When A and B are of different order given the $\det(AB)$,then calculate $\det(BA)$ Let 'A' be a $2 \times 3$ matrix where as B be a $3 \times 2$ matrix if $\det(AB) = 4$ the find value of the $\det(BA)$
My attempt: I took A = 
$$
        \begin{bmatrix}
        2 & 0  &0\\
        0 & 0  &2\\
        \end{bmatrix}
$$
 B=
$$
        \begin{bmatrix}
        1 & 0  \\
        0 & 0  \\
        0 & 1  \\
        \end{bmatrix}
$$
It satisfies given condition and I get $\det(BA)=0$ 
 But I have not proved it 
How do I prove that it is always zero
(background)I am 12th grader and I know about adjoint,inverse,determinant,rank of a matrix and the other basics. 
However I do NOT know about eigenvalues and eigenvectors.
 A: You will always get $\det(BA)=0$. The reason for this is very simple : $BA$ is a $3\times 3$ matrix with rank at most $2$; thus it is not invertible and thus it has $0$ determinant.
One possible definition of the rank is that it is the dimension of the subspace generated by the columns or the lines of the matrices. The lines of $BA$ are obtained by taking linear combinations of the lines of $A$ : for example, if
$$B=\begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31}& b_{32} \end{pmatrix},\ A= \begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}  \end{pmatrix},$$
then the first line of $BA$ is
$$b_{11}\cdot \begin{pmatrix}a_{11} & a_{12} & a_{13}  \end{pmatrix} + b_{12}\cdot \begin{pmatrix} a_{21} & a_{22} & a_{23}  \end{pmatrix}.$$
Since the lines of $BA$ are generated by only two lines (of $A$), the dimension of the generated subspace cannot be more than $2$.
This also means that the lines of $BA$ cannot be linearly independent; thus there must be some linear relation between them, which means that when you convert it to echelon matrix you will certainly get a line of $0$s.
A: Let $ T = BA $, then $ T : \mathbb{R}^3 \to \mathbb{R}^3 $ is a linear transformation with nontrivial kernel, as $ A : \mathbb{R}^3 \to \mathbb{R}^2 $ cannot be an injection. Thus, it is not invertible, and $ \det(T) = 0 $.
