Using linear algebra to solve algebra I feel like in my class all we do is memorize rules, lists of rules, terminology and more lists of rules. I have absolutely no clue how to apply any of these concepts. I have the matrix (which is representing a linear set of equations I think):
$$\left[\begin{array}{ccc|c}
      1 & 0 & 3 & 1       \\[0.55ex] 
     -1 & 1 & -2 & 4      \\[0.55ex]
\end{array}\right]$$
I am suppose to find a linear combination of the first 3 vectors that is a liner combination of the last one. I am not sure how to do this, I get the weird looking matrix:
$$\left[\begin{array}{ccc|c}
      1 & 0 & 3 & 1       \\[0.55ex] 
     0 & 1 & 1 & 5      \\[0.55ex]
\end{array}\right]$$
which gives me a really weird equation that I don't know how to work with. What can I do from here? I am not sure how to transform this into anything useable. 
 A: The second matrix is not at all weird: it's the row reduced form of your original matrix and it gives all the information you need: the dominant columns are the first and the second; moreover the last column is
$$
\begin{bmatrix}
1\\
5
\end{bmatrix}
=
1\begin{bmatrix}
1\\
0
\end{bmatrix}
+
5\begin{bmatrix}
0\\
1
\end{bmatrix}
$$
so that, returning to the original matrix, you can use the same coefficients in front of the first and second column:
$$
\begin{bmatrix}
1\\
4
\end{bmatrix}
=
1\begin{bmatrix}
1\\
-1
\end{bmatrix}
+
5\begin{bmatrix}
0\\
1
\end{bmatrix}
$$
A: You are on the right way.
Yes, the matrix represents in all likelyhood a system of linear equations, namely
$$\ \ \ 1\cdot x+0\cdot y+3\cdot z=1$$
$$-1\cdot x+1\cdot y-2\cdot z=4$$
but it suffices to just remember the coefficients as written in the matrix. 
The row operation you have performed corresponds to a combination of the euqations, namely their sum. 
The system you get is easier to work with, because there are as many zeros as possible. Since you have only two equations for three variables you have at least one degree of freedom, so we can just choose $z=0$ (because $0$ makes things easier and $z$ because the column has the most non-trivial entries).
To solve you now have to choose $x=1$ and $y=5$. You are done.
