I know this sounds like such a vague question, but seeing as we all know the formula for the determinant of a general nxn matrix, I want to know exactly why we define it as such. I know that determinants are used to define whether or not a matrix has a unique solution, but surely there much be other methods and other equations that we can use to determine this. So again, why is it that we use the current equation and the current definition?

A simpler related question: do you know why we might be interested in the determinant as it is defined for the $2\times 2$ case? Do you know what $\det = 0$ tells us and why? – Derek Allums Nov 02 '12 at 14:09

2You might have a look at some of the answers here: [What's an intuitive way to think about the determinant?](http://math.stackexchange.com/questions/668/whatsanintuitivewaytothinkaboutthedeterminant). – Martin Sleziak Nov 02 '12 at 14:13

6If you want to interpret the determinant as $n$dimensional volume (with sign as orientation), then you get some basic properties of volume (which seem obvious from geometrical viewpoint). Determinant is uniquely determined by this basic properties (namely the properties that is is multilinear, $I=1$ and interchanging rows changes the sign). See, for example, this question: [The determinant function is the only one satisfying the conditions](http://math.stackexchange.com/questions/211481/thedeterminantfunctionistheonlyonesatisfyingtheconditions). – Martin Sleziak Nov 02 '12 at 14:17

Well I have some notion as to that. For a 2x2 case, the determinant is the area of the rhombus that is defined by the entries of the matrix. Thus, when the det=0, that means the rhombus is a straight line. From that, I assume that when it becomes a straight line, it means that there is one unique solution. And I asssume that this is true for the 3x3 matrix and, with a stretch of the imagination, also applicable to the nxn case, which is a shape in the nth dimension. So is that the answer? Or am I at least in the ball park? – Additional Pylons Nov 02 '12 at 15:09

2You have it backwards. It's when the determinant is *not* zero that the (system of homogeneous equations determined by the) matrix has a unique solution. – Gerry Myerson Nov 03 '12 at 05:52
1 Answers
Historically the determinant was introduced by Gerolamo Cardano as an expression to determine if a linear $2\times 2$ system has a solution. Later this was generalized by Leibniz for larger linear systems.
You are right if you ask, why we use this particular notion in order to determine if a linear system has a solution or not. It is possible to define a different function that is zero if and only if the matrix induces a linear system that has a solution (say you can multiply the determinant with a scalar, or you take the absolute value). However, the determinant has other nice geometrically properties, which are helpful in many different applications.
Here is just one example. A $n\times n$ matrix $T$ decodes a linear transformation $t \colon \mathbb{R}^n \to \mathbb{R}^n$. If you consider the volume of a $n$simplex and its image in $t$, then the relative change of the simplex volume is $\det(T)$. In particular, if $\det(T)<0$ then the orientation of the simplex in the image is flipped.
So it makes sense to define the determinant this way, because it comes in handy for many applications.
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