# Prove $\begin{pmatrix} a&b\\ 2a&2b\\ \end{pmatrix} \begin{pmatrix} x\\y\\ \end{pmatrix}=\begin{pmatrix} c\\2c \\ \end{pmatrix}$

Prove that $$\begin{pmatrix} a & b \\ 2a & 2b \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} c \\ 2c \\ \end{pmatrix}$$ has solution $$\begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} \frac{c}{a} \\ 0 \\ \end{pmatrix} +t \begin{pmatrix} -b \\ a \\ \end{pmatrix}$$ and $t$ is real number.

If I plug some numbers, all $y$ variables are vanish and the matrix has infinitely many solutions, but I have no idea how the solution could be into like that. Also I haven't reached Vector chapter.

Any help is appreciated!

• Do you mean to ask why the solution is that or why there are infinitely many? – Christopher Liu Aug 6 '15 at 16:29
• @ChristopherLiu I asked why the solution is that. I can't make good words to the question so feel free to edit my question. – làntèrn Aug 6 '15 at 16:30
• did you solve it and reached a solution ? if so what what the presentation of the solution you have reached ? – d_e Aug 6 '15 at 16:33
• @d_e The solution is formula. I tried to plug in all values, solved it with standard elimination method and all $y$ variables are vanished. I don't know how to prove the formula. – làntèrn Aug 6 '15 at 16:38
• What if $a=0$?. – Michael Galuza Aug 6 '15 at 16:43

The fast approach, as I see it is (in case $a \neq 0$): you have $ax+by = c$ , just plug in $y=0+ta$, and you get $$ax + bta = c$$ $$x= \frac ca - bt$$.
alternatively, in case you don't have the solution in advance, use same approach and you will get: $$x = \frac ca - \frac ba y$$, so the general solution will be $$(x,y) = (\frac ca - \frac bat , t)$$ $$(x,y) = (\frac ca, 0) + (-\frac ba,1)t$$ where $t$ is free. if you plug $t=a$ you will end up with the same result.
• So, using $t$ here is redundant? – làntèrn Aug 6 '15 at 16:58
• I've edited my answer, I should have you $t$ ,, $y$ is somewhat confusing. – d_e Aug 6 '15 at 17:03
• So $\begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} \frac{c}{a} \\ 0 \\ \end{pmatrix}+t \begin{pmatrix} -b \\ a \\ \end{pmatrix}$ would be same with $\begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} \frac{c}{a} \\ 0 \\ \end{pmatrix}+t \begin{pmatrix} \frac{-b}{a} \\ 1 \\ \end{pmatrix}$ – làntèrn Aug 6 '15 at 17:27