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If I got the invertible matrix $A$, I can calculate the inverse matrix $A^{-1}$, so that $A \cdot A^{-1} = E$, where $E$ is the identity matrix.

Wikipedia says that not only $A\cdot A^{-1} = E$ must be fulfilled, but also $A^{-1} \cdot A = E $. Can someone explain to me why this is not a contradiction to the fact that matrix multiplication is not commutative ? Is the inverse matrix really defined as a matrix which fulfills both?

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  • $\begingroup$ Yes. $ { } { } { } $ $\endgroup$ – Workaholic Feb 2 '15 at 16:57
  • $\begingroup$ Not commutative in general, some matrices however might commute. $\endgroup$ – mvw Feb 2 '15 at 16:58
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The inverse of a matrix is defined as the matrix that satisfies both relationships.

For square matrices $A$ and $B$, $$ B\mbox{ is the inverse of }A:=B\mbox{ such that } AB{}={}BA{}={}I\,. $$

Incidentally, this also means that $A$ is the inverse of $B$.

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They are inverses of each other by definition,

therefore $AA^{-1} = E$ holds

now take $A= (A^{-1})^{-1}$ then $A^{-1}(A^{-1})^{-1} = E = A^{-1}A$ also holds as they are inverses of each other

commutativity of matrix multiplication holds in certain cases for example $AE =EA$ where $E$ is the identity matrix

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Let $A$ be a matrix.

The inverse of $A$ is by definition both a left and a right inverse. Hence by definition, a matrix $B$ is an inverse of $A$, if and only if $$ AB=BA=I. $$

The fascinating aspect of matrices is, that one of the defining equation implies also the other. That is, if $B$ satisfies $AB=E$ only, then it follows $BA=E$.

To see this, use the rank theorem to show that $AB=E$ implies that $A$ has full rank, thus is invertible. Thus, there is a matrix $A^{-1}$ such that $AA^{-1}=A^{-1}A=E$. Multiplying $AB=E$ from the left mit $A^{-1}$ yields $B=A^{-1}$. This implies that the matrix $B$ is the inverse, and also satisfies $BA=E$.

An alternative proof would be to use determinants: The existence of $B$ with $AB=E$ implies $\det(A)\ne0$, hence $A$ is invertible.

In the light of the above comments, we can compensate the lack of commutativity of matrix multiplication by other matrix specific tools (rank, determinant) to prove this result.

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