Eq(5.8) showed that axioms 1 and 2 alone can be used to define a determinant function. This raises the question of uniqueness of such functions.
Let d be a function satisfying all 3 axioms for a determinant function of order n. If f is another function satisfying axioms 1 and 2, then for every n ´ n matrix A, we have
Hence, if f also satisfies axioms 3, then so that d is unique.
Since both f and d satisfy axioms 1 and 2, so does g. However, using , we have
so that g does not satisfy axiom 3. Thus, eq(5.8) becomes
Putting this into eq(a), we recover eq(5.9). QED.