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")|`

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.