If $P$ is a $2\times 3$ matrix, $Q$ is a $3\times 2$ matrix and
$\det(PQ)=2019,$ then what is $\det(QP) $?
What I tried:
assume $$P = \begin{pmatrix}a&& b&&c\\ d&&e&&f\\\end{pmatrix}$$
and $$Q=\begin{pmatrix}u&&v\\w&&x\\y&&z\end{pmatrix}$$
$$\det(PQ)=\begin{vmatrix}au+bw+cy&&av+bc+cz\\ du+ew+fy&& dv+ex+fz\end{vmatrix}=2019\cdots (1)$$
and $\det(QP)=\begin{vmatrix}au+dv&&bu+ev&&cu+vf\\aw+dx&&bw+ex&&cw+fx\\ay+dz&&by+ex&&cy+fz\end{vmatrix}$
How do i solve it? Help me please
Note that $\ker P$ is non-trivial because $\dim \ker P = 3-\dim \text{im}P\ge 1$ by the dimension theorem. Also, we have $$ \ker (QP)\ge \ker P, $$ which implies that $QP$ is not invertible. This yields $\det(QP)=0$. Determinant of $PQ$ is irrelevant.