Why is det(AB)=det(A)det(B) true for all matrices?

Question

Neither did our professor nor book give a proof of this fact, but just in case I need to know that.

Answer 1

First you prove this for all $B$ and all elementary matrices $A$ (an elementary matrix is a matrix that is obtained from $I$ by using one elementary row operation).

Next, you show that multiplying by $A$ is the same as applying said elementary row operation.

After that, you study the effect of each elementary row operation on the determinant. Then you have proved the theorem for: arbitrary $B$, and elementary matrix $A$.

After that, you show that every invertible matrix is a product of elementary matrices. This ensures that the theorem is true when $A$ is invertible.

When $A$ is not invertible, treat that case separately. (You have to show that the determinant is zero, once again, by row-reducing.)

Answer 2

The result known as Cauchy-Binet Formula and holds in a more general context for $A$ m-by-n and $B$ n-by-m matrices and of course also in the special case of $m=n$ for square matrices.