Why is the det(A) computed as it is? I'm looking at how to compute the determinant of a matrix A, and it is mentioned that for a $2\times2$ matrix with elements [a b][c d] , it is given by ad - bc. Now about the $3\times3$ case, I understood how you compute it. But why are the signs - +? And why does this become - + - for a $4\times4$ case, and - + - + for the $5\times5$ case, and etc? Specifically, I'm wondering what is the basis for the signs in computing the det(A).
 A: The rules the OP posted are practical rules in the sense that these are the rules to explicitly calculate the determinant of $A$. To understand the signs, however, it might be convenient to use the definition of determinant of a square matrix $A=(a_{ij})$. Here $A$ is assumed to be a $n\times n$ matrix. This is more elaborate formula, and it reads:
\begin{eqnarray}
\det A = \sum_{\sigma \in S_{n}}\epsilon_{\sigma}\prod_{i=1}^{n}a_{i,\sigma(i)}\tag{1}\label{1}
\end{eqnarray}
You can find a more extensive discussion here but let me explain what are these factors and why it explains the signs on the OP question. Because you are dealing with $n\times n$ matrices, consider the set $I_{n}:=\{1,...,n\}$. A permutation of this set is simply a bijective function $\sigma: I_{n}\to I_{n}$. The set of all such permutations is denoted by $S_{n}$ (if you know what a group is, $S_{n}$ is a group with composition as its product). Thus, the first sum on (\ref{1}) is a sum over all permutations of $I_{n}$ or, alternatively, a sum over all permutations $\sigma \in S_{n}$.
Now, take $a\neq b$ with $a,b \in I_{n}$ and let $\tau$ be a permutation defined as follows: $\tau(a) = b$, $\tau(b) =a$ and $\tau(k) = k$ if $k \not\in \{a,b\}$. Intuitively, this is a permutation that only exchanges $a$ and $b$. Such a permutation (i.e. a permutation that exchanges only two different elements of $I_{n}$) is called a transposition. It is a known result that every permutation $\sigma$ has a decomposition in transpositions, that is, given $\sigma \in S_{n}$ we can also write $\sigma$ as a product (actually composites of) transpositions $\sigma = \tau_{1}\cdots\tau_{k}$. Is this decomposition unique? The answer is no; however, the parity of $k$ is. What this means is that if a permutation $\sigma$ can be written as a product of an even (respectivelly odd) number of transpositions, then every decomposition of $\sigma$ into a product of transpositions contains an even (respectivelly odd) number of transpositions. Thus, we can characterize every permutation by its parity: permutations with $k$ even are called even and are associated to a number $1$; permutations with $k$ odd are called odd and are associated with a number $-1$. Generically, we write $\epsilon_{\sigma}$ to represent either $1$ or $-1$, according to the $\sigma$. Because $\epsilon_{\sigma}= \pm $, we call this number the sign of the permutation $\sigma$.
Now, to finally answer your question, note that together with $\sum_{\sigma \in S_{n}}$, (\ref{1}) has a weight $\epsilon_{\sigma}$ which can take values $\epsilon_{\sigma} = \pm$, according to what permutation $\sigma$ is being considered. This is the reason why the determinant formula has a bunch of plus and minus signs; this comes from its definition. However, because (\ref{1}) is usually very difficult to handle, it is more convenient to evaluate determinants using other (more practical, convenient) rules.
Note: I'm sorry if the answer was too long and too detailed, but this is the best way to understand these ideas in my opinion.
A: An $(n\times n)$-matrix $[A]$ numerically describes an affine transformation $A:\>{\mathbb R}^n\to{\mathbb R}^n$. Such a transformation maps any  set $S\subset{\mathbb R}^n$ to some image set $A(S)\subset{\mathbb R}^n$. It is a basic geometric fact  that the volumes of  all sets $S\subset {\mathbb R}^n$ are multiplied by the same factor $\lambda\geq 0$:
$${\rm vol}\bigl(A(S)\bigr)=\lambda \>{\rm vol(S)}\ .$$
This factor $\lambda$ depends on the map $A$, hence on the given matrix $[A]$.
It follows that there is a function $\lambda:\>A\mapsto \lambda(A)$ that assigns to each $(n\times n)$-matrix $[A]$ its "volume scaling factor". Thinking a long time about this problem one finds that  there is a function $${\rm det}:\quad{\mathbb R}^{n\times n}\to{\mathbb R},\qquad [A]\mapsto {\rm det}\bigl[A]$$
that does the trick. This function is polynomial in the variables $a_{ik}$ forming the elements of the matrix $[A]$, with lots of symmetries, and sign rules. One then has
$$\lambda(A)=\bigl|\,{\rm det}[A]\,\bigr|\ .$$
There is no one line explanation of this result. One has to go through the proof given in any linear algebra textbook.
