Let $$A = \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix}$$. If $$B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} A \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}$$, then the sum of all the elements of the matrix $$\sum_{n=1}^{50} B^n$$ is equal to

We need to find the sum of all elements of the matrix $$\sum_{n=1}^{50} B^n$$.

Let $$P = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}$$ and $$Q = \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}$$. Then $$B = PAQ$$.

We verify that $$PQ = QP = I$$ (so $$Q = P^{-1}$$):

$$PQ = \begin{bmatrix} 1(-1)+2(1) & 1(-2)+2(1) \\ -1(-1)+(-1)(1) & -1(-2)+(-1)(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$

Therefore $$B^n = PA^nQ$$.

$$A = \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix}$$ is an upper triangular matrix with ones on the diagonal, so:

$$A^n = \begin{bmatrix} 1 & \frac{n}{51} \\ 0 & 1 \end{bmatrix}$$

$$\sum_{n=1}^{50} A^n = \begin{bmatrix} 50 & \frac{1}{51}\sum_{n=1}^{50} n \\ 0 & 50 \end{bmatrix} = \begin{bmatrix} 50 & \frac{1275}{51} \\ 0 & 50 \end{bmatrix} = \begin{bmatrix} 50 & 25 \\ 0 & 50 \end{bmatrix}$$

First: $$P \cdot \begin{bmatrix} 50 & 25 \\ 0 & 50 \end{bmatrix} = \begin{bmatrix} 50 & 125 \\ -50 & -75 \end{bmatrix}$$

Then: $$\begin{bmatrix} 50 & 125 \\ -50 & -75 \end{bmatrix} \cdot \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 75 & 25 \\ -25 & 25 \end{bmatrix}$$

$$75 + 25 + (-25) + 25 = 100$$

The correct answer is Option D: $$100$$.