Consider the linear transformation $\ltdefn{T}{S_{22}}{P_2}$ defined by \begin{align*} \lt{T}{\begin{bmatrix}a&b\\b&c\end{bmatrix}} &= \left(a+b+c\right) + \left(-a+2c\right)x + \left(2a+3b+6c\right)x^2 \end{align*}
$T$ is invertible, which you are able to verify, perhaps by determining that the kernel of $T$ is trivial and the range of $T$ is all of $P_2$. This will be easier once we have Theorem RPNDD, which appears later in this section.
By Theorem ILTIS we know $\ltinverse{T}$ exists, and it will be critical shortly to realize that $\ltinverse{T}$ is automatically known to be a linear transformation as well (Theorem ILTLT). To determine the complete behavior of $\ltdefn{\ltinverse{T}}{P_2}{S_{22}}$ we can simply determine its action on a basis for the domain, $P_2$. This is the substance of Theorem LTDB, and an excellent example of its application. Choose any basis of $P_2$, the simpler the better, such as $B=\set{1,\,x,\,x^2}$. Values of $\ltinverse{T}$ for these three basis elements will be the single elements of their preimages. In turn, we have \begin{align*} \preimage{T}{1}:&\\ &&\lt{T}{\begin{bmatrix}a&b\\b&c\end{bmatrix}} &=1+0x+0x^2\\ &&\begin{bmatrix} 1 & 1 & 1 & 1\\ -1 & 0 & 2 & 0\\ 2 & 3 & 6 & 0 \end{bmatrix} &\rref \begin{bmatrix} 1 & 0 & 0& -6 \\ 0 & 1 & 0& 10 \\ 0 & 0 & 1& -3 \end{bmatrix} \\ &\text{(preimage)}&\preimage{T}{1} &=\set{\begin{bmatrix}-6&10\\10&-3\end{bmatrix}}\\ &\text{(function)}&\lt{\ltinverse{T}}{1} &= \begin{bmatrix}-6&10\\10&-3\end{bmatrix}\\ \preimage{T}{x}:&\\ &&\lt{T}{\begin{bmatrix}a&b\\b&c\end{bmatrix}} &=0+1x+0x^2\\ &&\begin{bmatrix} 1 & 1 & 1 & 0\\ -1 & 0 & 2 & 1\\ 2 & 3 & 6 & 0 \end{bmatrix} &\rref \begin{bmatrix} 1 & 0 & 0& -3 \\ 0 & 1 & 0& 4 \\ 0 & 0 & 1& -1 \end{bmatrix} \\ &\text{(preimage)}&\preimage{T}{x} &=\set{\begin{bmatrix}-3&4\\4&-1\end{bmatrix}}\\ &\text{(function)}&\lt{\ltinverse{T}}{x} &= \begin{bmatrix}-3&4\\4&-1\end{bmatrix}\\ \preimage{T}{x^2}:&\\ &&\lt{T}{\begin{bmatrix}a&b\\b&c\end{bmatrix}} &=0+0x+1x^2\\ &&\begin{bmatrix} 1 & 1 & 1 & 0\\ -1 & 0 & 2 & 0\\ 2 & 3 & 6 & 1 \end{bmatrix} &\rref \begin{bmatrix} 1 & 0 & 0& 2 \\ 0 & 1 & 0& -3 \\ 0 & 0 & 1& 1 \end{bmatrix} \\ &\text{(preimage)}&\preimage{T}{x^2} &=\set{\begin{bmatrix}2&-3\\-3&1\end{bmatrix}}\\ &\text{(function)}&\lt{\ltinverse{T}}{x^2} &= \begin{bmatrix}2&-3\\-3&1\end{bmatrix} \end{align*}
Theorem LTDB says, informally, “it is enough to know what a linear transformation does to a basis.” Formally, we have the outputs of $\ltinverse{T}$ for a basis, so by Theorem LTDB there is a unique linear transformation with these outputs. So we put this information to work. The key step here is that we can convert any element of $P_2$ into a linear combination of the elements of the basis $B$ (Theorem VRRB). We are after a “formula” for the value of $\ltinverse{T}$ on a generic element of $P_2$, say $p+qx+rx^2$. \begin{align*} \lt{\ltinverse{T}}{p+qx+rx^2} &= \lt{\ltinverse{T}}{p(1)+q(x)+r(x^2)} &&\knowl{./knowls/theorem.VRRB.knowl}{\text{Theorem VRRB}}\\ &= p\lt{\ltinverse{T}}{1}+ q\lt{\ltinverse{T}}{x}+ r\lt{\ltinverse{T}}{x^2} &&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}}\\ &= p\begin{bmatrix}-6&10\\10&-3\end{bmatrix}+ q\begin{bmatrix}-3&4\\4&-1\end{bmatrix}+ r\begin{bmatrix}2&-3\\-3&1\end{bmatrix}\\ &= \begin{bmatrix} -6p-3q+2r & 10p+4q-3r \\ 10p+4q-3r & -3p -q + r \end{bmatrix} \end{align*}
Notice how a linear combination in the domain of $\ltinverse{T}$ has been translated into a linear combination in the codomain of $\ltinverse{T}$ since we know $\ltinverse{T}$ is a linear transformation by Theorem ILTLT.
Also, notice how the augmented matrices used to determine the three pre-images could be combined into one calculation of a matrix in extended echelon form, reminiscent of a procedure we know for computing the inverse of a matrix (see Example CMI). Hmmmm.