$$ A= \begin{bmatrix} 1 & & & & \\ & 1 & & (-1) & \\ & & 1 & & \\ & (0) & & \ddots & \\ & & & & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -1 & \dots & -1 \\ 0 & 1 & -1 & \dots & -1 \\ 0 & 0 & 1 & \dots & -1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \end{bmatrix}$$ Prove that $A$ is invertible, and find $A^{-1}$
What I did so far:
My idea was to use the Gauss Elimination Method to find the reduced echelon form of $A$ by performing (left-multiplication) some elementary operations (permutation of lines, multiplying a line by a scalar, ...) $T_1, T_2,\dots, T_k$ so that $T_1T_2\dots T_kA=I_n$ which I can easily calculate by performing the operation on $I_n$
From this, we find $A^{-1} = T_1T_2\dots T_k$
I found two methods:
Method 1:
Add line $L_2$ to $L_1$ to get $\begin{bmatrix}1&0&-2&-2&(-2)\end{bmatrix}$
Add line 2$L_3$ to $L_1$ to get $\begin{bmatrix}1&0&0&-4&(-4)\end{bmatrix}$
Add line 4$L_4$ to $L_1$ to get $\begin{bmatrix}1&0&0&0&(-8)\end{bmatrix}$
$\vdots$
Add line $2^{i-1}L_i$ to $L_1$ to get $\begin{bmatrix}1&0&0&0&(-2^i)\end{bmatrix}$
Then $A^{-1} = \begin{bmatrix} 1 & 1 & 2 & 4 & \dots & 2^{n-2} \\ 0 & 1 & 1 & 2 & \dots & 2^{n-3}\\ 0 & 0 & 1 & 1 & \dots & 2^{n-4}\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \dots & 1 \end{bmatrix}$
The Problem:
For our class, this is not considered as a valid proof. I have to be as rigorous as possible. And I have to prove that these operations can reduce the matrix.