Answer
See below
Work Step by Step
Since $A$ is $3 \times 1$ matrix, then $R \rightarrow R^3$
Obtain: $Ker(T)=\{x \in R: T(x)=0\}\\
=\{ x \in R: Ax=0\}\\
=\{x \in R: \begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}[x]=0\}\\
=\{x \in R: \begin{bmatrix}
x\\
2x \\
3x
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\ 0
\end{bmatrix}\}\\
=\{0\}$
Hence, $T$ is one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim R\\
0+\dim [Rng(T)]=1\\
\dim [Rng(T)]=1$
Since, $\dim R^3=3 \ne 1=\dim [Rng(T)]$
$\rightarrow T$ is not onto.
Hence, $T^{-1}$ is not exist.
