Answer
See below
Work Step by Step
Since $A$ is $2 \times 3$ matrix, then $R^3 \rightarrow R^2$
Obtain: $Ker(T)=\{x \in R^3: T(x)=0\}\\
=\{ x \in R^3: Ax=0\}\\
=\{x \in R: \begin{bmatrix}
1 & 2 & -1\\
2 & 5 & 1
\end{bmatrix}\begin {bmatrix} x \\ y \\ z\end{bmatrix}=0\}\\
=\{x \in R: \begin{bmatrix}
x+2y-z\\
2x+5y+z
\end{bmatrix}=\begin{bmatrix}
0 \\ 0
\end{bmatrix}\}\\
=\{(7z,-3z,z):z \in R\}\\
=\{z(7,-3,1): z \in R\}\\
=span \{(7,-3,1)\}$
Hence, $T$ is not one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim R\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)]=2$
$\dim R^2=2=\dim [Rng(T)]$
$\rightarrow T$ is onto.
Since $T$ is not one-to-one, $T^{-1}$ does not exist.
