IMC 1997 · Problem 2

Day 24th IMC · Plovdiv, Bulgaria

Statement

Let MM be an invertible matrix of dimension 2n×2n2n \times 2n, represented in block form as

M=[ABCD]andM1=[EFGH].M = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \quad \text{and} \quad M^{-1} = \begin{bmatrix} E & F \\ G & H \end{bmatrix}.

Show that detMdetH=detA\det M \cdot \det H = \det A.

Official solution

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