Consider the linear transformation \begin{equation*} \ltdefn{R}{P_3}{M_{22}},\quad \lt{R}{a+bx+cx^2+x^3}= \begin{bmatrix} a +b - c + 2d & 2a + 3b - 2c + 3d\\ a + b + 2 d & -a + b + 2c - 5d \end{bmatrix} \end{equation*}
If we wish to quickly find a formula for the inverse of $R$ (presuming it exists), then choosing “nice” bases will work best. So build a matrix representation of $R$ relative to the bases $B$ and $C$, \begin{align*} B&=\set{1,\,x,\,x^2,\,x^3}\\ C&=\set{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},\, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} } \end{align*}
Then, \begin{align*} \vectrep{C}{\lt{R}{1}} &=\vectrep{C}{\begin{bmatrix} 1 & 2 \\ 1 & -1 \end{bmatrix}} =\colvector{1\\2\\1\\-1}\\ \vectrep{C}{\lt{R}{x}} &=\vectrep{C}{\begin{bmatrix} 1 & 3 \\ 1 & 1 \end{bmatrix}} =\colvector{1\\3\\1\\1}\\ \vectrep{C}{\lt{R}{x^2}} &=\vectrep{C}{\begin{bmatrix} -1 & -2 \\ 0 & 2 \end{bmatrix}} =\colvector{-1\\-2\\0\\2}\\ \vectrep{C}{\lt{R}{x^3}} &=\vectrep{C}{\begin{bmatrix} 2 & 3 \\ 2 & -5 \end{bmatrix}} =\colvector{2\\3\\2\\-5} \end{align*}
So a representation of $R$ is \begin{equation*} \matrixrep{R}{B}{C}= \begin{bmatrix} 1 & 1 & -1 & 2 \\ 2 & 3 & -2 & 3 \\ 1 & 1 & 0 & 2 \\ -1 & 1 & 2 & -5 \end{bmatrix} \end{equation*}
The matrix $\matrixrep{R}{B}{C}$ is invertible (as you can check) so we know for sure that $R$ is invertible by Theorem IMR. Furthermore, \begin{equation*} \matrixrep{\ltinverse{R}}{C}{B} =\inverse{\left(\matrixrep{R}{B}{C}\right)} = \inverse{ \begin{bmatrix} 1 & 1 & -1 & 2 \\ 2 & 3 & -2 & 3 \\ 1 & 1 & 0 & 2 \\ -1 & 1 & 2 & -5 \end{bmatrix} } = \begin{bmatrix} 20 & -7 & -2 & 3 \\ -8 & 3 & 1 & -1 \\ -1 & 0 & 1 & 0 \\ -6 & 2 & 1 & -1 \end{bmatrix} \end{equation*}
We can use this representation of the inverse linear transformation, in concert with Theorem FTMR, to determine an explicit formula for the inverse itself, \begin{align*} \lt{\ltinverse{R}}{\begin{bmatrix}a&b\\c&d\end{bmatrix}} &=\vectrepinv{B}{\matrixrep{\ltinverse{R}}{C}{B}\vectrep{C}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}}&&\knowl{./knowls/theorem.FTMR.knowl}{\text{Theorem FTMR}}\\ &=\vectrepinv{B}{\inverse{\left(\matrixrep{R}{B}{C}\right)}\vectrep{C}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}}&&\knowl{./knowls/theorem.IMR.knowl}{\text{Theorem IMR}}\\ &=\vectrepinv{B}{\inverse{\left(\matrixrep{R}{B}{C}\right)}\colvector{a\\b\\c\\d}}&&\knowl{./knowls/definition.VR.knowl}{\text{Definition VR}}\\ &=\vectrepinv{B}{ \begin{bmatrix} 20 & -7 & -2 & 3 \\ -8 & 3 & 1 & -1 \\ -1 & 0 & 1 & 0 \\ -6 & 2 & 1 & -1 \end{bmatrix} \colvector{a\\b\\c\\d}}&&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}}\\ &=\vectrepinv{B}{ \colvector{ 20a - 7b - 2c + 3d\\ -8a + 3b + c -d\\ -a + c\\ -6a + 2b + c - d}}&&\knowl{./knowls/definition.MVP.knowl}{\text{Definition MVP}}\\ &=(20a - 7b - 2c + 3d)+(-8a + 3b + c -d)x\\ &\quad\quad +(-a + c)x^2+(-6a + 2b + c - d)x^3 &&\knowl{./knowls/definition.VR.knowl}{\text{Definition VR}} \end{align*}