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Show that any two-dimensional vector can be expressed in the form $a \begin{pmatrix} 3 \\ -1 \end{pmatrix} + b \begin{pmatrix} 2 \\ 7 \end{pmatrix}$, where $a$ and $b$ are real numbers.

I was able to simplify this down to: $\binom{3a+2b}{-a+7b}=\binom{x}{y}\\$, where $\binom{x}{y} = a\binom{3}{-1} + b \binom{2}{7}$, but I am not sure what to do from here.

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    Please edit the title of the question, it doesn't make any sense since $\displaystyle a\binom{3}{-1}+b\binom{2}{7}$ is not a statement and so it can't be true in the usual sense. – 2017-02-15
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    The result (not yet) in the title holds because these two vectors are a free family and they are because the determinant $$\begin{vmatrix}3&2\\-1&7\end{vmatrix}=23$$ is not zero. So, one does not need to solve the system to show $(a,b)$ exists. But isn't all this in your notes? – 2017-02-15

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You are almost there.

You have $\binom{3a+2b}{-a+7b}=\binom{x}{y}\\ $.

This is the set of equations $3a+2b = x$ and $-a+7b = y$.

Just solve these for $a$ and $b$ in terms of $x$ and $y$.

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    Oh, wow! Thanks, didn't know it was that simple. – 2017-02-15
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Note that for all $x,y \in \mathbb{R}$, if we solve $3a+2b=x$ and $-a+7b=y$, we get that $$a=\frac{7x}{23}-\frac{2y}{23}$$$$b=\frac{x}{23}+\frac{3y}{23}$$ We have then that $$a \begin{pmatrix} 3 \\ -1 \end{pmatrix} + b \begin{pmatrix} 2 \\ 7 \end{pmatrix}=\begin{pmatrix} 3a+2b \\ -a+7b \end{pmatrix}=\begin{pmatrix} x \\ y \end{pmatrix}$$ Which follows from calculations.

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    But aren't you just looping around the problem instead of proving there is actually such values of a and b? – 2017-02-17
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    @ReginaDea I proved that there exists such $a$ and $b$ in $\frac{7x}{23}-\frac{2y}{23}$ and $\frac{x}{23}+\frac{3y}{23}$. – 2017-02-17
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    But you are relying on that there is a x and y value. – 2017-02-17
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    @ReginaDea Yes, $x,y \in \mathbb{R}$, so they exist. They're just arbitrary reals. – 2017-02-17
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    Oh, I see. Thanks for explaining! – 2017-02-17