 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/digital-systems-linear-algebra-and-introduction-to-formal-proof/6/g/solution

The determinant is invariant under transpose, while multiplying all

n

rows by

- 1

multiplies it by

(- 1)^{n}

. Thus, for odd integer

n

det A = det A^{⊤} = det (- A) = (- 1)^{n} det A = - det A .

Over the real numbers, this forces $det A = 0$ , so

A

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