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I need to make sure I'm understanding this correctly. I skipped a few steps to reduce typing, but let me know if I need to clarify something.

Question asks:

Find $N(A)$ for $A$ = \begin{bmatrix} -3 & 6 & -1 & 1 & -7 \\ 1 & -2 & 2 & 3 & -1\\ 2 & -4 & 5 & 8 & -4 \\ \end{bmatrix}

First thing I did was put the augmented matrix into reduced echelon row:

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

$(1)$

So then...

$x=\begin{bmatrix} x_1\\ x_2 \\ x_3\\ x_4\\ x_5\\ \end{bmatrix} = \begin{bmatrix} 2x_2 + x_4 - 3x_5\\ x_2 \\ -2x_4 + 2x_5\\ x_4\\ x_5\\ \end{bmatrix}$

$(2) $

Since $x_2, x_4$ and $x_5$ are free variables..

$ x_2 \begin{bmatrix} 2\\ 1 \\ 0\\ 0\\ 0\\ \end{bmatrix} + x_4 \begin{bmatrix} 1\\ 0 \\ -2\\ 1\\ 0\\ \end{bmatrix} + x_5 \begin{bmatrix} -3\\ 0 \\ 2\\ 0\\ 1\\ \end{bmatrix}$

$(3)$

Resulting in..

$N(A)= \left( \begin{bmatrix} 2\\ 1 \\ 0\\ 0\\ 0\\ \end{bmatrix} , \begin{bmatrix} 1\\ 0 \\ -2\\ 1\\ 0\\ \end{bmatrix} , \begin{bmatrix} -3\\ 0 \\ 2\\ 0\\ 1\\ \end{bmatrix} \right)$

$(4)$

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1 Answer

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As mentioned in the comments, provided your arithmetic is accurate, this is the correct response. The idea behind the null space of a matrix is that it is precisely those vectors in the domain being sent to the $\mathbf{0}$ vector in the codomain. So, what you have (correctly) done, is determined the solution set of $A\mathbf{x}=\mathbf{0}$.

You did this by finding the null space of a reduced row echelon form of $A$, which has the same null space as $A$. That is, if $B$ is the reduced row echelon form for $A$ that you found, $A\mathbf{x}=\mathbf{0}$ if and only if $B\mathbf{x}=\mathbf{0}$. So, $N(B)=N(A)$.

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