Using properties of determinants, prove that:
`|(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^2+b^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1 + a^2 + b^2)^3`
Solution
`|(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^2+b^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1 + a^2 + b^2)^3`
Let
`Delta = |(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^2+b^2, 2a),(2b, -2a, 1-a^2-b^2)|`
We shall try to introduce zeros at a many places as possible keeping in mind that we have to introduce the factor `1 + a^2 + b^2`
Applying `C_1 -> C_1 - bC_3` and `C_2 -> C_2 + aC_3` we get
`Delta = |(1+a^2+b^2, 0, -2b),(0, 1+a^2+b^2, 2a), (b(1+a^2 +b^2), -a(1+a^2+b^2), 1-a^2 - b^2)|`
`=> Delta = (1 + a^2 +b^2)^2 |(1,0,-2b),(0,1,2a),(b, -a, -a^2-b^2)|` [Taking `(1+ a^2 + b^2)` common from both `C_1 and C_2`]
`=> Delta = (1+ a^2 + b^2)^2 |(1,0,-2b),(0,1, 2a),(0,0, 1+z^2+b^2) |` [Applying `R_3 -> R_3- bR_1 + aR_2`]
`=> Delta= (1 + a^2 + b^2)^2 xx 1 xx |(1, 2a),(0, 1+a^2 + b^2)|` [Expanding along `C_1`]
`=> Delta= (1 + a^2 + b^2 )^3`
