Answer
See below
Work Step by Step
Obtain: $Ker (T)=\{x \in R^2: T(x)=0\}\\
=\{\begin{bmatrix}
-x-y & 0 & 2x+2y\\
0 & 3x+3y & -9x-9y
\end{bmatrix}=\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}\}$
We have the system:
$-x-y=0\\
2x+2y=0\\
3x+3y=0\\
-9x-9y=0$
thus, $y=-x$
Thus, $Ker (T)=\{(x,-x):x\in R\}\\
=\{x(1,-1):x \in R\}\\
=span\{(1,-1)\}$
$\dim [Ker(T)]=1$
$Rng(T)=\{T(x): x \in R^2\}\\
=\{\begin{bmatrix}
-x-y & 0 & 2x+2y\\
0 & 3x+3y & -9x-9y
\end{bmatrix}:x,y \in R\}\\
=\{\begin{bmatrix}
-x & 0 & 2x\\
0 & 3x & -9x
\end{bmatrix}+\begin{bmatrix}
-y & 0 & 2y\\
0 & 3y & -9y
\end{bmatrix}:x,y \in R\}\\
=\{x\begin{bmatrix}
-1 & 0 & 2\\
0 & 3 & -9
\end{bmatrix}+y\begin{bmatrix}
-1 & 0 & 2\\
0 & 3 & -9
\end{bmatrix}: x,y \in R\}\\
=\{(x+y)\begin{bmatrix}
-1 & 0 & 2\\
0 & 3 & -9
\end{bmatrix}:x,y \in R\}\\
=span\{\begin{bmatrix}
-1 & 0 & 2\\
0 & 3 & -9
\end{bmatrix}\}$
Since $\{\begin{bmatrix}
-1 & 0 & 2\\
0 & 3 & -9
\end{bmatrix}\}$ is the basic for $Rng(T)$,
then $\dim [Rng(T)]=1$
