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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.

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  • $\begingroup$ Do you know about free variables? $\endgroup$ – Tojrah Apr 16 at 5:45
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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 $Ax=0$

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Columns 1 and 3 have the pivots. So the other two columns (2 and 4) correspond to the free variables.

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.

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