 Solution

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

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

A = (a_{ij}) \in M_{n} (C)

det A = \sum_{σ \in S_{n}} sgn (σ) \prod_{i = 1}^{n} a_{i, σ (i)} .

For

A = (M 0 X N),

any nonzero term in the Leibniz formula for determinants must match every row in the lower block to a column in the

N

block. The remaining upper rows must then match the

M

columns. The sum consequently factors into the determinant sums for the two diagonal blocks:

det A = (det M) (det N) .

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