Answer
$T$ is invertible and
$T^{-1}(x,y)=\left[ {\begin{array}{*{20}{c}} { - \frac{x}{2}\\ \frac{y}{2}} \end{array}} \right]$
Work Step by Step
Take the standard basis for $R^2,$ which is $\{(1,0),(0,1)\}=\{e_1,e_2\}$
We are given
$T(x,y)=(-2x,2y)$
Then, we have
$T(1,0)=(-2,0)\\ T(0,1)=(0,2)$
Therefore, the matrix corresponding to the linear transformation $T$ is given by:
$A=[T(e_1)$ $T(e_2)]$ $=\left[ {\begin{array}{*{20}{c}} { - 2}&0\\ 0&{ 2} \end{array}} \right]$
Since $A$ is invertible, then $T$ is invertible.
Therefore, the matrix corresponding to the linear transformation $T^{-1}$ is given by:
$A^{-1}=\left[ {\begin{array}{*{20}{c}} { - \frac{1}{2}}&0\\ 0&{ \frac{1}{2}} \end{array}} \right]$
Hence, the linear transformation $T^{-1}$ is defined as:
$T^{-1}(x,y)=A^{-1}(x,y)$
$=\left[ {\begin{array}{*{20}{c}} { - \frac{1}{2}}&0\\ 0&{ \frac{1}{2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { x\\ y} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} { - \frac{x}{2}\\ \frac{y}{2}} \end{array}} \right]$
