 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/3/b/i/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

The given symmetric matrices satisfy

E_{j}^{T} E_{j} = E_{j}^{2} = I

, so both are orthogonal matrices. Put

J = E_{1} E_{2}

c = cos (θ /2)

and

s = sin (θ /2)

. Anticommutation gives

J^{T} = E_{2} E_{1} = - J, J^{2} = E_{1} E_{2} E_{1} E_{2} = - I .

Thus

A = c I + s J

and

A^{T} = c I - s J

A A^{T} = (c^{2} + s^{2}) I = I

A - A^{T} = 2 s J

. Since

J^{2} = - I

makes

J

invertible, $A$ is symmetric exactly when $sin (θ /2) = 0$ .

 New to topics? Read the docs here!