 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-1/1f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Move the first

n - k

columns of

A = (0 N M 0)

past its last

k

columns. This requires

k (n - k)

column interchanges and produces the block diagonal matrix

(M 0 0 N) .

Each interchange reverses the determinant's sign, so part (a) gives

det A = (- 1)^{k (n - k)} (det M) (det N) .

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