Formulation of the matrix determinant How was the idea (and the equation) for the determinant of a square matrix formulated, and why does it work?
All I've learned is that the determinant of a matrix is 0 when some row is a linear combination of other rows. However, how does that complicated equation of the determinant do this?
If you guys want to go above and beyond for me, how come the area of a triangle expressed in cartesian coordinates and barycentric coordinates is the determinant of such a nice matrix?
 A: I don't know who thought it up, or how they arrived at it, but your statement about rows actually implies a bunch of things, like "if you swap two rows, the determinant negates" and "if two rows are the same, the det is zero" and "if you add a multiple of one row to another, the determinant doesn't change." (And the same things apply to columns, but that's not important right now). 
In fact, if you think of a matrix whose rows are all fixed except the first, then the determinant can be thought of as a function of that first row...and the statements above turn out to show that it's a LINEAR function of that first row. The same goes for any row. 
In short, the determinant is an "n-linear" (linear in each column) function of the $n$ row-vectors, and it's "alternating" (swapping rows negates the answer). 
If you have two alternating n-linear functions $f$ and $g$, then their sum is alternating n-linear, and so is any constant multiple of $f$, so these alternating n-linear functions actually form a vector space. With a bit more work...you can show that this vector space has dimension 1: they're all multiples of one specific function. If you choose, as this specific function , the one that gives the value "1" on the standard basis vectors, you get the a function that we call the determinant. 
It turns out -- the proof's not TOO awful -- that the "sum of products of matrix entries, one from each row and each column", with a carefully chosen sign" formula happens to have the right properties, so it must ALSO be the determinant function. Hence...you've proved the formula matches the thing. 
I know this is a lot of "deferring to stuff I'm not gonna write out," but I wanted you to get the idea of the proof: show something's one-dimensional, use that to show that a certain formula has to match something you define a completely different way, and you're done. 
A: The determinant of a matrix is least attractive when seen in equation form. Look up how to compute determinants via row column expansion, and also the method of dodgeson condensation for a more intuitive look into calculating a determinant.
A: Determinants were originally connected with systems of linear equations. Given $n$ equations in $n$ unknowns how could we 'determine' the solution, or determine that there will be no solution? Leibniz explored the cases $n=2,3$ in some unpublished manuscripts. In this way he stumbled on a summation form of what we now call the determinant. I don't know if he generalized this for any $n$ or if it was only for $n=2,3.$ 
