Linear Combinations I am currently trying to learn linear combinations and am stuck on two different problems. 
The first states:
Im given x1 = (4,3) x2 = (1,1) and x3 = (5,4). Find two linear combinations that get z = (1,0).

So does this mean adding and subtracting these together to find ways of get (1,0). If so, I can not find any way to do this because I keep getting (1,1) or (0,0). Am I having the correct approach?
The second problem of this kind is:
x1 = $\begin{bmatrix}1\\3\\4\end{bmatrix}$ x2 = $\begin{bmatrix}3\\1\\6\end{bmatrix}$ x3 = $\begin{bmatrix}7\\-3\\10\end{bmatrix}$
It says find a linear combination of the vectors that give the zero vector? Would I do the say as above and try to get it to all zeros?
Thanks a lot
 A: We need to find two solutions to $$a(4,3)+b(1,1)+c(5,4)=(1,0)$$ for $a,b,c \in \mathbb{R}$ (I presume).  This is equivalent to finding two solutions to the system of equations
\begin{align*}
4a+b+5c &= 1 \\
3a+b+4c &= 0.
\end{align*}
If there's one solution, there will be infinitely many (since there will be a free variable).
For this problem in particular, there's a way to "cheat": (a) try $a=0$ and find values of $b$ and $c$ that work by inspection, and (b) try $c=0$ and find values of $a$ and $b$ that work by inspection.
The second question is similar, we want to find $a,b,c \in \mathbb{R}$ (not all zero) such that
$$a\begin{bmatrix}1\\3\\4\end{bmatrix} +b \begin{bmatrix}3\\1\\6\end{bmatrix}+c  \begin{bmatrix}7\\-3\\10\end{bmatrix}=\begin{bmatrix}0\\0\\0\\\end{bmatrix}$$
This time, it's probably easier to solve the system of equations
\begin{align*}
a+3b+7c &=0 \\
3a+b-3c &=0 \\
4a+6b+10c &=0.
\end{align*}
using Gaussian elimination.
A: A linear combination means adding scalar multiples of the two vectors.
If your (real) vectors have dimension $2$ then if you have two vectors (a) neither of which is the zero vector, and (b) one is not a scalar multiple of the other, then you can create a linear combination of them to get any other two-dimensional (real) vectors.
So, for example, $\vec{x_1} - 3 \vec{x_2} = \vec{z}.$  Here, $z$ is a linear combination of $x_1, x_2.$
A more general way of solving this is to look at each component individually as its own equation.  So we're looking for scalars $a,b$ such that:
$$a \cdot \vec{x_1} + b \cdot \vec{x_2} = \vec{z}.$$
This implies
$$a \cdot 4 + b \cdot 1 = 1 \\ a \cdot 3 + b \cdot 1 = 0$$
or
$$4a + b = 1 \\ 3a + b = 0$$
Subtracting the second equation from the first gives $a=1$ and then resubstituting gives $b=-3$ which is what we found above.
Doing this for three-dimensional vectors works exactly the same way, except you'll have three equations.
