Answer
See below
Work Step by Step
Since $A$ is $2 \times 2$ matrix, then $R^2 \rightarrow R^2$
Obtain: $Ker(T)=\{x \in R: T(x)=0\}\\
=\{ x \in R: Ax=0\}\\
=\{x \in R: \begin{bmatrix}
4 & 2\\
1 & 3
\end{bmatrix}\begin {bmatrix} x \\ y\end{bmatrix}=0\}\\
=\{x \in R: \begin{bmatrix}
4x -2y\\
x- 3y
\end{bmatrix}=\begin{bmatrix}
0 \\ 0
\end{bmatrix}\}\\
=\{(0,0)\}$
Hence, $T$ is onto.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim R\\
0+\dim [Rng(T)]=2\\
\dim [Rng(T)]=2$
Since, $\dim R^2=2=\dim [Rng(T)]$
$\rightarrow T$ is onto.
$\begin{vmatrix}
4 & 2 \\
1 & 3
\end{vmatrix}=4\times 3 - 1 \times 2=10 \ne 0$
Hence, $T^{-1}$ is exist.
We have: $T^{-1}x=A^{-1}x\\
\rightarrow A^{-1}=\frac{1}{\det A}\begin{bmatrix}
3 & -1\\
-2 & 4
\end{bmatrix}=\frac{1}{10}\begin{bmatrix}
3 & -1\\
-2 & 4
\end{bmatrix}=\begin{bmatrix}
\frac{3}{10} & -\frac{1}{10}\\
\frac{-1}{5} & \frac{2}{5}
\end{bmatrix}\\
\rightarrow \begin{bmatrix}
\frac{3}{10} & -\frac{1}{10}\\
\frac{-1}{5} & \frac{2}{5}
\end{bmatrix}\begin{bmatrix}
x\\y
\end{bmatrix}=(\frac{3}{10}x-\frac{1}{5}y,-\frac{1}{10}x+\frac{2}{5}y)$
