Answer
See below
Work Step by Step
Let $x=(x,y)\in R^2$
We have: $T(x)=ax\rightarrow T(x,y)=\begin{bmatrix}
3 & 6\\
1 & 2
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix}=(3x+6y,x+2y)$
Obtain: $Ker (T)=\{x \in R^2: T(x)=0\}\\
=\{(x,y)\in R^2: T(x,y)=(0,0)\\
=\{(x,y)\in R^2:(3x+6y,x+2y)=(0,0)$
We have the system: $3x+6y=0\\
x+2y=0\\
\rightarrow x=-2y$
Thus, $Ker (T)=\{(-2y,y):y \in R\}\\
=\{y(-2,1): y \in R\}\\
=span \{(-2,1)\}\\
\rightarrow \dim [Ker (T)]=1$
$Ker(T)$ is the line passing through the origin $(0,0)$ with the direction $(-2,1)$
Then, $Rng(T)=\{T(x):x \in R^2\}\\
=\{T(x,y):x,y \in R\}\\
=\{(3x+6y,x+2y):x,y \in R\}
=\{ x(3,1)+y(6,2): x,y \in R\}
=span \{(3,1),(6,2)\}\\
=span \{(3,1)\}\\
\rightarrow dim [Rng(T)]=1$
Verify a Rank Nullity Theorem:
$\dim [Ker (T)]+\dim[Rng(T)]=1+1=2=\dim R^2$
