# Proof that the inverse of a square matrix is unique

From my textbook

... if a 2×2 matrix $A$ is invertible then its inverse is unique.

I wonder, how can one prove this? Also can one extend this proof to larger square matrices of order $n$? Thanks

• Take a look at this question. Don't let the fancy word “monoid” scare you. It just means you have an associative product. – Harald Hanche-Olsen Aug 1 '14 at 8:49

Notice that $\operatorname{GL}_n(\Bbb C)$ is a group and the proof of unicity of the inverse of a matrix is the same proof in any group. Let $A$ a given invertible matrix and denote $B$ and $C$ two inverses of $A$. Then: $$B=BI=B(AC)=(BA)C=IC=C$$

• Thanks matrix multiplication is associative right? – user167391 Aug 1 '14 at 8:58
• Yes. You're welcome. – user63181 Aug 1 '14 at 9:03

Hint:

Assume that there exists two inverses of $A$. This means $AB=BA=I=AC=CA$. Now, what happens if I multiply the equality $AB=I$ by $C$ from the left?

• ABC=C right?... hmm then B(AC)=C then B=C right? – user167391 Aug 1 '14 at 8:49
• @user167391 No, you don't get $ABC$, matrix multiplication is not commutative. – 5xum Aug 1 '14 at 8:51
• C(AB)=IC right? but then I can't get anywhere – user167391 Aug 1 '14 at 8:51
• Ah yes: $C = IC = ( BA ) C = B ( AC ) = BI = B$. Matrix multiplication is associative right? – user167391 Aug 1 '14 at 8:54
• If course, multiplication is indeed asociative. Your proof is OK now. Do you think it holds for $n=2$ or is it more general? – 5xum Aug 1 '14 at 9:03