I should find the inverse matrix of the matrix: \begin{pmatrix} n & n & \cdots & n & n \\ n-1 & n-1 & \cdots & n-1 & 0 \\ \vdots & & & & \vdots \\ 2 & 2 & 0 &\cdots & 0 & \\ 1 & 0 & 0 &\cdots & 0 & \\ \end{pmatrix}
for all integers $n$.
Now I could do this by basic Gauss elimination.. taking a unit matrix as our right side and working our way down but this matrix seems pretty interesting as if we look at the rows we can always factor something out.
From the first row we can factor out $n$ and we will get $1 1 1...1$, from the second we can factor out $(n - 1)$ and we will get $1 1 1...1 0$, third $(n - 2) 1 1 1...1 0 0$ and so on..
So we should be able to just factor out $n!$ and simplify the matrix a lot.
My question is: what´s the best way to work with this matrix now, when you want to find the inverse? How do you proceed when you have something factored out from the left side? Is it even a step in the right direction here?
Thank you.
You can write $A = D B J$ where $D$ is diagonal with diagonal entries $(n,n-1, \ldots, 1)$, $J$ the "antidiagonal" matrix with $J_{i,n+1-i} = 1$ for $i=1\ldots n$ and $0$'s elsewhere, $B$ the matrix with $1$'s on and above the diagonal, $0$'s elsewhere. Note that $J^{-1} = J$, while $B^{-1}$ has $1$ on the main diagonal, $-1$ on the superdiagonal (i.e. $(B^{-1})_{i,i+1} = -1$). Thus $$A^{-1} = J B^{-1} D^{-1} = \pmatrix{ 0 & 0 & \ldots & 0 & 1\cr 0 & 0 & \ldots & 1/2 & -1\cr 0 & 0 & \ldots & -1/2 & 0\cr \ldots & \ldots & \ldots & \ldots & \ldots\cr 0 & 1/(n-1) & \ldots & 0 & 0\cr 1/n & -1/(n-1) & \ldots & 0 & 0\cr }$$ i.e. $(A^{-1})_{i,n+1-i} = 1/i$, $(A^{-1})_{i,n+2-i} = -1/(i-1)$, all other elements $0$.