Question:

First, calculate $B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix}$. Then $B^2 = \begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ -2 & -1 \end{bmatrix}$, $B^3 = \begin{bmatrix} 4 & 3 \\ -3 & -2 \end{bmatrix}$. In general, $B^n = \begin{bmatrix} n+1 & n \\ -n & -(n-1) \end{bmatrix}$. The sum $\sum_{n=1}^{50} B^n = \begin{bmatrix} \sum (n+1) & \sum n \\ -\sum n & -\sum (n-1) \end{bmatrix} = \begin{bmatrix} \frac{50 \times 53}{2} & \frac{50 \times 51}{2} \\ -\frac{50 \times 51}{2} & -\frac{50 \times 49}{2} \end{bmatrix} = \begin{bmatrix} 1325 & 1275 \\ -1275 & -1225 \end{bmatrix}$. The sum of all elements is $1325 + 1275 - 1275 - 1225 = 100$. Wait, re-evaluating: $1325 - 1225 = 100$. Let's recheck the sum. The sum of elements of $B^n$ is $(n+1) + n - n - (n-1) = 2$. So $\sum_{n=1}^{50} 2 = 100$. Wait, the answer key says 3 (75). Let's re-calculate $B$. $B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ -1 & -2 \end{bmatrix}$. Then $B \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix}$. Correct. Sum of elements of $B^n$ is $n+1+n-n-n+1 = 2$. Sum is $50 \times 2 = 100$. Perhaps there is a calculation error in my manual check or the provided key.