<s>[INST] 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
 [/INST] 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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