How to prove inverse of Matrix If you have an invertible upper triangular matrix $M$, how can you prove that $M^{-1}$ is also an upper triangular matrix? I already tried many things but don't know how to prove this. Please help!

An upper triangular $n\times n$ matrix can be written $A=D+N$ with $D$ diagonal, $N$ strictly upper triangular (null diagonal). Then $A$ is invertible if and only if $D$ is invertible. In this case, we have
$$
A=D(I_n+D^{-1}N).
$$
Now $D^{-1}N$ is strictly upper triangular, so it is nilpotent. Concretely, $(D^{-1}N)^n=0$. So $I_n+D^{-1}N$ is invertible with inverse given by the (finite) Neumann series
$$
(I_n+D^{-1}N)^{-1}=\sum_{k=0}^{n-1}(-D^{-1}N)^k.
$$
Since upper triangularity is stable under products, we see that $(I_n+D^{-1}N)^{-1}$ is upper triangular, and finally 
$$
A^{-1}=(I_n+D^{-1}N)^{-1}D^{-1}
$$
is upper triangular.
