First year linear algebra level.
A is a matrix
$$
\begin{bmatrix}
-1&1&1&3\\
2&-2&1&-3\\
3&-3&-2&-8\\
\end{bmatrix}
$$
Then describe the kernel of A as a span of linearly independent vectors.
I reduced it to RREF:
$$
\begin{bmatrix}
1&-1&0&-2\\
0&0&1&1\\
0&0&0&0\\
\end{bmatrix}
$$
But I can't connect it to how it becomes the answer, which is:
$$
span(
\begin{bmatrix}
2\\0\\-1\\1
\end{bmatrix},
\begin{bmatrix}
1\\1\\0\\0\\
\end{bmatrix}
)
$$
Also I'm not sure if the 2 is a typo on the answer sheet.
Best Answer
Call $C_k$ the columns of matrix $R$ (the RREF transform)
You see that :
$$C_1+C_2=0 \ \iff \ (1)C_1+(1)C_2+(0)C_3+(0)C_4=0 \tag{1}$$
Collect the coefficients (between parentheses) of this null linear combination : you will get the second of your vectors (presented in a line instead of a column).
For the other one, it is the same :
$$(2)C_1+(0)C_2+(-1)C_3+(1)C_4=0$$
Explanation : Relationship (1) can be written under the form :
$$\underbrace{\left(C_1|C_2|C_3|C_4\right)}_{R}\underbrace{\begin{pmatrix}1\\1\\0\\0\end{pmatrix}}_{V}=\underbrace{\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}_0 \ \ \implies \ \ V \in \ker R$$