Proving $\operatorname{rank}A =\operatorname{rank}B$ when $AB = 2A + 3B$

Question

$A$ and $B$ are two square matrices such that $AB = 2A + 3B$. Show that $\operatorname{rank}A =\operatorname{rank}B$.

I managed to prove that the matrices $A-3I$ and $B-2I$ are invertible and that $AB=BA$.

Also if $A$ is invertible then $B$ is invertible because otherwise determinat of $2A$ would be $0$ which is false.

I don't know what to do when $A$ is not invertible.

Answer 1

Let $x \in \ker(B)$. Then $0=ABx=2Ax+3Bx=2Ax$, hence $x \in \ker(A).$

Thus $\ker(B) \subset \ker(A).$

Similar arguments give: $\ker(A) \subset \ker(B).$

Conclusion: $\ker(B)=\ker(A).$

The rank - nullity theorem gives now the result.

Answer 2

We have

$$A(B-2I) = 3B$$

Since $B-2I$ is invertible, it follows that $\operatorname{rank} A = \operatorname{rank}(3B) = \operatorname{rank} B $.