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