For a matrix $A = \begin{bmatrix} 1 & 2r-1 \\ 0 & 1 \end{bmatrix}$, the value of $\prod_{r=1}^{50} \begin{bmatrix} 1 & 2r-1 \\ 0 & 1 \end{bmatrix}$ is equal to -
$\begin{bmatrix} 1 & 100 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 1 & 4950 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 1 & 5050 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 1 & 2500 \\ 0 & 1 \end{bmatrix}$
Step-by-Step Solution
Key Concept: The product of matrices of the form [[1, a], [0, 1]] and [[1, b], [0, 1]] is [[1, a+b], [0, 1]]. Thus, the product of 50 such matrices is [[1, sum(2r-1 for r=1 to 50)], [0, 1]]. The sum of the first n odd numbers is n^2. For n=50, 50^2 = 2500.
Let $A_r = \begin{bmatrix} 1 & 2r-1 \\ 0 & 1 \end{bmatrix}$. Then $A_1 A_2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1^2+2^2 \\ 0 & 1 \end{bmatrix}$ is incorrect. Actually, $\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a+b \\ 0 & 1 \end{bmatrix}$. Therefore, $\prod_{r=1}^{50} A_r = \begin{bmatrix} 1 & \sum_{r=1}^{50} (2r-1) \\ 0 & 1 \end{bmatrix}$. The sum $\sum_{r=1}^{50} (2r-1) = 2 \frac{50(51)}{2} - 50 = 50(51) - 50 = 50(50) = 2500$. Thus, the result is $\begin{bmatrix} 1 & 2500 \\ 0 & 1 \end{bmatrix}$.
Correct Answer: 4