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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

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  • $\begingroup$ Take a look at this question. Don't let the fancy word “monoid” scare you. It just means you have an associative product. $\endgroup$ – Harald Hanche-Olsen Aug 1 '14 at 8:49
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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: $$B=BI=B(AC)=(BA)C=IC=C$$

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

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

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