# Basic way to show for $n\times n$ matrices $A$ and $B$, that $(AB)^{-1} = (B^{-1})(A^{-1})$

In looking at matrix inverses, I know the following works (I is the identity matrix):

If $AB$ are nxn matrices and are invertible, then $(AB)C = I$, and therefore $C = (AB)^{-1}$.

I can show that $B^{-1}A^{-1}$ substituted for C results in I (by way of matrices multiplied by their inverses), and thereby it must be equal to $(AB)^{-1}$, but what I am not seeing is how (if any way) to show directly that $(AB)^{-1}$ = $B^{-1}A^{-1}$. The approach of just using these as equivalent substitutions and then showing both result in $I$ seems a bit like guess-and-check.

Keep in mind I'm at the beginning of intro linear algebra, so I might be stepping beyond the scope of what's been taught already. So far I've covered vector math, and now am onto matrix multiplication (no eigenvalues, determinants, mappings, or subspaces, or other such advanced concepts yet).

## 2 Answers

Suppose $A$ and $B$ are two invertible $n \times n$ matrices and define $C = B^{-1} A^{-1}$. Then by the associative property of matrix multiplication we have $$(AB)C = AB(B^{-1} A^{-1}) = A(BB^{-1}) A^{-1} = A A^{-1} = I$$ and $$C(AB) = (B^{-1} A^{-1})AB = B^{-1} (A^{-1} A) B = B B^{-1} = I$$

Thus $C$ is by definition an inverse of $AB$. The result follows because a matrix inverse, if it exists, is unique.

Using associativity of matrix multiplication, and definition of matrix inverse: $$(AB)^{-1}=(AB)^{-1}AA^{-1}=(AB)^{-1}A(BB^{-1})A^{-1}=(AB)^{-1}(AB)(B^{-1}A^{-1})=B^{-1}A^{-1}$$