Solving Ax=b Using the Basis of the Nullspace Let B = \begin{bmatrix}2&3&1&-1\\1&2&1&2\\3&5&2&1\\4&7&3&3\end{bmatrix}
Find the complete solution to the nonhomogenous system Bx=\begin{bmatrix}6\\-4\\2\\-2\end{bmatrix} by first computing a basis for the nullspace of B.
x = $c_1$ + $sc_2$ + $tc_3$
Where $c_1$, $c_2$ and $c_3$ are vectors.
I know how to normally solve an Ax=b equation, by taking the inverse of A and multiplying by b, but I'm really not understanding the method that's being used here.
 A: Start with the reduced row echelon form:
$$
\begin{align}
 \mathbf{B} &\mapsto \mathbf{E}_{\mathbf{B}} \\
%
\left[
\begin{array}{rrrr}
 2 & 3 & 1 & -1 \\
 1 & 2 & 1 & 2 \\
 3 & 5 & 2 & 1 \\
 4 & 7 & 3 & 3 \\
\end{array}
\right]
%
 &\mapsto
%
\left[
\begin{array}{rrrr}
 \color{blue}{1} & 0 & -1 & -8 \\
 0 & \color{blue}{1} & 1 & 5 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right]
\end{align}
$$
The message is that the fundamental columns are the first two, highlighted by blue coloring.
Your question is look for the values of $\alpha$ and $\beta$ which solve
$$
\left[
\begin{array}{r}
 6 \\ -4 \\ 2 \\ -2
\end{array}
\right]
%
=
%
\alpha
\left[
\begin{array}{r}
 2 \\ 1 \\ 3 \\ 4
\end{array}
\right]
%
+
%
\beta
\left[
\begin{array}{r}
 3 \\ 2 \\ 5 \\ 7
\end{array}
\right]
$$
We are look for two unknowns, so solve using two rows. Let's pick rows 1 and 3, liberating us from the chance of sign errors. The problem is
now
$$ 
%
\left[
\begin{array}{rr}
 2 & 3 \\
 3 & 5
\end{array}
\right]
%
\left[
\begin{array}{c}
 \alpha \\
 \beta
\end{array}
\right]
%
=
\left[
\begin{array}{c}
 6 \\
 2
\end{array}
\right]
%%
\quad \Rightarrow \quad
%
\left[
\begin{array}{c}
 \alpha \\
 \beta
\end{array}
\right]
=
%
\left[
\begin{array}{rr}
 5 & -3 \\
 -3 & 2
\end{array}
\right]
%
\left[
\begin{array}{c}
 6 \\
 2
\end{array}
\right]
%
=
%
\left[
\begin{array}{r}
  24 \\
 -14
\end{array}
\right]
$$
It doesn't seem that we need to resolve the nullspace. But it is a byproduct of the augmented reduction producing $\mathbf{E}_{\mathbf{B}}$:
$$
  \mathcal{N} \left( \mathbf{B}^{\mathrm{T}} \right) =
\text{span } \left\{ \,
%
\left[
\begin{array}{r}
  8 \\
 -5 \\
  0 \\
  1
\end{array}
\right], \,
%
\left[
\begin{array}{r}
 -1 \\
 -1 \\
  1 \\
  0
\end{array}
\right]
%
\, \right\}
$$
Solution
The solution to
$$
\begin{align}
\mathbf{B} x &= y \\
%%
\left[
\begin{array}{rrrr}
 2 & 3 & 1 & -1 \\
 1 & 2 & 1 & 2 \\
 3 & 5 & 2 & 1 \\
 4 & 7 & 3 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 x_{1} \\
 x_{2} \\
 x_{3} \\
 x_{4} 
\end{array}
\right]
%
&=
%
\left[
\begin{array}{r}
  6 \\
 -4 \\
  2 \\
 -2 
\end{array}
\right]
%
\end{align}
$$
is
$$
%
\left[
\begin{array}{c}
 x_{1} \\
 x_{2} \\
 x_{3} \\
 x_{4} 
\end{array}
\right]
%
=
\left[
\begin{array}{r}
  24 \\
 -14 \\
  0 \\
  0
\end{array}
\right]
%
+
%
s
\left[
\begin{array}{r}
  8 \\
 -5 \\
  0 \\
  1
\end{array}
\right]
%
+
%
t
\left[
\begin{array}{r}
 -1 \\
 -1 \\
  1 \\
  0
\end{array}
\right]
$$
where the constants $s$ and $t$ are arbitrary. 
To emphasize:
$$ 
\left[
\begin{array}{rrrr}
 2 & 3 & 1 & -1 \\
 1 & 2 & 1 & 2 \\
 3 & 5 & 2 & 1 \\
 4 & 7 & 3 & 3 \\
\end{array}
\right]
\left(
\left[
\begin{array}{r}
  24 \\
 -14 \\
  0 \\
  0
\end{array}
\right]
%
+
%
s
\left[
\begin{array}{r}
  8 \\
 -5 \\
  0 \\
  1
\end{array}
\right]
%
+
%
t
\left[
\begin{array}{r}
 -1 \\
 -1 \\
  1 \\
  0
\end{array}
\right]
\right)
%
=
%
\left[
\begin{array}{r}
 6 \\
 -4 \\
 2 \\
 -2 
\end{array}
\right]
$$
