Determine basis for solution space when there are more unknowns than equations Let's say we have
$$3x_1+x_2+x_3+x_4=0$$
$$5x_1-x_2+x_3-x_4=0$$
Then I would have matrix $\begin{bmatrix} 3 &1&1&1\cr8&0&2&0 \end{bmatrix}$ which simplifies to $\begin{bmatrix}-1&1&0&1\cr4&0&1&0\end{bmatrix}$
then $4x_1+x_3=0$ and $-x_1+x_2+x_4=0$ after substitution we get $\frac{1}{4}x_3+x_2+x_4=0$
and I'm not sure what to do from here.
 A: \begin{bmatrix}-1&1&0&1\cr4&0&1&0\end{bmatrix}

Multiply 1st row by $4$ and add it to 2nd row to get : 
\begin{bmatrix}-1&1&0&1\cr 0&4&1&4\end{bmatrix} 
Let $x_3=t$ and $x_4=s$ and hence you have : $-x_1+x_2=-s$ and $4x_2=-t-4s$ and hence $-x_1=-s+t/4+s=t/4\implies x_1=-t/4$
$(x_1,x_2,x_3,x_4)^T=(-t/4,-t/4-s,t,s)^T=s(0, - 1,0,1)^T+t(-1/4,-1/4, 1,0)^T$

Basis of solution space (Null space in this particular case) is set $\{(0, -1,0,1)^T, (-1/4,-1/4, 1,0)^T\} $

A: Add the two equations to find:  $$8 x_1 + 2 x_3 = 0$$
or
$$x_3 = -4 x_1$$
And so....
A: You already have all the information, namely the echelon form, you need to write out the general solution.
The matrix $\begin{bmatrix}-1&1&0&1\cr4&0&1&0\end{bmatrix}$ you got tells you that
you can have $x_1$ and $x_2$ as the free variables and any solution can be written as
$$
x = a(1,0,-4,1)^T+b(0,1,0,-1)^T
$$
Let me know if you need more elaboration.
A: Take it to reduced row-echelon form. You have
$$\begin{bmatrix}-1&1&0&1\\4&0&1&0\end{bmatrix} \sim \begin{bmatrix}\color{red}1&0&\frac{1}{4}&0\\0&\color{red}1&\frac{1}{4}&1\end{bmatrix}.$$
From here, identify the pivots, coloured in red, and the columns without pivots in them. In this case, we have the third and fourth columns. This means it is safe to consider the third and fourth variables (i.e. $x_3$ and $x_4$) as free variables. The other two variables, $x_1$ and $x_2$, will have a unique solution given fixed values of $x_3$ and $x_4$.
Our equations are
\begin{align*}
x_1 + \frac{1}{4} x_3 &= 0 \\
x_2 + \frac{1}{4} x_3 + x_4 &= 0,
\end{align*}
or equivalently,
\begin{align*}
x_1 &= -\frac{1}{4} x_3 \\
x_2 &= -\frac{1}{4} x_3 - x_4 \\
x_3 &= x_3 \\
x_4 &= x_4.
\end{align*}
This gives us
\begin{align*}
(x_1, x_2, x_3, x_4) &= \left(-\frac{1}{4}x_3, -\frac{1}{4}x_3 - x_4, x_3, x_4\right) \\
&= \left(-\frac{1}{4}x_3, -\frac{1}{4}x_3, x_3, 0\right) + (0, -x_4, 0, x_4) \\
&= x_3\left(-\frac{1}{4}, -\frac{1}{4}, 1, 0\right) + x_4(0, -1, 0, 1).
\end{align*}
If you're up to bases yet, this tells you that
$$\left\{\left(-\frac{1}{4}, -\frac{1}{4}, 1, 0\right) + (0, -1, 0, 1)\right\}$$
forms a basis for your solution space.
