Answer the following question:

Without expanding determinant show that

`|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" + "b", "ab", "a"^2"b"^2)|` = 0

#### Solution

L.H.S. = `|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" + "b", "ab", "a"^2"b"^2)|`

Taking bc, ca, ab common from R_{1}, R_{2}, R_{3} respectively, we get

L.H.S. = `("bc")("ca")("ab")|(("b" + "c")/"bc", 1, "bc"),(("c" + "a")/"ca", 1, "ca"),(("a" + "b")/"ab", 1, "ab")|`

Taking abc common from C_{3}, we get

L.H.S. = `("a"^2"b"^2"c"^2)("abc")|(1/"c" + 1/"b", 1, 1/"a"),(1/"a" + 1/"c", 1, 1/"b"),(1/"b" + 1/"a", 1, 1/"c")|`

Applying C_{1} → C_{1} + C_{3}, we get

L.H.S. = `"a"^3"b"^3"c"^3|(1/"a" + 1/"b" + 1/"c", 1, 1/"a"),(1/"a" + 1/"b" + 1/"c", 1, 1/"b"),(1/"a" + 1/"b" + 1/"c", 1, 1/"c")|`

Taking `(1/"a" + 1/"b" + 1/"c")` common from C_{1}, we get

L.H.S. = `"a"^3"b"^3"c"^3 (1/"a" + 1/"b" + 1/"c")|(1, 1, 1/"a"),(1, 1, 1/"b"),(1, 1, 1/"c")|`

= `"a"^3"b"^3"c"^3 (1/"a" + 1/"b" + 1/"c")(0)` ...[∵ C_{1} and C_{2} are identical]

= 0

= R.H.S.