Consider the two linear transformations, \begin{align*} \ltdefn{T}{\complex{2}}{P_2}&\quad \lt{T}{\colvector{a\\b}}=(-a + 3b)+(2a + 4b)x+(a - 2b)x^2\\ \ltdefn{S}{P_2}{M_{22}}&\quad\ \lt{S}{a+bx+cx^2}= \begin{bmatrix} 2a + b + 2c & a + 4b - c\\ -a + 3c & 3a + b + 2 c\end{bmatrix} \end{align*} and bases for $\complex{2}$, $P_2$ and $M_{22}$ (respectively), \begin{align*} B&=\set{\colvector{3\\1},\,\colvector{2\\1}}\\ C&=\set{1-2x+x^2,\,-1+3x,\,2x+3x^2}\\ D&=\set{ \begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix},\, \begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix},\, \begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix},\, \begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix} } \end{align*}
Begin by computing the new linear transformation that is the composition of $T$ and $S$ (Definition LTC, Theorem CLTLT), $\ltdefn{\left(\compose{S}{T}\right)}{\complex{2}}{M_{22}}$, \begin{align*} &\lt{\left(\compose{S}{T}\right)}{\colvector{a\\b}}=\lt{S}{\lt{T}{\colvector{a\\b}}}\\ &\quad\quad=\lt{S}{(-a + 3b)+(2a + 4b)x+(a - 2b)x^2}\\ &\quad\quad= \begin{bmatrix} 2(-a + 3b) + (2a + 4b) + 2(a - 2b) & (-a + 3b) + 4(2a + 4b) - (a - 2b)\\ -(-a + 3b) + 3(a - 2b) & 3(-a + 3b) + (2a + 4b) + 2(a - 2b) \end{bmatrix}\\ &\quad\quad= \begin{bmatrix} 2a + 6b & 6a + 21b\\ 4a - 9b &a + 9b \end{bmatrix} \end{align*}
Now compute the matrix representations (Definition MR) for each of these three linear transformations ($T$, $S$, $\compose{S}{T}$), relative to the appropriate bases. First for $T$, \begin{align*} \vectrep{C}{\lt{T}{\colvector{3\\1}}}&=\vectrep{C}{10x+x^2}\\ &=\vectrep{C}{28(1-2x+x^2)+28(-1+3x)+(-9)(2x+3x^2)}\\ &=\colvector{28\\28\\-9}\\ \vectrep{C}{\lt{T}{\colvector{2\\1}}}&=\vectrep{C}{1+8x}\\ &=\vectrep{C}{33(1-2x+x^2)+32(-1+3x)+(-11)(2x+3x^2)}\\ &=\colvector{33\\32\\-11}\\ \end{align*}
So we have the matrix representation of $T$, \begin{equation*} \matrixrep{T}{B}{C}= \begin{bmatrix} 28 & 33 \\ 28 & 32 \\ -9 & -11 \end{bmatrix} \end{equation*}
Now, a representation of $S$, \begin{align*} &\vectrep{D}{\lt{S}{1-2x+x^2}}=\vectrep{D}{\begin{bmatrix}2 & -8 \\ 2 & 3 \end{bmatrix}}\\ &\quad\quad=\vectrep{D}{ (-11)\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+ (-21)\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+ 0\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+ (17)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix} }\\ &\quad\quad=\colvector{-11\\-21\\0\\17}\\ &\vectrep{D}{\lt{S}{-1+3x}}=\vectrep{D}{\begin{bmatrix}1 & 11 \\ 1 & 0\end{bmatrix}}\\ &\quad\quad=\vectrep{D}{ 26\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+ 51\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+ 0\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+ (-38)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix} }\\ &\quad\quad=\colvector{26\\51\\0\\-38}\\ &\vectrep{D}{\lt{S}{2x+3x^2}}=\vectrep{D}{\begin{bmatrix}8 & 5 \\ 9 & 8\end{bmatrix}}\\ &\quad\quad=\vectrep{D}{ 34\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+ 67\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+ 1\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+ (-46)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix} }\\ &\quad\quad=\colvector{34\\67\\1\\-46}\\ \end{align*}
So we have the matrix representation of $S$, \begin{equation*} \matrixrep{S}{C}{D}= \begin{bmatrix} -11 & 26 & 34 \\ -21 & 51 & 67 \\ 0 & 0 & 1 \\ 17 & -38 & -46 \end{bmatrix} \end{equation*}
Finally, a representation of $\compose{S}{T}$, \begin{align*} &\vectrep{D}{\lt{\left(\compose{S}{T}\right)}{\colvector{3\\1}}}=\vectrep{D}{\begin{bmatrix}12 & 39\\3 &12\end{bmatrix}}\\ &\quad\quad=\vectrep{D}{ 114\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+ 237\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+ (-9)\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+ (-174)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix} }\\ &\quad\quad=\colvector{114\\237\\-9\\-174}\\ &\vectrep{D}{\lt{\left(\compose{S}{T}\right)}{\colvector{2\\1}}}=\vectrep{D}{\begin{bmatrix}10 & 33\\-1 & 11\end{bmatrix}}\\ &\quad\quad=\vectrep{D}{ 95\begin{bmatrix} 1 & -2\\ 1 & -1 \end{bmatrix}+ 202\begin{bmatrix} 1 & -1\\ 1 & 2\end{bmatrix}+ (-11)\begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix}+ (-149)\begin{bmatrix} 2 & -3 \\ 2 & 2\end{bmatrix} }\\ &\quad\quad=\colvector{95\\202\\-11\\-149}\\ \end{align*}
So we have the matrix representation of $\compose{S}{T}$, \begin{equation*} \matrixrep{\compose{S}{T}}{B}{D}= \begin{bmatrix} 114 & 95 \\ 237 & 202 \\ -9 & -11 \\ -174 & -149 \end{bmatrix} \end{equation*}
Now, we are all set to verify the conclusion of Theorem MRCLT, \begin{align*} \matrixrep{S}{C}{D}\matrixrep{T}{B}{C} &= \begin{bmatrix} -11 & 26 & 34 \\ -21 & 51 & 67 \\ 0 & 0 & 1 \\ 17 & -38 & -46 \end{bmatrix} \begin{bmatrix} 28 & 33 \\ 28 & 32 \\ -9 & -11 \end{bmatrix}\\ &= \begin{bmatrix} 114 & 95 \\ 237 & 202 \\ -9 & -11 \\ -174 & -149 \end{bmatrix}\\ &=\matrixrep{\compose{S}{T}}{B}{D} \end{align*}
We have intentionally used nonstandard bases. If you were to choose “nice” bases for the three vector spaces, then the result of the theorem might be rather transparent. But this would still be a worthwhile exercise — give it a go.