<s>[INST] How to find $ker(A)$ How can I find $ker(A)$ for
$A=
\begin{pmatrix}
2 & 1 & 1  \\
4 & 2 & 2 \\
3 & 0 & 1  
\end{pmatrix}
$?
What does this represent?
Edit: OK, using the Hints i get
$2x+y+z=0$
$3x+z=0$
$y-x=0$
$y=x$
And now?
 [/INST] Straight from wikipedia:
$$\ker(A) = \left\{\mathbf{v}\in V : A\cdot \mathbf{x} = \mathbf{0}\right\}$$
In your case $V=\mathbb{R}^3$ so you want to find all vectors in the usual 3 space which make this true:
$$A\bf{x}=
\begin{pmatrix}
2 & 1 & 1  \\
4 & 2 & 2 \\
3 & 0 & 1  
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$
The science here is getting $A$ to reduced row echelon form. If that reduced matrix is $B$, then $\ker(A) = \ker(B)$.
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