Answer
Preimage of $(1,1)$ is $(\sqrt{2},0)$
Work Step by Step
Let $(x,y)$ be the preimage of $(1,1)$. Then
$\begin{array}{l}
T\left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right]\\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
{x\cos ({{45}^ \circ })}&{ - y\sin ({{45}^ \circ })}\\
{x\sin ({{45}^ \circ })}&{y\cos ({{45}^ \circ })}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right]\\
\Rightarrow \left[ {\begin{array}{*{20}{c}}
{x(\frac{1}{{\sqrt 2 }})}&{ - y(\frac{1}{{\sqrt 2 }})}\\
{x(\frac{1}{{\sqrt 2 }})}&{y(\frac{1}{{\sqrt 2 }})}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right]
\end{array}$
$ \Rightarrow x(\frac{1}{{\sqrt 2 }}) - y(\frac{1}{{\sqrt 2 }}) = 1 \text{ and }x(\frac{1}{{\sqrt 2 }}) + y(\frac{1}{{\sqrt 2 }}) = 1$
Adding both equations, we have
$x\frac{2}{\sqrt{2}}=2\\
\Rightarrow x=\sqrt{2}$
Put this in equation $(1)$. We get
$\sqrt{2}\frac{1}{\sqrt{2}}-y\frac{1}{\sqrt{2}}=1\\
1-y\frac{1}{\sqrt{2}}=1\\
\Rightarrow y=0$
Hence, the preimage of $(1,1)$ is $(\sqrt{2},0)$
