Answer
See below
Work Step by Step
Let $x=(x,y)\in R^2$
We have: $T(x)=ax\rightarrow T(x,y)=\begin{bmatrix}
3 & 2 & -6 & 3\\
-2 & 3 & -1 &2
\end{bmatrix}\begin{bmatrix}
x\\
y\\z \\ v \end{bmatrix}=(3x+2y-6z+3v,-2x+3y-z+2v)$
Obtain: $Ker (T)=\{x \in R^4: T(x)=0\}\\
=\{(x,y,z,v)\in R^4: T(x,y,z)=(0,0)\\
=\{(x,y,z,v)\in R^4:(3x+2y-6z+3v,-2x+3y-z+2v)=(0,0)$
We have the system: $3x+2y-6z+3v=0\\
-2x+3y-z+2v=0\\
\rightarrow x=\frac{5}{3}x-\frac{16}{9}y\\
z=\frac{4}{3}x-\frac{5}{9}y$
Thus, $Ker (T)=\{(x,y,\frac{4}{3}x-\frac{5}{9}y,\frac{5}{3}x-\frac{16}{9}y):x,y \in R\}\\
=\{x(1,0,\frac{4}{3},\frac{5}{3})+y(0,1,-\frac{5}{9},-\frac{16}{9}):x,y \in R\}\\
=span\{(1,0,\frac{4}{3},\frac{5}{3});(0,1,-\frac{5}{9},-\frac{16}{9})\}\\
\rightarrow \dim [Ker (T)]=span \{(1,0,\frac{4}{3},\frac{5}{3});(0,1,-\frac{5}{9},-\frac{16}{9})\}$
Then, $Rng(T)=\{ T(x):x \in R^4\}\\
=span \{(3,-2);(2,3);(-6,-1);(3,2))\}$
