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Question
Using properties of determinant show that
`|(1, log_x y, log_x z),(log_y x, 1, log_y z),(log_z x, log_z y, 1)|` = 0
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Solution
L.H.S. = `|(1, log_x y, log_x z),(log_y x, 1, log_y z),(log_z x, log_z y, 1)|`
= `|(log_"e" x/log_"e" x,log_"e" y/log_"e" x,log_"e" z/log_"e" x),(log_"e" x/log_"e" y,log_"e" y/log_"e" y,log_"e" z/log_"e" y),(log_"e" x/log_"e" z,log_"e" y/log_"e" z,log_"e" z/log_"e" z)| ...[because log_"e" "b" = log_"e" "b"/log_"e" "c"]`
Taking `1/log_"e" x, 1/log_"e" y, 1/log_"e" z` common from R1, R2, R3 respectively, we get
L.H.S. = `1/(log_"e" x*log_"e" y*log_"e" z) |(log_"e" x, log_"e" y, log_"e" z),(log_"e" x, log_"e" y, log_"e" z),(log_"e" x, log_"e" y, log_"e" z)|`
= `1/(log_"e" x*log_"e" y*log_"e" z)(0)` ...[∵ R1, R2, R3 are identical]
= 0
= R.H.S.
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