If `a^2+b^2+c^2=1,` then prove that `|[a^2 +(b^2+c^2)cosθ, ab(1−cosθ), ac(1−cosθ)],[ba(1−cosθ),b^2(c^2+a^2)cosθ,bc(1−cosθ)],[ca(1−cosθ),cb(1−cosθ),c^2+(a^2+b^2)cosθ]|` independent of a,b,c?

Explanation

Multiplying `C_(1) " by " a C_(2)` by b and `C_(3)` by c we have `Delta = (1)/(abc)|{:(a^(3)+a(b^(2)+c^(2))cos phi ,,ab^(2)(1-cos phi) ,,ac^(2)(1-cos phi)),(ba^(2)(1-cos phi),,b^(3)+b(c^(2)+a^(2))cos phi ,,bc^(2)(1-cos phi)),(ca^(2)(1-cos phi),,cb^(2)(1-cos phi),,c^(3)+c(a^(2)+b^(2))cos phi)):}|`
Now take a,b and c common from `R_(1), R_(2) " and "R_(3)` respectively to give
`Delta =(abc)/(abc)|{:(a^(2)+(b^(2)+c^(2)) cos phi,,b^(2)(1-cos phi),,c^(2)(1-cos phi)),(a^(2)(1-cos phi),,b^(2)+(c^(2)+a^(2)) cos phi,,c^(2)(1-cos phi)),(a^(2)(1-cos phi),,b^(2)(1-cos phi),,c^(2) + (a^(2)+b^(2))cos phi):}|`
Applying `C_(1) to C_(1)+C_(2) +C_(3)` we get
`Delta =|{:(1,,b^(2)(1-cos phi),,c^(2)(1-cos phi)),(1,,b^(2)+(c^(2)+a^(2))cos phi,,c^(2)(1-cos phi)),(1,,b^(2)(1-cos phi),,c^(2)+(a^(2)+b^(2))cos phi):}|`
Applying `R_(2) to R_(2) - R_(1) " and " R_(3) -R_(1)` we get
`Delta = |{:(1,,b^(2)(1-cos phi),,c^(2)(1-cos phi)),(0,,(a^(2)+b^(2)+c^(2))cos phi,,0),(0,,0,,(a^(2)+b^(2)+c^(2))cos phi):}|`
Expading along `C_(1)` we get `Delta =|{:(0,,cos phi),(cos phi,,0):}|=cos^(2)phi`