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There are many ways to define and interpret determinants. The one I'm more interested right now is the one that better describes its name: a number that can determinate if a system of linear equations has a solution or not. If we start with a system:

\begin{cases} ax + by = n \\ cx + dy = h \end{cases}

And, with Gauss elimination, we have, if I remember right

$$x = \frac{nd-hb}{ad-bc}\\y = \frac{ha-nc}{ad-bc}$$

We basically found the solution to $x$ and $y$, which will always work as long as $$ad-bc \ne 0$$ for obvious reasons.

I know that the determinant describes this behavior for system of $n$ linear equations, but I've never seen a general proof of it. Also, the proof I've seen for the $2\times2$ case supposes no term is $0$, but at some point, there is division by $a$, or $b$, or $c$, or $d$. Sal does this in this video.

So, I've seen a lot of books teaching how to compute the determinant of a $n\times n$ matrix, but I've never seen a proof of why that mechanism works. I think this may be a long of complicated proof, but I wanted to know who first developed it, and I wanted to learn it. Could somebody help me? Is there a book about it?

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We need a little linear algebra here: a system like the one you wrote can be expressed in matrix form:

$$\begin{cases} ax + by = n \\ cx + dy = h \end{cases}\;\iff\begin{pmatrix}a&b\\c&d\end{pmatrix}\binom xy=\binom nh $$

The above coefficients matrix, let's call it $\;A\;$ , can be seen as a linear map (in fact, a linear operator), and supposing we're working on the reals: $\; A:\Bbb R^2\to\Bbb R^2\;$ .

From linear algebra we know the map $\;A\;$ is an isomorphism iff it is injective iff its determinant is non zero . From here that

== If $\;\binom nh=\binom 00\;$ , the system has only the trivial solution iff $\;\det A\neq 0\;$ , since then $\;A\;$ is injective and thus $\;Av=0\iff v=0\;$ . Otherwise, i.e. $\;\det A=0\;$, it has non-trivial solutions: exists $\;0\neq v\;\;s.t.\;\;Av=0\;$

== If $\;\binom nh\neq\binom 00\;$, then the system has a unique solution iff $\;\det A\neq 0\;$, otherwise it either has none or several.

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The Wikipedia article on Cramer's rule has a proof.

Cramer's rule says that if a system of $n$ linear equations has a solution, the solution is a quotient of two order-$n$ determinants, and that the denominators in the quotients are the quantity you are asking about.

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The way I learned it in abstract algebra class was that $det(r_1,r_2,\ldots,r_n)$ is an alternating multilinear function of the rows of the matrix (which means that if two arguments of the function are the same, then the result is 0). This tells you that the determinant is invariant under row operations, because if $a$ is a real number, then for example $det(r_1+ar_2,r_2,\ldots,r_n)=det(r_1,r_2,\ldots,r_n)+det(ar_2,r_2,r_3,\ldots,r_n)=det(r_1,r_2,\ldots,r_n)+adet(r_2,r_2,r_3,\ldots,r_n)=det(r_1,r_2,\ldots,r_n)+0=det(r_1,r_2,\ldots,r_n)$.

Since it's multilinear, the property of it being nonzero is preserved by multiplying a row by a nonzero scalar. The matrix is invertible if and only if these operations can transform the matrix into the identity matrix, which has determinant $1$.

Conversely, if the determinant is nonzero that means no linear combination of the rows can be used to express a different row, because otherwise the alternating property would make the function zero, which means the vectors are linearly independent, which is equivalent to the matrix being invertible.

We showed in class that a nontrivial alternating multilinear function exists and is equal to the determinant up to a scalar. We used Artin's textbook.

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  • $\begingroup$ The problem is that this book starts defining determinants, rather than discovering them as a way to discover if the system ahs an unique solution. I need to know how to find it. $\endgroup$ – Guerlando OCs Nov 2 '14 at 3:38

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