First recall that in Chapter 14 we demonstrated (with some effort it must be said) that the generic form of a 2 × 2 Unitary matrix, M, is: Where a and b ∈ ℂ, θ is an angle with 0 ≤ θ < 2π and a^{2} + b^{2} = 1. We can use this to form our modified matrix, M – λI as follows: We then recall the requirement that the determinant of this matrix is zero: Again the determinant of a 2 × 2 matrix with entries {a, b, c, d} is simply ad – bc, so we have:

FROM CHAPTER 21
Back in Chapter 17, we showed that the CayleyHamilton Theorem for a 3 × 3 matrix, M_{3}, implies that:
M_{3}^{3} – tr(M_{3})M_{3}^{2} + ½[tr(M_{3})^{2} – tr(M_{3}^{2})]M_{3} – det(M_{3})I_{3} = 0
Where I_{3} is the 3 × 3 identity matrix.
If we consider a matrix S ∈ su(3), then by definition its trace is zero, so the terms involving tr(S) in the above drop out and we get:
S^{3} – ½tr(S^{2})S – det(S)I_{3} = 0
or
S^{3} = ½tr(S^{2})S + det(S)I_{3} (1)
Back in Chapter 18, we noted that:
tr(M + N) = tr(M) + tr(N) (2)
and
tr(αM) = α tr(M) (3)
Where α is a scalar.
If we take the trace of both sides of (1), we get:
tr(S^{3}) = tr[½tr(S^{2})S + det(S)I_{3}]
Using (2) we note that:
tr(S^{3}) = tr[½tr(S^{2})S] + tr[det(S)I_{3}]
Using (3) and noting that the trace of a matrix is a scalar we have that:
tr(S^{3}) = ½tr(S^{2})tr(S) + det(S)tr(I_{3})
Again tr(S) is zero and obviously tr(I_{3}) = 3, so:
tr(S^{3}) = 3det(S)
or
det(S) = tr(S^{3})/3