I feel like I almost understand the solution I've come up with, but a step in the logic is missing. I'll post the question and my solution in LaTeX form.(adsbygoogle = window.adsbygoogle || []).push({});

Paraphrasing of text question below in LaTeX. Text question can be seen in its entirety via this imgur link: http://i.imgur.com/41fvDRN.jpg

[tex]

\ if \begin{pmatrix}

a\\

b

\end{pmatrix} \ is \ a \ multiple \ of \begin{pmatrix}

c\\

d

\end{pmatrix} \ with \ abcd \neq 0, show \ that \begin{pmatrix}

a\\

c

\end{pmatrix} \ is \ a \ multiple \ of \begin{pmatrix}

b\\

d

\end{pmatrix}

[/tex]

My solution so far

[tex]

assume \ \lambda \begin{pmatrix}

c\\

d

\end{pmatrix} = \begin{pmatrix}

a\\

b

\end{pmatrix} \rightarrow \begin{matrix}

a = \lambda c\\

b = \lambda d

\end{matrix}

[/tex]

[tex]

now \ assume \ \gamma \begin{pmatrix}

b\\

d

\end{pmatrix} = \begin{pmatrix}

a\\

c

\end{pmatrix} \rightarrow \begin{matrix}

a = \gamma b\\

c = \gamma d

\end{matrix}

[/tex]

So I'm making two assumptions

Let's take the assumptions and put them into a system of four equations

[tex]1: \ a = \lambda c \ \ \ \ 2: \ b = \lambda d \\ 3: \ a = \gamma b \ \ \ \ 4: \ c = \gamma d[/tex]

Now if we sub 3 into 1 to get A, and sub 2 into A to get B and then sub 4 into B to get C

[tex]C \rightarrow \lambda \gamma d = \lambda \gamma d[/tex]

Similarly if we, sub 2 into 3 to get A, and 1 into A to get B, and 4 into B to get C

[tex]C \rightarrow \lambda \gamma d = \lambda \gamma d[/tex]

I want to stop trying all the possible ways to get C now, because I want to look for a generalized way to show that they will all end up at the same point.

But more than this...what is the step of logic that connects the final equation C to proving the first two assumptions. I feel like this should prove the assumptions, but I don't know how exactly, or how exactly to express it.

Thanks :)

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# Proving dependent columns when the rows are dependent

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