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Matrices & Determinants: Matrix Pmatrix Pmatrix Value Pmatrix Pmatrix Equal

JEE Maths question with a full step-by-step solution.

Question
For a matrix A=(12r−101)A = \begin{pmatrix}1 & 2r-1\\ 0 & 1\end{pmatrix}, the value of ∏r=150(12r−101)\displaystyle\prod_{r=1}^{50}\begin{pmatrix}1 & 2r-1\\ 0 & 1\end{pmatrix} is equal to:
A(110001)\begin{pmatrix}1 & 100\\ 0 & 1\end{pmatrix}
B(1495001)\begin{pmatrix}1 & 4950\\ 0 & 1\end{pmatrix}
C(1505001)\begin{pmatrix}1 & 5050\\ 0 & 1\end{pmatrix}
D(1250001)\begin{pmatrix}1 & 2500\\ 0 & 1\end{pmatrix}correct
Solution
Step 1: Establish the multiplication rule
(1p01)(1q01)=(1p+q01)\begin{pmatrix}1 & p\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & q\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & p+q\\ 0 & 1\end{pmatrix}
 By induction, the product of nn such matrices accumulates the upper-right entries:
∏r=150(12r−101)=(1∑r=150(2r−1)01)\prod_{r=1}^{50}\begin{pmatrix}1 & 2r-1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & \displaystyle\sum_{r=1}^{50}(2r-1)\\ 0 & 1\end{pmatrix}
 Step 2: Evaluate the sum
∑r=150(2r−1)=1+3+5+⋯+99=502=2500\sum_{r=1}^{50}(2r-1) = 1+3+5+\cdots+99 = 50^2 = 2500
∏r=150(12r−101)=(1250001)\prod_{r=1}^{50}\begin{pmatrix}1 & 2r-1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 2500\\ 0 & 1\end{pmatrix}
 Answer: (4)
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