Let $$A = \begin{pmatrix} 2 & -1 \\ 0 & 2 \end{pmatrix}$$. If $$B = I - {}^5C_1(\text{adj } A) + {}^5C_2(\text{adj } A)^2 - \ldots - {}^5C_5(\text{adj } A)^5$$, then the sum of all elements of the matrix $$B$$ is:

We are given $$A = \begin{pmatrix} 2 & -1 \\ 0 & 2 \end{pmatrix}$$ and $$B = I - {}^5C_1(\text{adj } A) + {}^5C_2(\text{adj } A)^2 - \ldots - {}^5C_5(\text{adj } A)^5$$. For a 2x2 matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$, the adjugate is $$\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$, so $$\text{adj } A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$$.

Recognizing the binomial expansion, we write $$B = \sum_{k=0}^{5} {}^5C_k (-\text{adj } A)^k = (I - \text{adj } A)^5$$. Then $$I - \text{adj } A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}$$.

Let $$M = \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix} = -\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$. Since $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$$, we have $$M^5 = (-1)^5 \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & -5 \\ 0 & -1 \end{pmatrix}$$, hence $$B = \begin{pmatrix} -1 & -5 \\ 0 & -1 \end{pmatrix}$$. Summing its entries gives $$-1 + (-5) + 0 + (-1) = -7$$.

Therefore, the answer is Option C: $$\textbf{-7}$$.