Answer
$T^{-1}(x_{1},x_{2})=(7x_{1}+9x_{2},4x_{1}+5x_{2})$
Work Step by Step
The standard matrix of $T$ is $A=\left[\begin{array}{ll}
-5 & 9\\
4 & -7
\end{array}\right]$.
$\det A=ad-bc=(-5)(-7)-9\cdot 4=-1\neq 0,\quad$ so A is invertible.
By Theorem 9, $T$ is also invertible and the standard matrix for $\mathrm{T}^{-1}$is $\mathrm{A}^{-1}.$
$\displaystyle \mathrm{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{ll}
d & -b\\
-c & a
\end{array}\right]=(-1)\displaystyle \left[\begin{array}{ll}
-7 & -9\\
-4 & -5
\end{array}\right]=\left[\begin{array}{ll}
7 & 9\\
4 & 5
\end{array}\right]$
$T^{-1}(x_{1},x_{2})=\left[\begin{array}{ll}
7 & 9\\
4 & 5
\end{array}\right]\left[\begin{array}{l}
x_{1}\\
x_{2}
\end{array}\right]$
$T^{-1}(x_{1},x_{2})=(7x_{1}+9x_{2},4x_{1}+5x_{2})$
