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I'm probably missing something obvious, but how would I go about solving this?

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    $\begingroup$ Use determinants... $\endgroup$
    – cjferes
    Sep 11, 2014 at 16:50

1 Answer 1

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Note that

$$\begin{array}{rcl} \det(CD)&=&\det(-DC)\\ \det(C)\cdot\det(D)&=&(-1)^n\det(D)\cdot\det(C)\qquad/n\,\mathtt{odd}\\ \det(C)\cdot\det(D)&=&-\det(D)\cdot\det(C)\\ \end{array}$$

So, either $\det(C)=0$ or $\det(D)=0$, so either $C$ or $D$ has no inverse.

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    $\begingroup$ The argument fails if $\ 2 = 0,\,$ i.e. over fields of characteristic $\,2.\ \ $ $\endgroup$ Sep 11, 2014 at 17:29
  • $\begingroup$ as the OP tags the question as linear algebra and matrices, I assumed that it needed only the basics, and there was no need of deeper maths. ;) $\endgroup$
    – cjferes
    Sep 11, 2014 at 21:13
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    $\begingroup$ Nowadays linear is often done over finite fields (even in applied math), and it's not usually consider deeper math. $\endgroup$ Sep 11, 2014 at 21:17

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