Engineering

Mathematics

Introduction to Determinants

Properties of Logarithm

Question

$|\begin{matrix} 1 & \log_{b} a \\ \log_{a} b & 1 \end{matrix}| =$

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Let $A = ∣ \begin{array}{cc} 1 & \log_{b} a \\ \log_{a} b & 1 \end{array} ∣$

Expanding the determinant,

A = 1 – log_ab × log_ba = 1 – 1 = 0
(because

\log_{a} b = \frac{1}{\log_{b} a}

)