Answer
See below
Work Step by Step
Let $x=(x,y)\in R^2$
We have: $T(x)=ax\rightarrow T(x,y)=\begin{bmatrix}
1 & -1 & 0\\
0 & 1 & 2 \\
2 & -1 & 1
\end{bmatrix}\begin{bmatrix}
x\\
y \\
z\end{bmatrix}=(x-y,y+2z,2x-y+z)$
Obtain: $Ker (T)=\{x \in R^3: T(x)=0\}\\
=\{(x,y,z)\in R^2: T(x,y,z)=(0,0,0)\\
=\{(x,y,z)\in R^2:(x-y,y+2z,2x-y+z)=(0,0,0)$
We have the system: $x-y=0\\
y+2z=0\\
2x-y+z=0
\rightarrow x=y=0\\
\rightarrow z=-\frac{1}{2}y=0\rightarrow z=0$
Thus, $Ker (T)=\{(0,0,0)\}\\
\rightarrow \dim [Ker (T)]=0$
$Ker(T)$ is the origin $(0,0,0)$
Then, $Rng(T)=\{T(x):x \in R^3\}\\
=\{T(x,y,z):x,y,z \in R\}\\
=\{(x-y,y+2z,2x-y+z):x,y,z \in R\}
=\{ x(1,0,2)+y(-1,1,-1)+z(0,2,1): x,y,z \in R\}
=span \{(1,0,2);(-1,1,-1);(0,2,1)\}$
Find a basic for $Rng(T)$
$\begin{bmatrix}
1 & -1 & 0\\
0 & 1 & 2 \\
2 & -1 & 1
\end{bmatrix}\approx\begin{bmatrix}
1 & 0 & 2\\
-1 & 1 & -1 \\
0 & 2 & 1
\end{bmatrix}\approx \begin{bmatrix}
1 & 0 & 2\\
0 & 1 & 1 \\
0 & 0 & -1
\end{bmatrix}\approx \begin{bmatrix}
1 & 0& 0\\
0 & 1 & 0 \\
0 & 0 & -1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}$
Hence, we have $\{(1,0,0);(0,1,0);(0,0,1)\}$ is a basic for $Rng(T)$
$\rightarrow Rng(T)=R^3\\
\rightarrow \dim [Rng(T)]=3$
Thus, $Rng(T)$ is a whole 3 dimensional space.
Verify a Rank Nullity Theorem:
$\dim [Ker (T)]+\dim[Rng(T)]=0+3=3=\dim R^3$
