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I need help with this problem:

If $(a,b)$ is a multiple of $(c,d)$ with $abcd\neq0$, show that $(a,c)$ is a multiple of $(b,d)$. This is suprisingly important: call it a challenge question. You could use numbers first to see how $a,b,c$ and $d$ are related. The question will lead to:

If A = $\left[\begin{array}{l}a&b\\c&d\end{array}\right]$ has dependent rows the it has dependent columns.

I tried to do it this way:

$(a,b)=x(c,d)\Rightarrow(a,b)=(xc,xd)$

$(a,c)=(xc,c)\Rightarrow c(x,1)$ and $(b,d)=(xd,d)\Rightarrow d(x,1)$

I don't know what to do after that, what should I do next?

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  • $\begingroup$ Do you mean that $a=\lambda c, b=\lambda d$ for some constant $\lambda$? But then $(a,c)=(\lambda c, c)=c\times (\lambda ,1) $ and $(b,d)=(\lambda d,d)=d\times (\lambda ,1)$ so... $\endgroup$
    – lulu
    Jan 27 '19 at 17:30
  • $\begingroup$ I think that I can solve oneof the ecuations for $(x,1)$, right? For example, if I solve $(b,d)=d(x,1)$, I end up with $(x,1)={(b,d)\over d}$, and the I can substitute that on the first ecuation like this: $(a,c)={c \over d}(b,d)$, am I right? $\endgroup$ Jan 27 '19 at 19:15
  • $\begingroup$ yes, that's right. $\endgroup$
    – lulu
    Jan 27 '19 at 19:17
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First note the following: $(a,b) = (a,(\frac{b}{a}) a)$. [As $abcd \not =0$ we can assume that $\frac{b}{a}$ exists]

So for some scalar $x$ we note: $(c,d) = x(a,b) = (xa,xb)$ $=(xa,x(\frac{b}{a}) a)$. Thus $c$ can be written $c=xa$ and $d$ can be written $d=x(\frac{b}{a}) a$.

Thus $(b,d) = (\frac{b}{a} a, x(\frac{b}{a})a) = \frac{b}{a}(a,xa) = \frac{b}{a}(a,c)$. So $(b,d) = y(a,c)$ where $y = \frac{b}{a}$.

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  • $\begingroup$ So I just need to make that algebraic trick, and that's all, right? $\endgroup$ Jan 27 '19 at 19:08
  • $\begingroup$ Yes that is correct. $\endgroup$
    – Mike
    Jan 28 '19 at 1:29
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Hint: If $ad=bc$ then $\dfrac ab=\dfrac cd$ and also $\dfrac ac=\dfrac bd$

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  • $\begingroup$ You can skip $ad=bc$, as it's not really the relevant hypothesis in this problem (is just important for the application mentioned afterwards). $\frac ac=\frac bd$ is the hypothesis, and $\frac ab=\frac cd$ is the conclusion $\endgroup$
    – Arthur
    Jan 27 '19 at 17:36
  • $\begingroup$ It just helps visually to see why the fractions are equal, as they only result from simple manipulations of this equation. Also OP went with determinant, and it relates well to ad-bc=0. $\endgroup$
    – zwim
    Jan 27 '19 at 17:41
  • $\begingroup$ You have to be careful lest you end up dividing by 0. $\endgroup$
    – Mike
    Jan 27 '19 at 17:42
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    $\begingroup$ @Mike OP specified $abcd\neq 0$. $\endgroup$
    – zwim
    Jan 27 '19 at 17:42
  • $\begingroup$ You are right, $abcd \not =0$ is assumed. But I also don't think it is clear (to the person assigning this nor to the person asking) that $ad=bc$ either $\endgroup$
    – Mike
    Jan 27 '19 at 17:44

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