Question

Without expanding evaluate the following determinant:

`|(1, "a", "b" + "c"),(1, "b", "c" + "a"),(1, "c", "a" + "b")|`

Answer 1

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