Simultaneous Equations and Vectors The question I am currently working on is, "...find $a$ and $b$ such that $\vec{v} = a \vec{u} + b \vec{w}$, where $ \vec{u} = \langle 1,2 \rangle$, $\vec{w} = \langle 1,-1 \rangle$, and $\vec{v} = \langle 2,1 \rangle$ Before I asked a question on this forum, I did a google search to see if it had been previously asked. Indeed, I found one on yahoo answers: http://answers.yahoo.com/question/index?qid=20080719094705AAiMeCG. However, it wasn't exactly helpful. For the best answer, I understand the solution up until the part, "comparing LHS and RHS." What did the person do, why can someone just split up the equation like the person did?
 A: $\alpha \vec{u} + \beta \vec{w}=\langle \alpha+\beta,2 \alpha-\beta\rangle=\langle2,1\rangle.$  and now $\alpha+\beta =2$ and $2\alpha-\beta=1$. Now it's simple. 
A: Putting in the values for $\vec{v}, \vec{u}$ and $\vec{w}$ in the equation, we get:
$$ \langle 2, 1 \rangle = a * \langle 1, 2 \rangle + b * \langle 1, -1 \rangle $$
Here, $a$ and $b$ are just scalar multipliers and nothing else. Using the property of vectors and their scalaar multiplication:
$$ f * \vec{v} = f * \left( v_1, v_2, \dots v_n \right) = \left( f*v_1, f*v_2, \dots f*v_n \right) $$
You'll arrive at two linear equations in $a$ and $b$ as follows:


*

*$$ a + b = 2 $$

*$$ 2a - b = 1 $$


Solving which, you get $a = 1$ and $b = 1$.
A: Your question is basically one of solving a system $y=Ax$ for a fixed $A$ and $y$. That's exactly what the multiplication of an $n\times n$ matrix with a vector of length $n$ is, a linear combination of the vectors in the matrix. 
We'll work explicitly in the 2 dimensional case, as that is the context of your problem. We're given two vectors $\vec v = <a,b>,$ and $\vec w = <c,d>$. We're trying to find $\alpha$ and $\beta$ such that $$\alpha \vec v + \beta \vec w = <e,f>$$
How do we do this? Well we use a wonderful piece of notation called a matrix. This can be represented by the following:
$$\left( \begin{array}{ccc}
a & c \\
b & d \end{array} \right)
\left( \begin{array}{ccc}
\alpha \\
\beta  \end{array} \right)=
\left( \begin{array}{ccc}
e \\
f  \end{array} \right)$$
or in your case:
$$\left( \begin{array}{ccc}
1 & 1 \\
2 & -1 \end{array} \right)
\left( \begin{array}{ccc}
\alpha \\
\beta  \end{array} \right)=
\left( \begin{array}{ccc}
2 \\
1  \end{array} \right)$$
Then you can invert the matrix using Cramer's rule, which solves the system $y=Ax$ by finding a matrix $A^{-1}$ such that $A^{-1}y=x$.  This only works when the determinant of the matrix is non-zero, because the formula for the inverse is: $$A^{-1} = \frac{1}{\det A}\operatorname{adj}A$$
Where $\det$ gives the determinant of $A$, and $\operatorname{adj}$ gives the adjugate matrix. Calculating these values allows us to find the inverse matrix:
$$\left( \begin{array}{ccc}
\frac{1}{3} & \frac{1}{3} \\
\frac{2}{3} & \frac{-1}{3} \end{array} \right)$$
And to find $\alpha, \beta$ you simply multiply this by the vector $\vec y$, and get $$\left( \begin{array}{ccc}
\alpha \\
\beta  \end{array} \right)
= \left( \begin{array}{ccc}
\frac{1}{3} & \frac{1}{3} \\
\frac{2}{3} & \frac{-1}{3} \end{array} \right)\left( \begin{array}{ccc}
2 \\
1  \end{array} \right)\\
 \left( \begin{array}{ccc}
\alpha \\
\beta  \end{array} \right)= \left( \begin{array}{ccc}
1 \\
1  \end{array} \right)$$
Which gives your solution. There are easier solutions for such a simple case, but this generalizes far better. 
