Matrices & Determinants
Properties of determinants
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/svg.image?\begin{vmatrix}&space;\log_e&space;a_1^r&space;a_2^k&space;&&space;\log_e&space;a_2^r&space;a_3^k&space;&&space;\log_e&space;a_3^r&space;a_4^k&space;\\&space;\log_e&space;a_4^r&space;a_5^k&space;&&space;\log_e&space;a_5^r&space;a_6^k&space;&&space;\log_e&space;a_6^r&space;a_7^k&space;\\&space;\log_e&space;a_7^r&space;a_8^k&space;&&space;\log_e&space;a_8^r&space;a_9^k&space;&&space;\log_e&space;a_9^r&space;a_{10}^k&space;\end{vmatrix}&space;=&space;0">. Then the number of elements in S, is :
(1) Infinitely many
(2) 4
(3) 10
(4) 2
Step-by-Step Solution
Key Concept: Since a_i are in G.P., let a_n = a*x^(n-1). The log terms become log(a*x^(n-1)^r * a*x^n^k) = log(a^(r+k) * x^((n-1)r + nrk)). This simplifies to a linear combination of log(a) and log(x). The rows of the determinant will be in arithmetic progression, making the determinant zero for all r, k.
Let a_n = a*x^(n-1). The general term in the determinant is log(a_n^r * a_(n+1)^k) = log(a^r * x^((n-1)r) * a^k * x^(nk)) = log(a^(r+k) * x^(r(n-1) + nk)) = (r+k)log(a) + (r(n-1) + nk)log(x). Let C1 = (r+k)log(a) and C2 = log(x). The term is C1 + (r(n-1) + nk)C2. Since the rows are in arithmetic progression, the determinant is zero for all r, k in N.
Correct Answer: 1