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Question
By using properties of determinants, prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0.
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Solution
L.H.S. = `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|`
Applying R1 → R1 + R2, we get
L.H.S. = `|(x + y + z, x + y + z, x + y + z),(z, x , y),(1, 1, 1)|`
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 and R3 are identical]
= 0
= R.H.S.
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