Prove that if a product$ AB$ of $n\times n$ matrices is invertible, so are the factors $A$ and $B$.

Solution 1:

First: your solution isn't actually a solution. By writing $A^{-1}$ you are assuming that $A$ is invertible, which is what you want to prove.

One way to do this would be to note that $$ \det(AB) = \det(A)\det(B). $$ If $AB$ is invertible, then $\det(AB)\neq 0$, and so $\det(A)\neq 0$ and $\det(B)\neq 0$.

Another way: If $AB$ is invertible, then there is a $C$ such that $$ (AB)C = I $$ That is $$ A(BC) = I $$ So $BC$ is a right inverse of $A$, and so $A$ is invertible. Likewise $B$ is invertible. (See this question/answers for a bit more: If $AB = I$ then $BA = I$).

Solution 2:

Suppose $AB$ is invertible. Then there is a matrix $C$ such that $C(AB)=I$ and $(AB)C=I$. Rearranging the parenthesis, we have $(CA)B=I$ and $A(BC)=I$, which shows $A$ and $B$ are invertible.