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