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Let $A$ be a $2 \times 3$ matrix with real entries and let $B$ be a $3 \times 2$ matrix with real entries.

If $\det(AB)=4$ then find the value of $\det(BA)$.

My attempt:

I am aware that $\det(AB)=\det(BA)$ when $A$ and $B$ are of same order. But how to do this?

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marked as duplicate by Arnaud D., Jeremy Rickard, Did, Parcly Taxel, Watson Sep 16 '16 at 16:15

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    $\begingroup$ What is the order of $BA$? What is its rank? $\endgroup$ – Guy Sep 16 '16 at 10:09
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$A: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ and hence the image of $A$ in $\mathbb{R}^3$ is at most a subspace of dim 2. Since $\det(AB) \neq 0$, it follows that the image of $A$ is of dimension 2. Since $B: \mathbb{R}^3 \rightarrow \mathbb{R}^2$, the image of $BA$ can not have dimension more than 2 and hence rank($BA$) < 3. Thus $\det(BA) = 0$

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    $\begingroup$ The answer is correct, I would just like to point out that $\det(BA)=0$ regardless of what $\det(AB)$ is, and in fact the equality holds for all matrices $A,B$ with given dimensions. $\endgroup$ – 5xum Sep 16 '16 at 10:31
  • $\begingroup$ Yes, I understand. If $A$ is $m \times n$ and $B$ is $n\times m$ where $m < n$, $\det(BA) = 0$. $\endgroup$ – user348749 Sep 16 '16 at 10:41

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