Matrices & Determinants
Geometric Progression
Grade Class 12
Question:
Let a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..., a<sub>10</sub> be in G.P. with a<sub>i</sub> > 0 for i = 1,2,..., 10 and S be the set of pairs (r, k), r, k ∈ N (the set of natural numbers) for which <br><img src="https://latex.codecogs.com/png.image?\dpi{150}\begin{vmatrix} \log_e a_1^r a_2^k & \log_e a_2^r a_3^k & \log_e a_3^r a_4^k \\ \log_e a_4^r a_5^k & \log_e a_5^r a_6^k & \log_e a_6^r a_7^k \\ \log_e a_7^r a_8^k & \log_e a_8^r a_9^k & \log_e a_9^r a_{10}^k \end{vmatrix} = 0">. Then the number of elements in S, is :
(1) Infinitely many
(2) 4
(3) 10
(4) 2
Step-by-Step Solution
Key Concept: The terms of a G.P. are a_n = a*R^(n-1). The log terms become linear combinations of log(a) and log(R). Specifically, log(a_n^r * a_{n+1}^k) = r*log(a_n) + k*log(a_{n+1}) = r*(log(a) + (n-1)log(R)) + k*(log(a) + n*log(R)) = (r+k)log(a) + (r(n-1) + kn)log(R). Since each row is a linear combination of two vectors, the determinant is zero for all r, k.
Let a_n = a*R^(n-1). Then log(a_n^r * a_{n+1}^k) = r*log(a_n) + k*log(a_{n+1}) = r(log a + (n-1)log R) + k(log a + n log R) = (r+k)log a + (r(n-1) + kn)log R. Let X = log a and Y = log R. The entries are of the form (r+k)X + (r(n-1)+kn)Y. Since each entry is a linear combination of X and Y, the rows are linearly dependent, making the determinant zero for all r, k.
Correct Answer: (1)