ddots & \ddots & \ddots & \vdots & d \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & c & d \\ 0 & \ldots & \ldots & 0 & e & a & c \\ f & f & \ldots & \ldots & f & e & a\end{pmatrix}$$ and with $a,b,c,d,e,f\in\mathbb{R}$ such that $M$ and every one of its $(i,i)$-minors is invertible. Show that the LU decomposition of $M$ is of the following form:
$$L=\begin{pmatrix}1 & 0 & \ldots & \ldots & \ldots & \ldots & 0 \\ e & 1 & 0 & \ddots & \ddots & \vdots & 0 \\ 0 & e & \ddots & \ddots & \ddots & \vdots & 0 \\ 0 & e & \ddots & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\ 0 & \ldots & \ldots & 0 & e & 1 & 0 \\ f & f & \ldots & \ldots & f & e & 1\end{pmatrix}, \;U=\begin{pmatrix}a & c & 0 & \ldots & \ldots & 0 & d \\ 0 & a & c & \ddots & \ddots & \vdots & d \\ \vdots & 0 & \ddots & \ddots & \ddots & \vdots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & c & d \\ \vdots & \ddots & \ddots & \ddots & 0 & a & c \\ 0 & \ldots & \ldots & \ldots & \ldots & 0 & a\end{pmatrix}$$
My problem is that, performing LU decomposition as usual, we get that $L_{2,1}=\frac{e}{a}$, since row $2$ minus $\frac{e}{a}$ times row $1$ gives us a $0$ in the first entry of row $2$. I'm guessing that one has to use the fact that $M$ and all its $(i,i)$-minors are invertible but I don't know how exactly I can use that fact here.
Edit:
You're saying in the comments that the point is to prove that $a=1$. It's impossible. Assume that it is possible, and multiply every entry in $M$ by $2$. Then every minor is multiplied by a power of $2$, so non-zero minors are still non-zero and every assumption is respected. But $a$ is not equal to $1$ any more.
Original answer:
It is immediate to see that the product of the provided $L,U$ matrices is not equal to $M$ unless $ea=e$, that is $e=0$ or $a=1$. The fact that all $(i,i)$ minors are invertible is unlikely to imply something as strong as $e=0$ or $a=1$ (and also it doesn't really make sense either, since you don't what $i$ is, and it clearly can't be true for all $i$, since there are zeroes in the matrix).