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Question
Without expanding evaluate the following determinant:
`|(1, "a", "b" + "c"),(1, "b", "c" + "a"),(1, "c", "a" + "b")|`
Solution
Let D = `|(1, "a", "b" + "c"),(1, "b", "c" + "a"),(1, "c", "a" + "b")|`
Applying C3 → C3 + C2, we get
D = `|(1, "a", "a" + "b" + "c"),(1, "b", "a" + "b" + "c"),(1, "c", "a" + "b" + "c")|`
Taking (a + b + c) common from C3, we get
D = `("a" + "b" + "c")|(1, "a", 1),(1, "b", 1),(1, "c", 1)|`
∴ D = (a + b + c)(0) ...[∵ C1 and C3 are identical]
∴ D = 0
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