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I could only get a part of the solution and not quite it. The problem is:

A=$\left[\begin{matrix} 1 & -1 & 2 &1 \\ 2 & 1 & 0 & 1 \end{matrix}\right]$

Which of the following sets is a basis for Null(A)?

The correct answer is :{$\left[\begin{matrix} 0\\ 1\\ 1\\ -1 \end{matrix}\right]$, $\left[\begin{matrix} -4\\ 5\\ 3\\ 3 \end{matrix}\right]$}

And what I got is : {$\left[\begin{matrix} -2\\ 4\\ 3\\ 0 \end{matrix}\right]$,$\left[\begin{matrix} -2\\ 1\\ 0\\ 3 \end{matrix}\right]$}

The 2nd vector of the correct answer is 2 of mine added together, I don't know why and I'm still missing the first one. I looked at this question and got the correct answer for it by trying to do it the same way I did mine. Find the basis for the null space of the given matrix and give nullity(A)?

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Both answers are correct. That space has infinitely many bases, of course.

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  • $\begingroup$ Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer. $\endgroup$ – Antoni Malecki Jan 9 at 16:07
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This looks like it came from a multiple-choice question. As has been pointed out, bases are not unique, so there’s no particular reason to expect that the basis that you computed is going to match any of the proposed bases in the question.

A better strategy would’ve been to go through the possible answers and eliminate those that can’t be a null space basis. First, it’s clear from inspection that $A$’s nullity is 2, so if any of the possible answers don’t consist of two vectors, you can throw those out. Next, the vectors in a basis must be linearly independent, which is trivial to check for a pair of vectors. Discard any of the remaining answers that are linearly dependent. For the remaining viable candidates, multiply each of the vectors by $A$ and check that you get zero for each one. Whatever’s left must be a basis for $N(A)$.

In this case, the two vectors in the given answer are obviously linearly independent, and their products with $A$ are both zero, so they are indeed a basis for $N(A)$ (as is the set you found).

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You probably went for the RREF of $A$, which is $$ \begin{bmatrix} 1 & 0 & -2/3 & -2/3 \\ 0 & 1 & 4/3 & 1/3 \end{bmatrix} $$ With $x_3=3$ and $x_4=0$ you get the first vector, with $x_3=0$ and $x_3=3$ you get the second one. Your solution is correct, good work.

If you try $x_3=1$ and $x_4=-1$, you find the first vector in the solution; with $x_3=3$ and $x_3=3$, you find the second one. I'm not sure how they got their vectors, to be honest.

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  • $\begingroup$ It was from a multiple choice so probably was done to cause confusion. $\endgroup$ – Antoni Malecki Jan 10 at 10:20

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