Answer
See below
Work Step by Step
Since $A$ is $4 \times 3$ matrix, then $R^3 \rightarrow R^4$
Obtain: $Ker(T)=\{x \in R^3: T(x)=0\}\\
=\{ x \in R^3: Ax=0\}\\
=\{x \in R: \begin{bmatrix}
0 & 1 & 2 \\
3 & 4 & 5\\
5 & 4 & 3\\
2 & 1 & 0
\end{bmatrix}\begin {bmatrix} x \\ y \\ z\end{bmatrix}=\begin {bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}\}\\
=\{x \in R: \begin{bmatrix}
y+2z\\
3x+4y+5z\\
5x+4y+3z\\
2x+y
\end{bmatrix}=\begin{bmatrix}
0 \\ 0 \\ 0 \\ 0
\end{bmatrix}\}\\
=\{(x,-2x,x): x\in R\}\\
=\{x(1,-2,1): x \in R\}\\
=span \{(1,-2,1)\}$
Hence, $T$ is not one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim R\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)]=2$
$\dim R^2=4 \ne 2=\dim [Rng(T)]$
$\rightarrow T$ is not onto.
Since $T$ is neither onto nor one-to-one, $T^{-1}$ does not exist.
