1F (Linear Algebra)

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) .

b

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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