# Basis for nullspace - Free variables and basis for $N(A)$

I'm trying to understand the solution for a textbook question where I am asked to find a basis for $$N(A)$$. I took a screenshot of the page, and I circled the portion I don't quite understand. I don't understand where the values for $$s$$ and $$t$$ came from? Like how did they get $$s = 1$$, and $$t = -2$$ for the first part in that tuple? I tried finding a correlation between these values and the RREF matrix but I just couldn't see the connection. Any help or explanation would be appreciated.

• Do you know about free variables? – Tojrah Apr 16 at 5:45

We already transform $$A$$ into $$\begin{pmatrix} 1&-1&0&2\\0&0&1&-1\end{pmatrix}$$ Since first and third column contains the pivot, so $$x_1$$ and $$x_3$$ are pivot variables. That is, $$x_2$$ and $$x_4$$ are free. The task now is to solve $$x_1-x_2+2x_4=0\\x_3-x_4=0$$ Set $$x_2=1$$ and $$x_4=0$$ to see $$\begin{pmatrix} 1\\1\\0\\0 \end{pmatrix}$$ is a special solution. Similarly set $$x_2=0$$ and $$x_4=1$$ to see $$\begin{pmatrix} -2\\0\\1\\1 \end{pmatrix}$$ is a special solution. Now the combination $$s\begin{pmatrix} 1\\1\\0\\0 \end{pmatrix}+t\begin{pmatrix} -2\\0\\1\\1 \end{pmatrix}$$ gives all solution s to $$Ax=0$$
Then call $$x_4=t$$. Then the last equation says $$x_3 - x_4 =0 \leftrightarrow x_3 = x_4= t$$
Call the other free variable $$x_2=s$$. Then the first equation becomes, after substiting what we know so far:
$$x_1 - s + 2t = 0$$ from which
$$x_1 = s -2t$$ follows.