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Is vector $\mathrm b = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ in the null space of $\mathrm A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6\end{pmatrix}$?

My solution:

I took matrix A and put it in reduced echelon form. I calculated this to be: $$\begin{pmatrix} 1 & 0 & | & 0 \\ 0 & 1 & | & 0 \\ 0 & 0 & | & 0\end{pmatrix}$$

So $x_1= 0$ and $x_2 = 0$. But from here I am not sure how to answer the question

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  • $\begingroup$ You're welcome! You can now look at the code that I used to format it like this so you can do it yourself next time! $\endgroup$ – Nigel Overmars Sep 18 '16 at 13:27
  • $\begingroup$ perfect thanks for the help do you know how to help me with this question? $\endgroup$ – guestuser Sep 18 '16 at 13:28
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    $\begingroup$ Well, for $b$ to be in the null space you need to have $Ab = 0$, so why don't you check that? $\endgroup$ – Nigel Overmars Sep 18 '16 at 13:29
  • $\begingroup$ so just multiply the two matrices Ab? $\endgroup$ – guestuser Sep 18 '16 at 13:30
  • $\begingroup$ sorry I made a mistake in by b matrix but I just edited it $\endgroup$ – guestuser Sep 18 '16 at 13:32
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We have that $b \in \operatorname{Null} (A)$ if and only if $Ab =0$, so in this case we have $$Ab = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} \neq \vec{0}$$ Hence $b \not \in \operatorname{Null}(A)$.

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  • $\begingroup$ awesome that makes sense to me, thanks for your time $\endgroup$ – guestuser Sep 18 '16 at 13:37

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