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 solve for 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 employ properties of logarithms and determinants. Notice the matrix is constructed using logarithmic relationships derived from distinct bases \(a, b, c\).
Our aim is to show the determinant equals 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 the 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 due to the subtraction of equivalent fractions. 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 of the matrix is zero.