I recently studied $\det (P Q)=\det (P)\cdot \det (Q)$, now as $\det$ is scalar quantity $\implies \det (Q P)=\det (P)\cdot \det (Q)=\det (P Q)$. As matrix $PQ$ is not necessarily equal to $QP$, why is there determinant equal?
Also consider $A$ to be $2\times 3$ and $B$ to be $3\times 2$ then,
Is such $2\times 3$ and $3\times 2$ matrix determinant always zero?
I'm very new to this please don't use higher level of Mathematics.
On
Determinants are not defined for $m\times n$ matrix where $m \ne n$. $Det(A)$ is interpreted as volume of parallelepiped.
In general, you cannot define $det(AB)$ and $det(BA)$ let alone their equality, when $A ~\&~ B$ are not square matrices. For square matrices why $det(AB)=det(BA)$ check this link for proof.
Luckily, for your specific choice of $ A $ and $B$ both $AB \& BA$ matrices are defined. But this doesn't hold in general. Take for example $A$ as $4\times 7$ and $B$ as $7 \times 3$. Your $AB$ matrix is defined as $4 \times 3$ however $BA$ is not defined.
First: $A$ and $B$ do not have determinants. Only square matrices do.
Second: You correctly calculated the matrix $BA$ but that matrix is not the product of the two matrices in the last line, one with a row and one with a column of $0$s.
Third: There is some truth in what you say. The matrix $B$, considered as a linear transformation, maps a two dimensional space to a three dimensional space. That means that when you think of $BA$ as a linear transformation it can't be surjective. So it is not invertible and has determinant $0$.
The third point is not very "advanced mathematics". If you haven't seen it yet in your study of linear algebra you will soon.