Linear combination $\{2,3\}$ I am trying to write \begin{bmatrix}
      2        \\[0.3em]
       3
     \end{bmatrix} as a linear combination of 
\begin{bmatrix}
      1         \\[0.3em]
       -1
     \end{bmatrix}
\begin{bmatrix}
      6         \\[0.3em]
       7
     \end{bmatrix}
I concluded that I couldn't. This is wrong but I don't see how I can combine these, what steps do I need to take to do this?
 A: I'll assume that these are vectors in $\mathbb{R}^2$.  Then what you want to do is find $a$ and $b$ so that $a\begin{bmatrix} 1\\ -1\end{bmatrix} + b\begin{bmatrix}6\\ 7\end{bmatrix} = \begin{bmatrix}2\\ 3\end{bmatrix}$ by solving the system of equations
$$\left\{\begin{aligned}a+6b &=2 \\ -a+7b &= 3\end{aligned}\right.$$
A: If it wants to be true, you should find $a,b\in\mathbb R$ such that $$a(1,-1)+b(6,7)=(2,3)$$ Equivalently, you should check if the following system has any solution or not: $$a+6b=2,~~-a+7b=3$$ Note that $\begin{vmatrix}
  1 & 6 \\
  -1 & 7 \\ \end{vmatrix}=7+6=13\neq0$
A: Note that
$\begin{bmatrix} 2 \\ 3 \end{bmatrix}
  = a \begin{bmatrix} 1 \\ -1 \end{bmatrix}
  + b \begin{bmatrix} 6 \\ 7 \end{bmatrix}
= \begin{bmatrix} 1 & 6 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}$,
and now solve a simple system of eqautions.
A: If you write $\begin{bmatrix}
      2        \\[0.3em]
       3
     \end{bmatrix}=a\begin{bmatrix}
      1        \\[0.3em]
       -1
     \end{bmatrix}+b\begin{bmatrix}
      6        \\[0.3em]
       7
     \end{bmatrix}$  you get $2=a+6b,3=-a+7b$  
Adding the two gives $5=13b$   and we find $b=\frac5{13},a=-\frac {4}{13}$
A: Set
$\begin{bmatrix} 2 \\ 3 \end{bmatrix} = a\begin{bmatrix} 1 \\ -1 \end{bmatrix} + b\begin{bmatrix} 6 \\ 7 \end{bmatrix}; \tag{1}$
then
$a + 6b = 2, \tag{2}$
and
$-a + 7b = 3. \tag{3}$
Adding these equations yields
$13b = 5, \tag{4}$
or
$b = \frac{5}{13}. \tag{5}$
Using this value of b in (2) we obtain
$a = 2 - \frac{30}{13} = -\frac{4}{13}. \tag{6}$
It is easily checked that these values of $a$ and $b$ satisfy (1), (2), and (3).
The point is, of course, that (1) is really just a simple linear system in the two variables $a$ and $b$.  
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
