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 $B$ byIt is an alternating multilinear map, because both summands are, and $F(I_n)=\det A+1-\det A=1$. Hence $F$ is a normalized determinantal function. By the asserted uniqueness of the determinant, $F(B)=\det B$ for every $B$. Rearranging yieldsThis argument also covers $\det A=0$ without dividing by it.