1. ## Nullspace Q

The question reads: "For the matrix A below, find a spanning set for its Nullspace.
$A=\left( \begin{array}{cccc} 1 & 3 & -2 & 1 \\ 2 & 1 & 3 & 2 \\ 3 & 4 & 5 & 6 \end{array}\right)$"

So... I want to solve for $x$ such that Ax = 0.

I row reduce; r2 -> r2-2r1, r3->r3-3r1

$A=\left( \begin{array}{cccc} 1 & 3 & -2 & 1 \\ 0 & -5 & 7 & 0 \\ 0 & -5 & 11 & 3 \end{array}\right)$

then r3->r3-r2

$A=\left( \begin{array}{cccc} 1 & 3 & -2 & 1 \\ 0 & -5 & 7 & 0 \\ 0 & 0 & 4 & 3 \end{array}\right)$

Giving $x_{1}+3x_{2}-2x_{3}+x_{4} = 0$
$-5x_{2}+7x_{3} = 0
$
$4x_{3}+3x_{4}$.

So we can say $x_{4} = \frac{4x_{3}}{3}$, $x_{3}=x_{3}$, $x_{2} = \frac {7x_{3}}{5}$, $x_{1} = \frac{-26x_{3}}{15}$.

So the Nullspace is spanned by $\left (\begin{array}{c} \frac{-26}{15} \\ \frac{7}{15} \\ 1 \\ \frac{4}{3} \end{array}\right)x_{3}$

Should Ax not then be zero?

$\left(\begin{array}{cccc} 1 & 3 & -2 & 1 \\ 0 & -5 & 7 & 0 \\ 0 & -5 & 11 & 3 \end{array}\right) \left (\begin{array}{c} \frac{-26}{15} \\ \frac{7}{15} \\ 1 \\ \frac{4}{3} \end{array}\right)x_{3} = \left(\begin{array}{c} -1 \\ \frac{8}{3} \\ \frac{29}{3} \end{array} \right)$

I'm obviously wrong somewhere... but I can't spot where...

2. ## starting off...

$
4 x_3 + 3 x_4 = 0
$

so shouldn't
$
x_4 = - 4 x_3 / 3
$

Note the negative sign, and keep on going...

3. Got it. Thank you.