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Question
Without expanding evaluate the following determinant:
`|(1, "a", "b" + "c"),(1, "b", "c" + "a"),(1, "c", "a" + "b")|`
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Solution
Let D = `|(1, "a", "b" + "c"),(1, "b", "c" + "a"),(1, "c", "a" + "b")|`
By C3 + C2, we get,
D = `|(1, "a", "a" + "b" + "c"),(1, "b", "a" + "b" + "c"),(1, "c", "a" + "b" + "c")|`
By taking (a + b + c) common from C3, we get,
D = `("a" + "b" + "c")|(1, "a", 1),(1, "b", 1),(1, "c", 1)|`
= (a + b + c) × 0 ...[ ∵ C1 ≡ C3]
= 0.
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