Answer
See below
Work Step by Step
Obtain: $Ker(T)=\{p \in P_1(R): T(p)=0\}\\
=\{ax+b: T(ax+b)=0\}\\
=\{ax+b:(2b-a)x+(b+a)=0\}$
We have the system: $2b-a=0\\
b+a=0\\
\rightarrow a=b=0\\
\rightarrow Ker(T)=\{(0\}$
Hence, $T$ is one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T_2T_1)]+\dim [Rng(T)=\dim P_1(R)\\
0+\dim [Rng(T)]=2\\
\dim [Rng(T)]=2$
$\dim P_1(R)=2=\dim [Rng(T)]\\
\rightarrow Rng=P_1(R)$
$\rightarrow T$ is onto.
Assume $T^{-1}$ be an inverse transformation of $T$
$T^{-1}T(p)=p\\
T^{-1}T(ax+b)=ax+b\\
T^{-1}[(2b-a)x+(a+b)]=ax+b$
Let $\alpha =2b-a\\
\beta =a+b\\
\rightarrow a=2b- \alpha\\
\rightarrow b=\frac{\alpha + \beta}{3}\\
\rightarrow \alpha=\frac{2\beta - \alpha}{3}$
Hence $T^{-1}(ax+b)=\frac{2\beta - \alpha}{3}x+\frac{\beta + \alpha}{3}$
or $T^{-1}(ax+b)=\frac{2b -a}{3}x+\frac{a+ b}{3}$
