Matrices & Determinants Question - None

Question

If $a_1, a_2, \ldots, a_n, \ldots$ form a G.P. and $a_i > 0$, for all $i \geq 1$, then $\begin{vmatrix} \log a_n & \log a_{n+1} & \log a_{n+2} \ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{vmatrix}$ is equal to

Accepted Answer

Step 1: In a G.P. with first term a₁ and common ratio r: aₙ = a₁·r^(n-1)

Step 2: Taking logarithms: log(aₙ₊ₖ) = log(a₁) + k·log(r)

So log(aₙ) = log(a₁), log(aₙ₊₁) = log(a₁) + log(r), log(aₙ₊₂) = log(a₁) + 2log(r), etc.

Step 3: The matrix becomes:

|log(a₁) log(a₁)+log(r) log(a₁)+2log(r)|

|log(a₁)+3log(r) log(a₁)+4log(r) log(a₁)+5log(r)|

|log(a₁)+6log(r) log(a₁)+7log(r) log(a₁)+8log(r)|

Step 4: Each row has entries in A.P. with common difference log(r). This means R₂ - R₁ and R₃ - R₂ are proportional (both equal multiples of log(r)).

Step 5: More precisely, R₂ = R₁ + 3log(r)·(1,1,1) and R₃ = R₂ + 3log(r)·(1,1,1), making the rows linearly dependent.

Step 6: When rows are linearly dependent, the determinant equals 0.

∴ Answer: A (0)

Question:

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