TITLE: Demystifying the Determinant
QUESTION [12 upvotes]: Preface: I’ve read a few of the posts on here about the determinant but none seem to put to rest the the questions I have, so I’ve decided to ask outright.
Question: In general, I would like to just have a deeper understanding of where the determinant comes from. It seems so fundamental to almost any field of mathematics. So here are my questions about it:

Who discovered the formula for the determinant? Was it noticed or was it built for a specific purpose? If so what was that motivation?
Why/how does it relate to consistent/inconsistent systems?
How does it relate to cross product? Why is it that two vectors aligned in a certain way produces a third orthogonal vector and how does the determinant make this happen?
Why is the determinant used in finding  eigenvalues and eigenvectors?

Any details would be very much appreciated!

REPLY [2 votes]: The beautiful interpretation of the determinant as a signed volume gets so much attention that sometimes I worry new students won't realize that the determinant can easily be discovered just by solving $Ax = b$ by hand, using high school algebra, in the case where $A$ is a $2 \times 2$ or $3 
\times 3$ matrix. For example, if $A$ is $2 \times 2$, we want to solve
\begin{align}
a_{11} x_1 + a_{12} x_2 &= b_1 \\
a_{21} x_1 + a_{22} x_2 &= b_2.
\end{align}
Multiplying both sides of the the first equation by $a_{21}$, and then multiplying both sides of the second equation by $a_{11}$ and subtracting, we find that
$$
(a_{11} a_{22} - a_{21} a_{12}) x_2 = a_{11} b_2 - a_{21} b_1.
$$
If the number $a_{11} a_{22} - a_{21} a_{12}$ is not zero, then we can solve for $x_2$. (And we find a similar formula for $x_1$.)
So immediately, we have discovered the determinant of a $2 \times 2$ matrix, and we see that if this special number is not zero then $Ax = b$ has a unique solution for any choice of $b$. The case where $A$ is $3 \times 3$ is not much harder. Someone who is sufficiently obsessed, as many mathematicians are, would not hesitate to work out the $4 \times 4$ case by hand if the pattern is not already clear. So, there's nothing difficult about discovering the determinant. (By the way, we have also discovered Cramer's rule in the process.)

If $\lambda$ is an eigenvalue of $A$, this means that there exists some nonzero vector $x$ such that $Ax = \lambda x$. Equivalently,
$$
\tag{1} (A - \lambda I) x = 0,
$$
where $I$ is the identity matrix.
But, as we have discovered above, if $\det (A - \lambda I) \neq 0$, then (1) has a unique solution, which is $x = 0$. So the fact that there is a connection between eigenvalues and the determinant is immediate. You might conjecture that if $\det (A - \lambda I) = 0$ then (1) has a nonzero solution, and it turns out to be possible to show that this is correct. So, $\lambda$ is an eigenvalue of $A$ if and only if $\det (A - \lambda I) = 0$.