linear algebra - proof: inverse of lower triangular identity matrix

As you know that is enough negating below of diagonal to inverse of lower triangular identity matrix.

example

$$A = \left(\begin{matrix}
1 & 0 & 0 & 0 \\
3 & 1 & 0 & 0 \\
-1 & 0 & 1 & 0 \\
2 & 0 & 0 & 1 \\
\end{matrix}\right)
$$

basically inverse of A

$$A' = \left(
\begin{matrix}
1 & 0 & 0 & 0 \\
-3 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
-2 & 0 & 0 & 1 \\
\end{matrix}\right)
$$

I just need to prove it.

my question is not related any software. it is general linear algebra question. It's not enough to say that "if $A'=$ (inversion of $A$), multiplication $A'$ and $A$ should be $I$ (identity matrix)". we cannot say for all case. I need a general proof.

it's related topic with Gauss Elimination - LU decomposition

thank you for any help.