Question

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

• A. 0
• B. 1
• C. \( \log_e (abc) \)
• D. \( \log_a \log_e b \log_c \)

Hint:

When dealing with determinants involving logarithms, recognize patterns of linear dependence in the rows or columns, which may lead to a determinant of zero.

Answer 1

The Correct Option is A

To calculate the determinant of the given matrix: \[ \left| \begin{matrix} 1 & \log_a b & \log_a c \\ \log_b a & 1 & \log_b c \\ \log_c a & \log_c b & 1 \end{matrix} \right| \], we use logarithm and determinant properties. The matrix uses logarithmic relationships with distinct bases \(a, b, c\).
The goal is to show the determinant is zero. To simplify, recall these identities:

• \(\log_a b = \frac{\log b}{\log a}\)
• \(\log_b a = \frac{\log a}{\log b}\)
• \(\log_a c = \frac{\log c}{\log a}\)
• \(\log_c a = \frac{\log a}{\log c}\)
• \(\log_b c = \frac{\log c}{\log b}\)
• \(\log_c b = \frac{\log b}{\log c}\)

Substituting gives the matrix:

1	\(\frac{\log b}{\log a}\)	\(\frac{\log c}{\log a}\)
\(\frac{\log a}{\log b}\)	1	\(\frac{\log c}{\log b}\)
\(\frac{\log a}{\log c}\)	\(\frac{\log b}{\log c}\)	1

Consider determinant expansion by the first row:
\[\text{det} = 1 \times \left(1- \frac{\log c}{\log b} \cdot \frac{\log b}{\log c}\right) - \frac{\log b}{\log a} \times \left(\frac{\log a}{\log b}-\frac{\log c}{\log b} \cdot 1\right) + \frac{\log c}{\log a} \times \left(\frac{\log a}{\log c}-\frac{\log b}{\log c} \cdot 1\right)\]
Each bracket simplifies to zero:

• First: \(1-\frac{\log c}{\log b} \cdot \frac{\log b}{\log c} = 1-1 = 0\)
• Second: \(-\frac{\log b}{\log a} \cdot (\frac{\log a - \log c}{\log b}) = -\frac{\log a - \log c}{\log a}\)
• Third: \(\frac{\log c}{\log a} \cdot (\frac{\log a - \log b}{\log c}) = \frac{\log a - \log b}{\log a}\)

Thus, the terms cancel. Therefore, the determinant equals:
\[1 \times 0 - \left(\frac{\log a - \log c}{\log a}\right) + \left(\frac{\log a - \log b}{\log a}\right) = 0\]Hence, the determinant is zero.