# Determine the null space of the following matrix

Determine the null space of the following matrix:

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

For this question, I reduced the row echelon form into $$\begin{bmatrix} 1 & 2 & -3& -1 \\ 0& 0 &0 &1 \end{bmatrix},$$ but then I have no idea how to determine the null space, because there's no relationship between $$x_1, x_2, x_3, x_4$$.

• you are encouraged to include your attempt. – Siong Thye Goh Jan 6 at 6:44
• For this question, I reduced the row echelon form into ( 1 2 -3 -1 0 0 0 1 ), but then I have no idea how to determine the null space, because there's no relationship between x1, x2, x3, x4 – Shadow Z Jan 6 at 6:45
• Include your attempt in the original post directly. – Siong Thye Goh Jan 6 at 6:46
• You require $$\begin{bmatrix} 1 & 2 & -3& -1 \\ 0& 0 &0 &1 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3\\x_4 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0\\0 \end{bmatrix}$$ Now just find the relations between $x_1,x_2,x_3,x_4$ as in the answer below. – Shubham Johri Jan 6 at 6:57
• See math.stackexchange.com/a/1521354/265466 for how to read a basis for the null space directly from the RREF. – amd Jan 6 at 8:13

Great that you have found a row echelon form.

From the second row, we can conclude that $$x_4=0$$.

Also, from there, and the first row of the row echelon form, we have

$$x_1+2x_2-3x_3=0$$

Now you have a relationship between the variables. Hopefully you can take it from here.

• Thanks! Then I set x2=α, x3= β, then x1 = 3β-2α. So the null space is spanned by α(-2 1 0 0)^T + β(3 0 1 0)^T, is that true? – Shadow Z Jan 6 at 6:56
• @ShadowZ Yes, that is correct – Shubham Johri Jan 6 at 6:59
• Congratulations, good job. try to pick up mathjax, it makes maths prettier and more importantly makes communication smoother. – Siong Thye Goh Jan 6 at 6:59