<s>[INST] Linear transformation ker and image Let $\varphi\colon \mathbb{R}^4 \rightarrow \mathbb{R}^3$ be described by $\varphi(X)=AX$ where 
$A=\begin{pmatrix}
 3 & 2 & 1 & 3 \\ 
 1 & 1 & 1 & 1 \\ 
 2 & 1 & 0 & 1
\end{pmatrix} $
. Find base vectors of $\ker \varphi$ and $\operatorname{Im} \varphi$.   In my opinion those vectors will be $[-1,1,0,0]$ and $[-1,0,1,0]$ for kernel and $[3,1,2]$  and  $[3,1,1] $ for image. Am I correct?
 [/INST] Start with the definitions:
$$\ker\varphi:=\{X\in\mathbb R^4:\varphi(X)=0\},$$ that is, the set of all $(x_1,x_2,x_3,x_4)$ such that
$$
\begin{bmatrix}
 3 & 2 & 1 & 3 \\ 
 1 & 1 & 1 & 1 \\ 
 2 & 1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x_1\\x_2\\x_3\\x_4
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\0
\end{bmatrix}.
$$
Can you find the solution set of this system? If so, you will be able to write down a basis for $\ker\varphi$. Note that row reducing yields
$$
\begin{bmatrix}
 3 & 2 & 1 & 3 \\ 
 1 & 1 & 1 & 1 \\ 
 2 & 1 & 0 & 1
\end{bmatrix}
\to
\begin{bmatrix}
 1 & 0 & -1 & 0 \\ 
 0 & 1 & 2 & 0 \\ 
 0 & 0 & 0 & 1
\end{bmatrix}
$$
so
\begin{align}
x_1-x_3&=0,\\
x_2+2x_3&=0,\\
x_4&=0
\end{align}
and thus
$x_4=0$, $x_2=-2x_3$, $x_1=x_3$ and $x_3$ is free. Hence a basis for $\ker\varphi$ is
$$
\left\{\begin{bmatrix} 1\\-2\\1\\0\end{bmatrix}\right\}.$$
As for $\text{Im}\,\varphi$, again go to the definition
$$
\text{Im}\,\varphi:=\{Y\in\mathbb{R}^3:\varphi(X)=Y\text{ for some }x\in\mathbb{R}^4\}.$$ So look at all possible outputs of $\varphi$ acting on an $X$ after row reducing:
\begin{align}
\begin{bmatrix}
 3 & 2 & 1 & 3 \\ 
 1 & 1 & 1 & 1 \\ 
 2 & 1 & 0 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
 1 & 0 & -1 & 0 \\ 
 0 & 1 & 2 & 0 \\ 
 0 & 0 & 0 & 1
\end{bmatrix}.
\end{align}
Since you have leading ones in columns 1, 2, and 4, use columns 1, 2, and 4 of the original matrix as the basis for $\text{Im}\,\varphi$.
</s>