# Is the product $AB$ invertible if $A$ is invertible and $B$ is non-invertible?

$$A$$ is an invertible matrix and $$B$$ is a non-invertible matrix. Can $$AB$$ be invertible? I have the following idea:

Sup. $$AB$$ is invertible, then:

$$B=IB=(A^{-1}A)B=A^-1(AB)$$, then apply inverse both sides:

$$B^{-1}=(AB)^{-1}(A^{-1})^{-1}=(AB)^{-1}A$$, but $$B$$ is non-invertible (hip). This leads to a contradiction, as we supposed $$AB$$ is invertible.

Therefore $$AB$$ is non-invertible.

I'm not sure if the step where I apply "inverse both sides" is right. Otherwise I'm not sure how to prove this.

Note 1: I CAN'T use $$(AB)^{-1}=B^{-1}A^{-1}$$ since the hypothesis for that theorem is $$A, B$$ invertible matrices and this is not the case.

Note 2: I CAN'T use determinants yet.

• A product of invertible matrices is invertible. Oct 9, 2020 at 13:59
• If $A$ is invertible and $B$ is not, then $AB$ is not invertible. Oct 9, 2020 at 13:59
• Can you use $\det AB=\det A\det B$? Oct 9, 2020 at 13:59
• Refer to the invertible matrix theorem which gives a long list of equivalent conditions for a matrix to be considered invertible. You should have seen at least some of these proven before. It may be easiest to use a specific one of these, be it related to determinants, or related to nullspaces or injectivity, or what have you. Oct 9, 2020 at 14:02
• @FabrizioGambelín you can use what I said to show, in a very simple way, that $B$ must be invertible if $A$ and $AB$ are Oct 9, 2020 at 14:02

Your argument is correct, though depending on the level, you may want to explain why $$B^{-1}=\big(A^{-1}(AB)\big)^{-1}=(AB)^{-1}(A^{-1})^{-1}.$$ Another approach would be to suppose that $$AB$$ is invertible with inverse $$C$$.Then $$(CA)B=C(AB)=I,$$ so $$CA$$ is the inverse of $$B$$, a contradiction.
Alternatively, you could note that if $$B$$ is non-invertible, then there exists some nonzero vector $$x$$ such that $$Bx=0$$. Then also $$ABx=0$$, and so $$AB$$ is non-invertible.