Gaussian elemination method question? Change rows if neccessary

Question

We have the following system.I have to solve this using Gaussian elemination.

We have here

x1+x2+x4=2

2x1+x2-x3+x4=2

4x1-x2-2x3+2x4=0

3x1-x2-x3+2x4=-3

The augmented matrix is \begin{matrix} 1 & 1 & 0 & 1 | 2\\ 2 & 1 & -1&1 |1\\ 4 & -1 & -2 & 2 |0 \\ 3 & -1& -1&2 |3 \end{matrix}

I multiply the first row with -2 and add it to the second row.Then I multiply the first row with -3 and add it to the last row.So I have

\begin{matrix} 1 & 1 & 0 & 1 | 2\\ 0 & -1 & -1&1 |-3\\ 4 & -1 & -2 & 2 |0 \\ 0 & -4& -1&1 |-3 \end{matrix}

How do I continue to solve this now?

Answer 1

First make zero all the entries on the first column behind the firsty entry:

$$\begin{pmatrix}1&1&0&1&2\\2&1&\!\!-1&1&2\\4&\!\!-1&\!\!-2&2&0\\3&\!\!-1&\!\!-1&2&\!\!-3\end{pmatrix}\stackrel{\begin{align}&R_2-R_1\\&R_3-4R_1\\&R_4-3R_1\end{align}}\longrightarrow\begin{pmatrix}1&1&0&1&2\\0&\!\!-1&\!\!-1&\!\!-1&\!\!-2\\0&\!\!-5&\!\!-2&\!\!-2&\!\!-8\\0&\!\!-4&\!\!-1&\!\!-1&\!\!-9\end{pmatrix}$$

Now the same with the entries on the second column below the $\;2-2\;$ entry (which is $\;-1\;$) , to get :

$$\begin{pmatrix}1&1&0&1&2\\0&\!\!-1&\!\!-1&\!\!-1&\!\!-2\\0&\;0&3&3&2\\0&0&3&3&\!\!-1\end{pmatrix}\stackrel{R_4-R_3}\longrightarrow\begin{pmatrix}1&1&0&1&2\\0&\!\!-1&\!\!-1&\!\!-1&\!\!-2\\0&\;0&3&3&2\\0&0&0&0&\!\!-3\end{pmatrix}$$

So what can you deduce from this?