Uniqueness of determinant

Question

In Artin Algebra 2nd edition page 22, the author proved the uniqueness of determinant by saying that any matrix $A$ can be written in reduced row-echelon form $A'$: $A'=E_1\cdots E_kA$ where $E_i$ are the elementary matrix. Then $A'$ is either $I$ or has a zero row. If $A'=I$, then $\delta(A')=1$. Otherwise, $\delta(A')=0$. In both cases, $\delta(A')$ is determined, and hence by $$\delta(A')=\delta(E_1)\cdots\delta(E_k)\delta(A)$$ $\delta(A)$ is determined uniquely.

However, as he himself pointed out immediately in the following paragraph, the sequence $E_1\cdots E_k$ is not unique. Then why is $\delta(A)$ uniquely determined?

Edit: The author defined determinant as a function $\delta(A)=d\in \mathbb{R}$ satisfying the following 3 conditions:

(i) $\delta(I)=1$

(ii) $\delta$ is linear in the rows of the matrix $A$

(iii) If two adjacent rows of $A$ are equal, then $\delta(A)=0$

He then proved that the above conditions imply some properties that all of us know, e.g.,

(a) Interchanging two rows reverses the sign

(b) If $A$ has a zero row, then $\delta(A)=0$

(c) Multiplying one row by a number and adding it to another row doesn't change the determinant

(d) $\delta(E)=\pm1$ or $c$

(e) $\delta(AB)=\delta(A)\delta(B)$

Then he proved that the function $\delta$ so defined is unique, as shown in the beginning of my post, which I don't understand

Answer 1

If the determinant $\delta$ exists, you can prove that

1. $\delta(E)=c$ if $E$ is the elementary matrix corresponding to multiplication of a row by $c$;

2. $\delta(E)=1$ if $E$ is the elementary matrix corresponding to summing a row to another multiplied by a constant;

3. $\delta(E)=-1$ if $E$ is the elementary matrix corresponding to switching two rows;

4. $\delta(A)=0$ if $A$ is not invertible;

5. $\delta(AB)=\delta(A)\delta(B)$.

If two determinant functions $\delta$ and $\delta'$ exist, then they are both zero on the noninvertible matrices, but, writing an invertible $A$ as $$ A=E_1E_2\dots E_k $$ a product of elementary matrices, we conclude that $$ \delta(A)=\delta(E_1)\delta(E_2)\dots\delta(E_k)=\delta'(A) $$

Note the initial if: we're assuming the existence, not proving it. Also, note that the decomposition into a product of elementary matrices doesn't depend on the determinant; so just use the same decomposition for computing $\delta(A)$ and $\delta'(A)$.

The fact that the decomposition as product of elementary matrices is not unique is indeed a problem, but with respect to the existence of the determinant. In principle you could find two decompositions that produce different values when computed with rules 1–5, but this would simply prove that the determinant doesn't exist.