 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/3/d/solution

Writing the matrix explicitly gives

M = (i x_{3} i x_{1} - x_{2} i x_{1} + x_{2} - i x_{3}) .

det M = x_{3}^{2} - (i x_{1} + x_{2}) (i x_{1} - x_{2}) = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} .

Since

det A_{1}^{†} = (det A_{1})^{*} = 1

det (A_{1}^{†} M A_{1}) = det M = ∣ x ∣^{2} .

The conjugation therefore preserves the squared Euclidean length encoded in

M

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