If the product of n matrices $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \dots \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$ is equal to the matrix $\begin{bmatrix} 1 & 378 \\ 0 & 1 \end{bmatrix}$ then the value of n is equal to -
Step-by-Step Solution
Key Concept: The product of matrices of the form [1 k; 0 1] is [1 sum(k); 0 1]. The sum of the first n integers is n(n+1)/2. Set n(n+1)/2 = 378 and solve for n.
Let $A_k = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$. Then $A_i A_j = \begin{bmatrix} 1 & i \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & j \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & i+j \\ 0 & 1 \end{bmatrix}$. By induction, the product of $n$ matrices is $\begin{bmatrix} 1 & \sum_{k=1}^n k \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & \frac{n(n+1)}{2} \\ 0 & 1 \end{bmatrix}$. Given $\frac{n(n+1)}{2} = 378$, we have $n^2 + n - 756 = 0$. Solving this quadratic equation, $(n+28)(n-27) = 0$. Since $n > 0$, $n = 27$.
Correct Answer: 2