Question
Determinant $\left|\begin{array}{cc}1 & \log _b a \\ \log _a b & 1\end{array}\right|=$

Answer

(B)
$
\begin{aligned}
1-\log _b a \times \log _a b & =1-\log _a a=1-1=0 \\
\log _a a & =1
\end{aligned}
$
Hence correct option is (B).

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