I am looking forward to an application of the universal property of the tensor product of vector spaces.
We DEFINE the determinant of a $n\times n$ matrix $A=(a_{ij})$ over a field $F$ by that for any fixed $i=1, 2, ..., n$, $$\det{A}=\sum_{j=1}^{n}(-1)^{i+j}a_{ij} |A_{ij}|,$$ where $|A_{ij}|$ is the minor of $A$ at position $(i, j)$.
The determinant is a multilinear form from $\overbrace{F^{n}\times F^n\times \cdots \times F^n}^{n \text{ times}}$ to $F$. By the universal property of tensor product of vector spaces, we have the following commutative diagram. $$ \begin{array}{rrl} {} \overbrace{F^{n}\times F^n\times \cdots \times F^n}^{n \text{ times}} & \longrightarrow & \overbrace{F^{n}\otimes F^n\otimes \cdots \otimes F^n}^{n \text{ times}} \\ & \det{}\searrow & \downarrow \theta \\ && F \end{array} $$ where $\theta$ is a linear transformation (functional).
Question. Can we prove the two important properties $\det{AB}=\det{A}\cdot \det{B}$ and $\det{A^t}=\det{A}$ via the linearity of $\theta$?
I want to use this question as a motivation of tensor product of vector spaces. So please don't use any concept which is deeper than tensor product, e.g., exterior product.
And avoid proving it by the elementary method (as possible), e.g., rank and elementary matrix. Because I hope this question can show the power of tensor product. That is, it should be more elegent than the elementary proof.