Answer
See below
Work Step by Step
Let $x=(x,y)\in R^2$
We have: $T(x)=ax\rightarrow T(x,y)=\begin{bmatrix}
0 & -\frac{2}{3} & \frac{1}{3} \\
\frac{3}{2} & -1 & 1\\
-\frac{1}{4} & \frac{2}{5} & -\frac{2}{5}\\
-\frac{1}{4} & \frac{17}{15} & \frac{8}{15}
\end{bmatrix}\begin{bmatrix}
x\\
y\\z \end{bmatrix}=(-\frac{2}{3} x+\frac{1}{3}z,\frac{3}{2} x-y+z,-\frac{1}{4} x+\frac{2}{5} y-\frac{2}{5} z,-\frac{1}{4}x +\frac{17}{15} y+\frac{8}{15} z)$
Obtain: $Ker (T)=\{x \in R^3: T(x)=0\}\\
=\{(x,y,z)\in R^3: T(x,y,z)=(0,0)\\
=\{(x,y,z)\in R^3:(-\frac{2}{3} x+\frac{1}{3}z,\frac{3}{2} x-y+z,-\frac{1}{4} x+\frac{2}{5} y-\frac{2}{5} z,-\frac{1}{4}x +\frac{17}{15} y+\frac{8}{15} z)=(0,0,0,0)$
We have the system: $-\frac{2}{3}y+\frac{1}{3}z=0\\
\frac{3}{2}x-y+z=0\\
-\frac{1}{4}x+\frac{2}{5}y-\frac{2}{5}z=0\\
-\frac{1}{4}x+\frac{17}{15}y+\frac{8}{15}z=0\\
\rightarrow x=\frac{-2}{3}y\\
z=2y\\
y=0\\
\rightarrow x=y=z=0$
Thus, $Ker (T)=\{(0,0,0)\}\\
\rightarrow \dim [Ker (T)]=0$
Apply Rank Nullity Theorem:
$\dim [Ker (T)]+\dim [Rng(T)]=\dim R^3\\
0+\dim [Rng(T)]=3\\
\dim [Rng(T)]=3$
Then, $Rng(T)= span\{(0,\frac{3}{2},-\frac{1}{4},-\frac{1}{4});(-\frac{2}{3},-1,\frac{2}{5},\frac{17}{15});(\frac{1}{3},1,-\frac{2}{5},\frac{8}{15})$
