This question already has an answer here:

The most common method to find the inverse of a invertible square matrix is to apply row operations to the matrix and reduce it to the identity matrix.

An row operation is applied to the matrix by doing a left multiplication by an elementary matrix.

Denote the elementary matrices as $E_1, E_2, ... , E_n$ , then

$(E_1, E_2, ... , E_n)A = I$ , where $E_1, E_2, ... , E_n$ is the inverse of $A$.

However, the procedure to check whether two matrices $A$ , $B$ are the inverse of each other is to check that $AB=I$ and $BA=I$.

Why do we claim that $(E_1, E_2, ... , E_n)$ is the inverse of $A$ without checking that $A(E_1, E_2, ... , E_n) = I$ ?


marked as duplicate by Claude Leibovici, Dominik, KittyL, C. Falcon, Namaste Dec 25 '16 at 13:51

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Because for square matrices existence of any one side inverse imply the existence of the other. For proof check this out If $AB = I$ then $BA = I$

  • $\begingroup$ Thank you. Merry christmas! $\endgroup$ – Little Rookie Dec 25 '16 at 9:36

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