TITLE: When A and B are of different order given the $\det(AB)$,then calculate $\det(BA)$
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: I think what you are looking for is an answer like this
$B=\begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31}& b_{32} \end{pmatrix}$, $A= \begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}  \end{pmatrix}$,
Then
\begin{align}BA & =\begin{pmatrix}b_{11}a_{11}+b_{12}a_{21} & b_{11}a_{12}+b_{12}a_{22} &  b_{11}a_{13}+b_{12}a_{23}\\ b_{21}a_{11}+b_{22}a_{21} & b_{21}a_{12}+b_{22}a_{22} &b_{21}a_{13}+b_{22}a_{23} \\b_{31}a_{11}+b_{32}a_{21} & b_{31}a_{12}+b_{32}a_{22} &  b_{31}a_{13}+b_{32}a_{23}  \end{pmatrix} \\ & =\begin{pmatrix}b_{11}a_{11}+b_{12}a_{21}+0.0 & b_{11}a_{12}+b_{12}a_{22}+0.0 &  b_{11}a_{13}+b_{12}a_{23}+0.0\\ b_{21}a_{11}+b_{22}a_{21}+0.0 & b_{21}a_{12}+b_{22}a_{22}+0.0 &b_{21}a_{13}+b_{22}a_{23}+0.0 \\b_{31}a_{11}+b_{32}a_{21}+0.0 &b_{31}a_{12}+b_{32}a_{22}+0.0 &  b_{31}a_{13}+b_{32}a_{23} +0.0 \end{pmatrix}\\ & =\begin{pmatrix}b_{11} & b_{12} & 0\\ b_{21} & b_{22} &0 \\b_{31} & b_{32} &  0 \end{pmatrix} \begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} &a_{23} \\0 & 0 &  0 \end{pmatrix}.\end{align}
Let $C=\begin{pmatrix}b_{11} & b_{12} & 0\\ b_{21} & b_{22} &0 \\b_{31} & b_{32} &  0 \end{pmatrix}$ and $D= \begin{pmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} &a_{23} \\0 & 0 &  0 \end{pmatrix}$.
Now, $BA=CD$. Therefore, $\det(BA)=\det(CD)=\det(C)\cdot \det(D)=0\cdot 0=0$.