We want to find the nullspace of $A$

$$$A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \end{bmatrix}$$$

$A$ and $U$ have the same nullspace. Elimination brings us to $U$

$\begin{equation*} A \implies \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & -1 & -2 & -3 \end{bmatrix} \implies \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} =U \end{equation*}$

We identify our pivot columns and our free colums. The pivot columns, $x_1$, and $x_2$ contain our pivots. The other columns, $x_3$ and $x_4$, are free. Now we have two choices: back subsitution, and solving by row reduced echelon form.

## Back Subsitution

For each free column, $x_n$:

1. Set that col to 1
2. Set all the other free columns to 0
3. Solve all equations for 0 with back subsitution

The upper triangular matrix above, $U$, becomes:

\begin{align} x_1 &+& x_2 &+& x_3 &+& x_4 &= 0 \\ && x_2 &+& 2x_3 &+& 3x_4 &= 0 \\ && && && 0 &= 0 \end{align}

Do back subsitution with $x_3 = 1$ and $x_4 = 0$:

\begin{align} 0 &= x_2 + 2 &\implies& x_2 = -2 \\ 0 &= x_1 -2 + 1 &\implies& x_1 = 1 \end{align} $\begin{equation*} A \begin{bmatrix} 1 \\ -2 \\ 1 \\ 0\end{bmatrix} = 0 \end{equation*}$

Do back subsitution with $x_3 = 0$ and $x_4 = 1$:

\begin{align} 0 &= x_2 + 3 &\implies& x_2 = -3 \\ 0 &= x_1 -3 + 1 &\implies& x_1 = 2 \end{align} $\begin{equation*} A \begin{bmatrix} 2 \\ -3 \\ 0 \\ 1 \end{bmatrix} = 0 \end{equation*}$

One way to write the complete solution:

$\begin{equation*} x_3 \begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 2 \\ -3 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} x_3 + 2x_4 \\ -2x_3 - 3x_4 \\ x_3 \\ x_4 \end{bmatrix} \end{equation*}$

## Row Reduced Echelon Form

However, it is quicker to "complete" elimination, and find the solution from rref. $A$, $U$, and $R$ have the same nullspace. Our example matrix, $A$, is very close to rref already (the pivots are already 1).

$\begin{equation*} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \implies \begin{bmatrix} 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}$

Notice how if we set $x_3 = 1$ and $x_4 = 0$ we can read the values $x_1$ and $x_2$ from the third column (if we reverse the sign)

$A \begin{bmatrix} 1 \\ -2 \\ 1 \\ 0\end{bmatrix} = 0$

You cannot do back subsitution on a matrix in rref. This yields incorrect results.