Question

If \( a, b, c \) are positive real numbers each distinct from unity, then the value of the determinant \[ \begin{vmatrix} 1 & \log_a b & \log_a c \\ \log_b a & 1 & \log_b c \\ \log_c a & \log_c b & 1 \end{vmatrix} \] is:

• A. \( 0 \)
• B. \( 1 \)
• C. \( \log_e(abc) \)
• D. \( \log_a e \cdot \log_b e \cdot \log_c e \)

Hint:

For log determinants: Convert logs using \( \log_a b = \frac{\ln b}{\ln a} \). Try row scaling to expose symmetry. Identical or proportional rows \( \Rightarrow \) determinant = 0.

Answer 1

The Correct Option is A

Concept: Use the identity: \[ \log_a b = \frac{\ln b}{\ln a} \] So every logarithm can be expressed in terms of natural logs. This often converts determinant rows into linearly dependent rows. Step 1: Convert logs using natural logarithm. Let: \[ x = \ln a, \quad y = \ln b, \quad z = \ln c \] Then: \[ \log_a b = \frac{y}{x}, \quad \log_a c = \frac{z}{x} \] \[ \log_b a = \frac{x}{y}, \quad \log_b c = \frac{z}{y} \] \[ \log_c a = \frac{x}{z}, \quad \log_c b = \frac{y}{z} \] Step 2: Rewrite determinant. \[ \begin{vmatrix} 1 & \frac{y}{x} & \frac{z}{x} \\ \frac{x}{y} & 1 & \frac{z}{y} \\ \frac{x}{z} & \frac{y}{z} & 1 \end{vmatrix} \] Step 3: Multiply rows to remove denominators.

• Row 1 by \( x \)
• Row 2 by \( y \)
• Row 3 by \( z \)

Determinant gets multiplied by \( xyz \). New determinant: \[ \begin{vmatrix} x & y & z \\ x & y & z \\ x & y & z \end{vmatrix} \] Step 4: Evaluate determinant. All rows are identical, so determinant = 0. Since we only multiplied by nonzero constants, original determinant is also: \[ 0 \]