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# proof of matrix inverse calculation by Gaussian elimination

Let $A$ be an invertible matrix, and $A^{{-1}}$ its inverse, whose columns are $A^{{-1}}_{1},\cdots,A^{{-1}}_{n}$. Then, by definition of matrix inverse, $AA^{{-1}}=I_{n}$. But this implies $AA^{{-1}}_{1}=e_{1},\ldots,AA^{{-1}}_{n}=e_{n}$, with $e_{1},\cdots,e_{n}$ being the first,$\ldots$,$n$-th column of $I_{n}$ respectively.

$A$ being non singular (or invertible), for all $k\leq n$, $AA^{{-1}}_{k}=e_{k}$ has a solution for $A^{{-1}}_{k}$, which can be found by Gaussian elimination of $[A\mid e_{k}]$.

The only part that changes between the augmented matrices constructed is the last column, and these last columns, once the Gaussian elimination has been performed, correspond to the columns of $A^{{-1}}$. Because of this, the steps we need to take for the Gaussian elimination are the same for each augmented matrix.

Therefore, we can solve the matrix equation by performing Gaussian elimination on $[A\mid e_{1}\cdots e_{n}]$, or $[A\mid I_{n}]$.

## Mathematics Subject Classification

15A09*no label found*

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