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Question
Answer the following question:
By using properties of determinant prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0
Solution
L.H.S. = `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|`
By R1 + R2 , we get,
L.H.S. = `|(x + y + z, x + y + z, x + y + z),(z, x, y),(1, 1, 1)|`
By taking (x + y + z) common from R1, we get,
L.H.S. = `(x + y + z)|(1, 1, 1),(z, x, y),(1, 1, 1)|`
=(x + y + z) × 0 ...[∵ R1 ≡ R3]
= 0
= R.H.S.
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