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**ThePerfectHacker** When you have the system $\displaystyle M\bold{x} = \bold{0}$ and you are solving for $\displaystyle \bold{x}$.

The matrix you end up with is actually:

$\displaystyle \begin{bmatrix} 1&0&1&-2&0 \\ 1&1&-3&4&0 \\ 3&1&-1&0&0 \\ 4&2&-4&4& 0 \\ 0&1&-4&6&0 \end{bmatrix}$

Now if you done the row-reductions (assuming you were correct!) we get:

$\displaystyle \begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&8&0\\0&0&0&16&0\\0&0&0&8&0 \end{bmatrix}$

You can bring the bottom three rows to:

$\displaystyle \begin{bmatrix} 1&0&1&-2&0\\0&1&-4&-2&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}$

However this matrix is still now in row-reduced echelon form.

We will add the (3rd row $\displaystyle \times$ 2) to 2nd row and 1st row.

$\displaystyle \begin{bmatrix} 1&0&1&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}$

Finally, add (2nd row $\displaystyle \times -\tfrac{1}{4}$) to 1st row:

$\displaystyle \begin{bmatrix} 1&0&0&0&0\\0&1&-4&0&0\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0 \end{bmatrix}$

This tells you that:

$\displaystyle x_1 = 0$

$\displaystyle x_2 - 4x_3 = 0$

$\displaystyle x_4 = 1$.

Thus, $\displaystyle x_1=0,x_4=1,x_2 = 4x_3$ (notice that $\displaystyle x_3$ is "free").

This means if $\displaystyle t\in \mathbb{R}$ then $\displaystyle \bold{x} = \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix}$ solves $\displaystyle M\bold{x} = \bold{0}$, and conversely.

Thus, the nullspace is $\displaystyle \left\{ \begin{bmatrix}0\\4t\\t\\1 \end{bmatrix} : t\in \mathbb{R} \right\} = \left\{ t \begin{bmatrix}0\\4\\1\\1 \end{bmatrix} : t\in \mathbb{R}\right\}$

We see the dimension is one, and so the nullity is one.