The scalar product, as defined in the book "Geometric-algebraic-for-computer-science" ,is directly based on this determinant. Is it possible for this definition to be independent of the determinant? I would like to understand its motives better (of course, I am only at chapter 4)
$ \newcommand\tr{\mathop{\mathrm{tr}}} $It is simply the scalar part of the geometric product: $$ A*B = \langle AB\rangle_0. $$
We can actually define it without referring to the exterior algebra structure at all. We consider a geometric algebra over a space of dimension $n$. Fix an element $A$ of the geometric algebra. The function $X \mapsto AX$ using the geometric product is linear so its trace is well-defined; call this the trace of $A$ and denote it $\tr(A)$. The normalized trace is gotten by dividing by the dimension of the algebra, and we the get a scalar product $$ A*B = \frac1{2^n}\tr(AB). $$
I claim that the normalized trace is simply the scalar part. To see this, let $e_i$ be an orthonormal vector basis. This generates a basis for the algebra of the form $$ e_{i_1}\dotsb e_{i_k},\quad 1\leq i_1 < \dotsb < i_k \leq n,\quad 0\leq k \leq n $$ where $k = 0$ is defined to give us the identity $1$. The only $E$ of the above form satisfying $E^2 = E$ is $E = 1$ so this is the only component that contributes to the trace. Because $\tr 1 = 2^n$ we get that $\tr A$ is $2^n$ times the $1$-component of $A$, otherwise known as the scalar part.