 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-i/3/a/v/solution

We have

D (I_{n}) = det A

. Define a function of the columns of

B

F (B) = D (B) + (1 - det A) det B .

It is an alternating multilinear map, because both summands are, and

F (I_{n}) = det A + 1 - det A = 1

. Hence

F

F (B) = det B

for every

B

. Rearranging yields

det (A B) = det A det B .

This argument also covers

det A = 0

without dividing by it.

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