TITLE: The definition of Determinant in the spirit of algebra and geometry
QUESTION [13 upvotes]: The concept of determinant is quite unmotivational topic to introduce. Textbooks use such "strung out" introductions like axiomatic definition, Laplace expansion, Leibniz'a permutation formula or something like signed volume.
Question: is the following a possible way to introduce the determinant?

Determinant is all about determing whether a given set of vectors are linearly independent, and a direct way to check this is to add scalar multiplications of column vectors to get the diagonal form:
$$\begin{pmatrix} 
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22}  & a_{23} & a_{24} \\
a_{31} & a_{32}  & a_{33} & a_{34} \\
a_{41} & a_{42}  & a_{43} & a_{44} \\
\end{pmatrix} \thicksim \begin{pmatrix} 
d_1 & 0 & 0 & 0 \\
0 & d_2  & 0 & 0 \\
0 & 0  & d_3 & 0 \\
0 & 0  & 0 & d_4 \\
\end{pmatrix}.$$
During the diagonalization process we demand that the information, i.e. the determinant, remains unchanged. Now it's clear that the vectors are linearly independent if every $d_i$ is nonzero, i.e. $\prod_{i=1}^n d_i\neq0$. It may also be the case that two columns are equal and there is no diagonal form, so we must add a condition that annihilates the determinant (this is consistent with $\prod_{i=1}^n d_i=0$), since column vectors can't be linearly independent.
If we want to have a real valued function that provides this information, then we simply introduce an ad hoc function $\det:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ with following properties:

$$\det (a_1,\ldots,a_i,\ldots,a_j,\ldots,a_n)=\det (a_1,\ldots,a_i,\ldots,k\cdot a_i+a_j,\ldots,a_n).$$
$$\det(d_1\cdot e_1,\ldots,d_n\cdot e_n)=\prod_{i=1}^n d_i.$$
$$\det (a_1,\ldots,a_i,\ldots,a_j,\ldots,a_n)=0, \space \space \text{if} \space \space a_i=a_j.$$


From the previous definition of determinant we can infer the multilinearity property:
$$[a_1,\ldots,c_1 \cdot u+c_2 \cdot v,\ldots,a_n]\thicksim diag[d_1,\ldots,c_1 \cdot d'_i+c_2 \cdot d''_i ,\ldots,d_n],$$ so $$\det[a_1,\ldots,c_1 \cdot u+c_2 \cdot v,\ldots,a_n]=\prod_{j=1:j\neq i}^n d_j(c_1 \cdot d'_i+c_2 \cdot d''_i)$$ $$=c_1\det(diag[d_1,\ldots, d'_i,\ldots,d_n])+c_2\det(diag[d_1,\ldots, d''_i,\ldots,d_n])$$ $$=c_1\det[a_1,\ldots,u,\ldots,a_n]+c_2\det[a_1,\ldots, v,\ldots,a_n].$$
Note that previous multilinearity together with property $(1)$ gives the property $(2)$, so we know from the literature that the determinant function $\det:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ actually exists and it is unique.

Obviously, the determinant offers information how orthogonal a set of vectors is. Thus, with Gram-Schmidt process we can form an orthogonal set of vectors form set $(a_1,\ldots, a_n)$, and by multilinearity and property $(2)$ the absolute value of determinant is the volume of parallelepiped spanned by the set of vectors.
Definition.
Volume of parallelepiped formed by set of vectors $(a_1,\ldots, a_n)$ is $Vol(a_1,\ldots, a_n)=Vol(a_1,\ldots, a_{n-1})\cdot |a_{n}^{\bot}|=|a_{1}^{\bot}|\cdots |a_{n}^{\bot}|$, where $a_{i}^{\bot} \bot span(a_1,\ldots, a_{i-1}).$

This approach to determinant works equally well if we begin with the volume of a parallelepiped (geometric approach) or with the search of invertibility (algebraic approach). I was motivated by the book Linear algebra and its applications by Lax on chapter 5:
Rather than start with a formula for the determinant, we shall deduce it from the properties forced on it by the geometric properties of signed volume. This approach to determinants is due to E. Artin.

$\det (a_1,\ldots,a_n)=0$, if $a_i=a_j$, $i\neq j.$
$\det (a_1,\ldots,a_n)$ is a multilinear function of its arguments, in the sense that if all $a_i, i \neq j$ are fixed, $\det$ is a linear function of the remaining argument $a_j.$
$\det(e_1,\ldots,e_n)=1.$

REPLY [3 votes]: The way I teach determinants to my students is to start with the case $n=2$, and to use the complex numbers and/or trigonometry in order to show that, for $(a,b), (c,d)$ vectors on the plane, the quantity
$$ad-bc=||(a,b)||\cdotp ||(c,d)|| \sin \theta$$
is the signed area between $(a,b)$ and $(c,d)$ (in this order).
Then, using the vector product and its properties (we have seen it before coming to the topic of determinants in full generality), we check that $3$ by $3$ determinants carry the meaning of signed volumes.
The next step is to introduce determinants as alternate multilinear functions.  We have seen examples of bilinear maps (inner products), trilinear maps, such as
$$(u,v,w)\mapsto (u\times v)\bullet w$$
and the quadrilinear maps $$(a,b,c,d)\mapsto (a\bullet c) (b\bullet d)-(b\bullet c) (a\bullet d),$$
$$(a,b,c,d)\mapsto (a \times b)\bullet (c\times d).$$
Now, when explaining multilinearity we did emphasise that the fact that the last two examples are equal can be proven if only we check equality for the case where $a,b,c,d$ are vectors of the canonical basis.
Then the time comes to define the determinant of $n$ vectors in $\mathbb{R}^n$, which is a new example of $n-$linear, alternate function. They check that the vector space of such maps is indeed $\left(^n_n\right).$  The students thus learn that the determinant is essentially the only possible such function, up to a multiple, in the same way they saw that more general multilinear maps depend exclusively on their values on vectors of a chosen basis (say, the canonical basis in our case).
Although I learnt to prove stuff such as $\det(AB)=\det(A) \det(B)$ by strict functoriality, in class we do define the map
$$L(X_1, \ldots , X_n)=\det(AX_1, \ldots , AX_n),$$ which by uniqueness is a constant multiple of the determinant function $T(X_1, \ldots , X_n)=\det(X_1, \ldots, X_n),$ and compute the constant by evaluating on the identity matrix, i.e. $X_i=e_i.$
Thus $\det(AB)=\det(A)\det(B).$
It is in T.W. Korner's book called Vectors, Pure and Applied that one can see a construction that uses elementary matrices and is rigorous.  The OP can check Korner's book to see a nice, slightly more down-to-earth exposition.
In op. cit. one can see how Korner uses the fact that an invertible matrix can be decomposed as a product of elementary matrices to obtain the formula $\det(AB)=\det(A)\det(B).$
Note:  I have been deliberately brief in my exposition, just so as not to repeat too much stuff that was already included in other answers.