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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

Myunghyun Song
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jacky
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1 Answers1

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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.

Myunghyun Song
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  • did not understand I mean what is $\dim$ and $\ker$ – jacky Mar 16 '19 at 02:19
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    Without those technical terms like $\dim$(dimension), $\ker$(kernel) and the dimension theorem, what I insist is essentially that there is a non-zero vector $x$ such that $Px=0$. This is true, since the dimension of row of $P$ is less than that of column (2<3.) So, $QPx=0$ also has a non-trivial solution $x$, which means that is is not invertible, thus has determinant zero. I hope this makes it a bit clearer. If there is still trouble, please feel free to ask. – Myunghyun Song Mar 16 '19 at 05:38