(Another approach to this solution would follow Example CIVLT.)
We are given that $R$ is invertible. The inverse linear transformation can be formulated by considering the pre-image of a generic element of the codomain. With injectivity and surjectivity, we know that the pre-image of any element will be a set of size one — it is this lone element that will be the output of the inverse linear transformation.
Suppose that we set $\vect{v}=\begin{bmatrix}x\\y\end{bmatrix}$ as a generic element of the codomain, $M_{21}$. Then if $\begin{bmatrix}r & s\end{bmatrix}=\vect{w}\in\preimage{R}{\vect{v}}$,
\begin{align*}
\begin{bmatrix}x\\y\end{bmatrix}&=\vect{v}=\lt{R}{\vect{w}}\\
&=
\begin{bmatrix}
r+3s\\
4r+11s
\end{bmatrix}
\end{align*}
So we obtain the system of two equations in the two variables $r$ and $s$,
\begin{align*}
r+3s&=x\\
4r+11s&=y
\end{align*}
With a nonsingular coefficient matrix, we can solve the system using the inverse of the coefficient matrix,
\begin{align*}
r&=-11x+3y\\
s&=4x-y
\end{align*}
So we define,
\begin{equation*}
\lt{\ltinverse{R}}{\vect{v}}
=\lt{\ltinverse{R}}{\begin{bmatrix}x\\y\end{bmatrix}}
=\vect{w}
=\begin{bmatrix}r & s\end{bmatrix}
=\begin{bmatrix}-11x+3y & 4x-y\end{bmatrix}
\end{equation*}