 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/6/b/solution

x^{T} Q x

Q = 503 0 - 3 0 30 - 3 .

The

y

axis has eigenvalue

- 3

. On the

x z

det (5 - ℓ 3 3 - 3 - ℓ) = (5 - ℓ) (- 3 - ℓ) - 9 = ℓ^{2} - 2 ℓ - 24 = (ℓ - 6) (ℓ + 4) .

Thus the orthogonal diagonalization from part (a) yields

λ = 6, μ = - 3, ν = - 4, 6 u^{2} - 3 v^{2} - 4 w^{2} = α .

For later use, one possible orthonormal basis is

e_{u} = \frac{( 3 , 0 , 1 )}{10}, e_{v} = (0, 1, 0), e_{w} = \frac{( 1 , 0 , - 3 )}{10} .

Then

u = (3 x + z) / 10

v = y

w = (x - 3 z) / 10

. This explicit choice also confirms that the Euclidean norm is preserved:

x^{2} + y^{2} + z^{2} = u^{2} + v^{2} + w^{2}

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