What is the best way of introducing determinants in a linear algebra course? I want to give real life examples of where the determinant is applied. It should have a real impact.
[Math] Best way of introducing determinants in a linear algebra course
determinanteducationlinear algebra
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I normally have two ways of viewing determinants without appealing to higher-level math like multilinear forms.
The first is geometric, and I do think that most vector calculus classes nowadays should teach this interpretation. That is that, given vectors $v_1, \ldots, v_n \in \mathbb{R}^n$ dictating the sides of an $n$-dimensional parallelepiped, the volume of this parallelepiped is given by $\det(A)$, where $A = [v_1 \ldots v_n]$ is the matrix whose columns are given by those vectors. We can then view the determinant of a square matrix as measuring the volume-scaling property of the matrix as a linear map on $\mathbb{R}^n$. From here, it would be clear why $\det(A) = 0$ is equivalent to $A$ not being invertible - if $A$ takes a set with positive volume and sends it to a set with zero volume, then $A$ has some direction along which it "flattens" points, which would precisely be the null space of $A$. Unfortunately, I'm under the impression that this interpretation is at least semi-modern, but I think this is one of the cases where the modern viewpoint might be better to teach new students than the old viewpoint.
The old viewpoint is that the determinant is simply the result of trying to solve the linear system $Ax = b$ when $A$ is square. This is most likely how the determinant was first discovered. To derive the determinant this way, write down the generic matrix and then proceed by Gaussian elimination. This means you have to choose nonzero leading entries in each row (the pivots) and use them to eliminate subsequent entries below. Each time you eliminate the rows, you have to multiply by a common denominator, so after you do this $n$ times, you'll end up with the sum of all the permutations of entries from different rows and columns merely by virtue of having multiplied out to get common denominators. The $(-1)^k$ sign flip comes from the fact that at each stage in Gaussian elimination, you're subtracting. So on the first step you're subtracting, but on the second step you're subtracting a subtraction, and so forth. At the very end, by Gaussian elimination, you'll obtain an echelon form (upper triangular), and one knows that if any of the diagonal entries are zero, then the system is not uniquely solvable; the last diagonal entry will precisely be the determinant times the product of the values of previously used pivots (up to a sign, perhaps). Since the pivots chosen are always nonzero, then it will not affect whether or not the last entry is zero, and so you can divide them out.
EDIT: It isn't as simple as I thought, though it will work out if you keep track of what nonzero values you multiply your rows by in Gaussian elimination. My apologies if I mislead anyone.
Linear transformations, if you mean linear applications, are fundamental in linear algebra. Actually, pretty much all the theorems in linear algebra can be formulated in terms of linear applications properties. Moreover, linear applications are morphisms which preserve the vector space structure and linear algebra is the study of vector spaces and for a big part the study of their endomorphisms. Endomorphisms are applications which are linear and associate vectors from one vector space to vectors in the same vector space. In general, every (good) algebra course talking about a certain structure (it could be groups, rings, fields, modules, linear representations, categories...) always start by defining the structure and its axioms, then defining sub-structures, and then morphisms that preserve that structure. In finite dimension, vector spaces are convenient because their scalars are elements of a field and they [the vector spaces] have a base, i.e. a family of vectors that are linearly free and generate any other vector. This property allows to represent endomorphisms as a table that gives you how you transform the vectors of that base into vectors of another base (this is theorem actually). Having this information is enough because you can reconstruct any other vector's image by linear combination and the properties of linearity of the endomorphism. Matrices thus definitely come after linear transformations as they are only a representation of them up to the choice of a base for the vector spaces. For linear applications that are from one vector space to another of different dimension (if it's the same dimension, the two vector spaces are isomorphic and you have an automorphism), the matrix is rectangular because the two bases don't have the same cardinality (i.e. not the same dimension).
Best Answer
I don't know that this answers your question in terms of addressing "real-life" applications of the determinant, but I think it addresses your title question:
Historically:
From what I understand, the calculation of determinants didn't make use of matrices; it was applied to systems of linear equations and treated as a property which measures the existence of unique solutions, providing, if you will, a "litmus test" for determining whether unique solutions exist. Only later did determinants become associated with matrices.
The origin of the concept and use of the determinant dates back, I believe, to the $3^{rd}$ century, when Chinese mathematicians used the determinants in their book The Nine Chapters on the Mathematical Art. (A summary of the text, in Engish, can be found here. See, in particular, the synopsis of Chapter 8.)
Geometrically:
It helps many students to "visualize" the determinant of a $2 \times 2$ matrix geometrically, to better understand the determinant, as representing the area of a parallelogram. (This can be found in many standard books.):
So you can clearly see that if $ad-bc=0$ then the parallelogram suddenly becomes a straight line (having no area).
For a $3\times 3$ matrix, we have that the value of the determinant is equal to the volume of a parallelepiped whose coordinates correspond to the matrix's entries. When such a determinant is zero the parallelepiped suffers a decrement by one dimension by one (hence, is without volume).
Resources that may be helpful:
Apart from reading about determinants in Wikipedia and pursuing any promising links there, you may want to read a very good article Making Determinants Less Weird by John Duggan.
Also see Determinants and Linear Transformations - a helpful website for tying together the concepts of determinants and linear transformations. At the bottom of this linked webpage, there are additional links relating to the determinant, e.g., following one such link elaborates on the relationship between determinants and area and volume.
Follow-up: I couldn't resist revisiting this post to include the link to an earlier (semi-related) math.se post and the answers provided there, particularly the answer provided by @I.J.Kennedy. There, you'll find a "Proof without words: A $2 \times 2$ determinant is the area of a parallelogram," by Solomon W. Golomb.