Question

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`

Answer 1

`|(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`