Matrix multiplications is in general non-commutative. But in the special case that $$AB=I_n$$ where $A$, $B$ are $n\times n$ square matrices.
How to prove that $$BA=I_n?$$
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Hint: think about determinants. The linked duplicate has answers using elementary linear algebra tools. – Ethan Bolker May 01 '17 at 00:01
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A simple solution : that $Rank (AB) \leq Rank(B).$ Since the each row of $AB$ can be written as the linear combination of rows $B.$ since $AB=I_n$ this implies $Rank(B)=n$ thus $B$ is invertible. Now by multiplying $B^{-1}$ to the both sides of $AB=I_n$ (From right) you will get $A=B^{-1}$ this implies $BA=I_{n}$ – Red shoes May 01 '17 at 00:11
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Generally the term "one-sided inverse" of an $n\times m$ matrix $A$ where $n\ge m,$ can mean an $m\times n$ matrix $B$ such that $BA = I_m.$ Or the same thing with left and right reversed. When $n\ge m,$ this "left inverse" of $A$ exists if and only if the columns of $A$ are linearly independent. (The rows of $A$ cannot be linearly independent when $n>m). \qquad$ – Michael Hardy May 01 '17 at 00:13