If `A+B+C=pi` and `e^(itheta)=costheta+isintheta` and `z=|[e^(2iA), e^(-iC), e^(-iB)] , [e^(-iC), e^(2iB), e^(-iA)], [e^(-iB), e^(-iA), e^(2iC)]|`, then

Options

A. 1
B. -1
C. -2
D. -4

Answer

D

Explanation

Since `A+B +C =pi " and " e^(ipi) =cos pi+ isin pi=-1`
`e^(i(B+C)) =e^(i(pi-A))=-e^(-iA)" and " e^(-i(B+C))=-e^(iA)`
By taking `e^(iA),e^(iB),e^(iC)` common from `R_(1),R_(2)" and " R_(3)` respectively we have
` Delta = |{:(e^(iA),,e^(-i(A+C)),,e^(-i(A+B))),(e^(-i(B+C)),,e^(iB),,e^(-(A+B))),(e^(-(B+C)),,e^(-(A+C)),,e^(iC)):}|`
`=- |{:(e^(iA),,-e^(iB),,-e^(iC)),(-e^(iA),,e^(iB),,-e^(iC)),(-e^(iA),,-e^(iB),,e^(iC)):}| `
By taking `e^(iA) ,e^(iB) ,e^(iC)` common from `C_(1),C_(2)" and " C_(3)` respectively
`Delta= |{:(1,,-1,,-1),(-1,,1,,-1),(-1,,-1,,1):}|=-4`