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