Question

Evaluate `|(cos alpha cos beta, cos alpha sin beta, -sin alpha),(-sin beta, cos beta, 0),(sin alpha cos beta, sin alpha sin beta,cos alpha )|`

Answer 1

Δ = `|(cosalphacosbeta, cosalphasinbeta,-sinalpha),(-sinbeta,cosbeta,0),(sinalpha cosbeta,sinalphasinbeta,cosalpha)|`

Expanding along C₃,

= `cos alphacosbeta |(cosbeta, 0), (sinalphasinbeta,cosalpha)| - cosalpha sinbeta |(-sinbeta, 0), (sinalphacosbeta, cosalpha)| - sinalpha|(-sinbeta, cosbeta), (sinalpha cosbeta, sin alpha sin beta)|`

we have:

Δ = −sin α(−sin α sin² β − cos² β sin α) + cos α(cos α cos² β + cos α sin² β)

= sin² α(sin² β + cos² β) + cos² α(cos² β + sin² β)

= sin² α(1) + cos² α(1)

= 1