Answer
See below
Work Step by Step
Assume $v \in V$
We know that $\{v_1,v_2\}$ is a basic for the vector space $V$ there exist scalars $\alpha, \beta$ such as:
$v=\alpha v_1+\beta v_2$
Obtain: $T(v)=T(\alpha v_1+\beta v_2)\\
=\alpha T(v_1)+\beta T(v_2)\\
=\alpha (v_1+2v_2)+\beta(2v_1-3v_3)\\
=(\alpha + 2\beta)v_1+(2\alpha -3\beta)v_2$
Hence, $Ker(T)=\{T(v)=0\}\\
=\{\alpha v_1+\beta v_2 \in V:( \alpha +2 \beta)v_1+(2\alpha -3\beta)v_2=0\}$
We have the system: $\alpha + 2\beta=0\\
2\alpha - 3\beta =0\\
\rightarrow \alpha =\beta=0$
Hence, $Ker(T)=\{0v_1+0v_2\}\\
=\{0\}$
$T$ is one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)=\dim V\\
0+\dim [Rng(T)]=2\\
\dim [Rng(T)]=2$
Since, $\dim V=2=\dim [Rng(T)]\\
\rightarrow Rng(T) =V$
$\rightarrow T$ is onto.
Since $T$ is both onto and one-to-one. $T^{-1}$ does exist.
With $v \in V \rightarrow T^{-1}T(v)=v\\
T^{-1}T(\alpha v_1+\beta v_2)=\alpha v_1+\beta v_2\\
T^{-1}[(\alpha +2\beta)v_1+(2\alpha -3\beta)v_2]=\alpha v_1+\beta v_2$
then we have a system: $a=\alpha +2\beta\\
b=2\alpha -3\beta\\
\rightarrow \beta=\frac{2a-b}{7}\\
\alpha=a-2\beta=\frac{3a+2b}{7}$
Thus, $T^{-1}(av_1+bv_2)=\frac{3a+2b}{7}v_1+\frac{2a-b}{7}v_2$
