# Find the inverse matrix $A^{-1}$ , using Gauss-Elimination without LU decomposition (and without Gauss-Jordan)?

Given a non-singular matrix A , is it possible to find the inverse matrix $A^{-1}$ , using Gauss-Elimination , without using LU decomposition and without using Gauss-Jordan ?

I know that I can use LU decomposition and then apply Gauss-elimination on $L$ and $U$ , this would require :

1. Finding $L$ and $U$

2. Calculate $L*Y = e(i)$ , from here I'd get $Y$

3. Calculate $U*(current-column) = Y$ , from here I'd get each time the column

Or , I can use Gauss-Jordan method (without LU decomposition) where I put the $I$ matrix on the right of $A$ , and then use the Gauss-Jordan elimination .

Both ways works great, but , is it possible to calculate the inverse of $A$ only with Gauss-elimination , without LU ... ?

Regards

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I will make a guess that the problem actually wants you to find an inverse of $A$ given a method to compute $A^{-1}b$ for any vector $b$, because that's what Gaussian elimination does.
Assume $A$ is a non-singular $n$-by-$n$ real matrix. Let $e_i$ be the $i$-th basis vector of the standard basis of $\mathbb R^n$. Use the given method (Gaussian elimination in this case) to solve for $x_i$ from $Ax_i = e_i$, $i = 1, 2, \ldots, n$. The matrix $[x_1 \ x_2 \ \ldots \ x_n]$ will be $A^{-1}$.
First triangulate the matrix using Gauss elimination with pivoting. Then solve $\textbf{Ax} = \textbf{I}$, $\textbf{I}$ being the first column of the identity matrix The solution $\textbf{I}$ will be the first column of $\textbf{A}^{-1}$ Do the same for the next column of the identity matrix Each successive solution for $\textbf{x}$ will be the next column of $\textbf{A}^{-1}$ You can do this for as big a matrix as you like.