If a1, a2, a3, ..., a9 are in G.P., then what is the value of the determinant \(\begin{vmatrix} \rm \ln a_1 & \rm \ln a_2 & \rm \ln a_3 \\ \rm \ln a_4 & \rm \ln a_5 & \rm \ln a_6 \\ \rm \ln a_7 & \rm \ln a_8 & \rm \ln a_9 \end{vmatrix}\)?

  1. 0
  2. 1
  3. -1
  4. 3

Answer (Detailed Solution Below)

Option 1 : 0
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NDA 01/2025: English Subject Test
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Detailed Solution

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Concept:

Geometric Progression:

  • The ratio between any two consecutive terms of a geometric progression is a fixed constant, called the common ratio (r) of the progression.


Logarithms:

  • log a - log b = \(\rm \log {a\over b}\).


Determinants:

  • A linear combination of the rows/columns does not affect the value of the determinant.
  • If two rows/columns of a given matrix are interchanged, then the value of the determinant gets multiplied by -1.
  • If a row/column of a given matrix is multiplied by a scalar k, then the value of the determinant is also multiplied by k.


Calculation:

Let the common ratio of the given G.P. be r.

Let D = \(\begin{vmatrix} \rm \ln a_1 & \rm \ln a_2 & \rm \ln a_3 \\ \rm \ln a_4 & \rm \ln a_5 & \rm \ln a_6 \\ \rm \ln a_7 & \rm \ln a_8 & \rm \ln a_9 \end{vmatrix}\).

Using C1 → C1 - C2 and C2 → C2 - C3, we get:

⇒ D = \(\begin{vmatrix} \rm \ln a_1 -\ln a_2& \rm \ln a_2 -\ln a_3& \rm \ln a_3 \\ \rm \ln a_4 -\ln a_5& \rm \ln a_5 -\ln a_6& \rm \ln a_6 \\ \rm \ln a_7-\ln a_8 & \rm \ln a_8-\ln a_9 & \rm \ln a_9 \end{vmatrix}\)

⇒ D = \(\begin{vmatrix} \rm \ln {a_1 \over a_2}& \rm \ln {a_2 \over a_3}& \rm \ln a_3 \\ \rm \ln {a_4 \over a_5}& \rm \ln {a_5 \over a_6}& \rm \ln a_6 \\ \rm \ln {a_7 \over a_8} & \rm \ln {a_8 \over a_9} & \rm \ln a_9 \end{vmatrix}\)

⇒ D = \(\begin{vmatrix} \rm \ln r& \rm \ln r& \rm \ln a_3 \\ \rm \ln r& \rm \ln r& \rm \ln a_6 \\ \rm \ln r & \rm \ln r & \rm \ln a_9 \end{vmatrix}\)

Using C1 → C1 - C2, we get:

⇒ D = \(\begin{vmatrix} 0& \rm \ln r& \rm \ln a_3 \\0& \rm \ln r& \rm \ln a_6 \\ 0 & \rm \ln r & \rm \ln a_9 \end{vmatrix}\)

Expanding along C1, we get:

⇒ D = 0.

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