 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-1/8a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

M

M^{T} M = I

. The planar rotation

R = (cos θ sin θ - sin θ cos θ)

obeys

R^{T} R = I

For

A = (a b b c)

, the off-diagonal entry of

R A R^{T}

\frac{1}{2} (a - c) sin 2 θ + b cos 2 θ

. Thus

(a - c) sin 2 θ + 2 b cos 2 θ = 0.

(a - c, b) \neq = (0, 0)

, equivalently

θ = \frac{1}{2} atan2 (- 2 b, a - c) + kπ /2

; if

A

Because

R A R^{T}

λ_{\pm} = \frac{tr A \pm ( tr A ) ^{2} - 4 d e t A}{2} .

C = (1221)

3, - 1

along

(1, 1)^{T}, (1, - 1)^{T}

. Raising its orthogonal diagonalization to the even power gives

C^{2 N} = \frac{1}{2} (3^{2 N} + 1 3^{2 N} - 1 3^{2 N} - 1 3^{2 N} + 1) .

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