Matrices & DeterminantsmediumFree

Matrices & Determinants — JEE Maths practice question

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

Question

If the product of

n

matrices

\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

\begin{pmatrix}1 & 378\\ 0 & 1\end{pmatrix}

, then

n

is equal to:

26

27

correct

377

378

Solution

Step 1: Determine the general product Using the identity

\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:

\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

n

\frac{n(n+1)}{2} = 378 \implies n(n+1) = 756 = 27\times 28 \implies n = 27

Answer: (2)

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