The set $W=\set{p(x)\in P_3\mid p(1)=0, p(3)=0}\subseteq P_3$ is a subspace of the vector space of polynomials $P_3$. This subspace has $C=\set{3-4x+x^2,\,12-13x+x^3}$ as a basis (check this!). Suppose we consider the linear transformation $\ltdefn{S}{P_3}{M_{22}}$ with values \begin{align*} \lt{S}{3-4x+x^2}=\begin{bmatrix}1&-3\\2&0\end{bmatrix}&& \lt{S}{12-13x+x^3}=\begin{bmatrix}0&1\\1&0\end{bmatrix} \end{align*}
By Theorem LTDB we know there is a unique linear transformation with these two values. To illustrate a sample computation of $S$, consider $q(x)=9-6x-5x^2+2x^3$. Verify that $q(x)$ is an element of $W$ (does it have roots at $x=1$ and $x=3$?), then find the scalars needed to write it as a linear combination of the basis vectors in $C$. Because \begin{equation*} q(x)=9-6x-5x^2+2x^3=(-5)(3-4x+x^2)+(2)(12-13x+x^3) \end{equation*}
The proof of Theorem LTDB gives us \begin{equation*} \lt{S}{q}=(-5)\begin{bmatrix}1&-3\\2&0\end{bmatrix} + (2)\begin{bmatrix}0&1\\1&0\end{bmatrix} = \begin{bmatrix}-5&17\\-8&0\end{bmatrix} \end{equation*}
And all the other outputs of $S$ could be computed in the same manner. Every output of $S$ will have a zero in the second row, second column. Can you see why this is so?