Answer
See below
Work Step by Step
Let $x=(x,y)\in R^2$
We have: $T(x)=ax\rightarrow T(x,y)=\begin{bmatrix}
-4 & 3 & 0\\
-5 & 2 & -3\\
15 & -7 & 6
\end{bmatrix}\begin{bmatrix}
x\\
y\\z \end{bmatrix}=(-4x+3y,-5x+2y-3z,15x-7y+6z)$
Obtain: $Ker (T)=\{x \in R^3: T(x)=0\}\\
=\{(x,y,z)\in R^3: T(x,y,z)=(0,0)\\
=\{(x,y,z)\in R^3:(-4x+3y,-5x+2y-3z,15x-7y+6z)=(0,0,0)$
We have the system: $-4x+3y=0\\
-5x+2y-3z=0\\
15x-7y+6z=0\\
\rightarrow x=\frac{4}{3}y\\
z=0\\
\rightarrow x=y=z=0$
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, $Rng(T)= R^3$
