# Is $b=\left(\begin{smallmatrix}2\\-1\end{smallmatrix}\right)$ in the null space of $A=\left(\begin{smallmatrix}1&2\\3&4\\5&6\end{smallmatrix}\right)$?

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

• 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! – Nigel Overmars Sep 18 '16 at 13:27
• perfect thanks for the help do you know how to help me with this question? – guestuser Sep 18 '16 at 13:28
• Well, for $b$ to be in the null space you need to have $Ab = 0$, so why don't you check that? – Nigel Overmars Sep 18 '16 at 13:29
• so just multiply the two matrices Ab? – guestuser Sep 18 '16 at 13:30
• sorry I made a mistake in by b matrix but I just edited it – guestuser Sep 18 '16 at 13:32

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 $b \not \in \operatorname{Null}(A)$.