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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 $x_2, x_3$ 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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    $\begingroup$ Assuming your arithmetic checks out in row reducing, then $N(A)$ is the span of those vectors, yes! $\endgroup$ – Santana Afton Feb 24 '17 at 1:52
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    $\begingroup$ Why bother to augment the matrix with a zero column? It’s going to remain zero through all of the manipulations and doesn’t contribute anything to the answer, anyway. You really only need to augment when you’re solving an inhomogeneous system, inverting a matrix, or the like? Also, you could checking your work yourself by multiplying each of the resulting vectors by $A$. If you don’t get zero, you’re making a mistake somewhere along the way. $\endgroup$ – amd Feb 24 '17 at 2:02
  • $\begingroup$ Yeah, I realized that as well after completing the problem. That was going to be my next question, if I actually needed to include the zero column. I had it written that way on my lecture notes. Lol. Thanks for that last tip, will do. @amd $\endgroup$ – cisco Feb 24 '17 at 2:21
  • $\begingroup$ It's easy enough to check that these are in the null space (just multiply the matrix $A$ times each) and are linearly independent. $A$ certainly has rank at least $2$ (if it just had rank $1$, the rows would all be scalar multiples of one row), so the null space can't have dimension more than $3$. $\endgroup$ – Robert Israel Feb 24 '17 at 2:30
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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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  • $\begingroup$ When determining the free variables, I was told the amount of free variables will be the "numbers of columns minus the number of rows". Do I have to take into account the column of zeros? For example, do I go off of the number of columns of the original A given in the question, or the augmented A (1) to find the number of free variables? $\endgroup$ – cisco Feb 24 '17 at 2:29
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    $\begingroup$ The number of free variables is the number of columns minus the number of nonzero rows in the echelon form. Yes, this includes any column of all $0$'s (but such a column would have been all $0$'s in the original matrix as well). $\endgroup$ – Robert Israel Feb 24 '17 at 2:33

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