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Matrices & Determinants: Product Matrices Pmatrix Pmatrix Pmatrix Pmatrix Pmatrix Pma

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

Question
If the product of nn matrices (1101)(1201)(1301)⋯(1n01)\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & 2\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & 3\\ 0 & 1\end{pmatrix}\cdots\begin{pmatrix}1 & n\\ 0 & 1\end{pmatrix} equals the matrix (137801)\begin{pmatrix}1 & 378\\ 0 & 1\end{pmatrix}, then nn is equal to:
A2626
B2727correct
C377377
D378378
Solution
Step 1: Determine the general product Using the identity (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}, the product accumulates as:
∏k=1n(1k01)=(1n(n+1)201)\prod_{k=1}^{n}\begin{pmatrix}1 & k\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & \frac{n(n+1)}{2}\\ 0 & 1\end{pmatrix}
 Step 2: Solve for nn
n(n+1)2=378  ⟹  n(n+1)=756=27×28  ⟹  n=27\frac{n(n+1)}{2} = 378 \implies n(n+1) = 756 = 27\times 28 \implies n = 27
 Answer: (2)
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