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Question
Answer the following question:
Without expanding determinant show that
`|(x"a", y"b", z"c"),("a"^2, "b"^2, "c"^2),(1, 1, 1)| = |(x, y, z),("a", "b", "c"),("bc", "ca", "ab")|`
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Solution
L.H.S. = `|(x"a", y"b", z"c"),("a"^2, "b"^2, "c"^2),(1, 1, 1)|`
By taking a, b, c common from C1, C2, C3 respectively, we get,
L.H.S. = `"abc"|(x, y, z),("a", "b", "c"),(1/"a", 1/"b", 1/"c")|`
= `|(x, y, z),("a", "b", "c"),("abc"/"a", "abc"/"b", "abc"/"c")|`
= `|(x, y, z),("a", "b", "c"),("bc", "ca", "ab")|`
= R.H.S.
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