Proof that the inverse of a square matrix is unique

Question

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 01 '14 at 08:49

score 6 · Accepted Answer · edited Jul 25 '17 at 11:50

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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: $$B=BI=B(AC)=(BA)C=IC=C$$

edited Jul 25 '17 at 11:50

José Carlos Santos

427,504

answered Aug 01 '14 at 08:49

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Thanks matrix multiplication is associative right? – user167391 Aug 01 '14 at 08:58
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Yes. You're welcome. – Aug 01 '14 at 09:03

score 2 · Answer 2 · answered Aug 01 '14 at 08:48

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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?

answered Aug 01 '14 at 08:48

5xum

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ABC=C right?... hmm then B(AC)=C then B=C right? – user167391 Aug 01 '14 at 08:49
@user167391 No, you don't get $ABC$, matrix multiplication is not commutative. – 5xum Aug 01 '14 at 08:51
C(AB)=IC right? but then I can't get anywhere – user167391 Aug 01 '14 at 08:51
Ah yes: $C = IC = ( BA ) C = B ( AC ) = BI = B$. Matrix multiplication is associative right? – user167391 Aug 01 '14 at 08:54
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 01 '14 at 09:03
more general... – user167391 Aug 01 '14 at 09:13
@user167391 So you answered the second part of your question as well. – 5xum Aug 01 '14 at 09:18

score 0 · Answer 3 · answered Oct 03 '20 at 13:34

We are treating matrices as linear transformation and a linear transformation is a function (T: $R^n$ -> $R^m$ ).

A function (f: x -> y) is invertible if it is one-one and onto i.e. for every x there is a unique y and for every y there is an x. So a function is not invertible when multiple x's map to the same y or when some y's are not produced by any x.

On applying a similar analogy to invertibility of matrices (Ax=b where x= $A^{-1}$b) then a matrix would not be invertible when

There are some b's for which A$x_1$=b and A$x_2$=b. Because in this case $A^{-1}$ will not give a unique value of x. (infinite solutions)
There are some b's which are not produced by any x. (No solution)

So a matrix is invertible only when there is a unique x for every b. (One solution)

Hello @Arup Das, welcome to MSE. Please use MathJax to typeset the math formulas in an answer. A tutorial can be found here. — Ernie060, Oct 03 '20 at 14:08

Proof that the inverse of a square matrix is unique

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