Matrix inverses - Why are they derived the way they are? Note that this is not a question of how, but why. I know the mechanics of it, but this is the first thing i've come across that truly seems like magic, rather than a rigorous mathematical process.
There are questions on SE about proofs for the inverse, but niether here nor anywhere else on the internet, can I find a decent explanation of why it works the way it does:
So, basically, suppose the matrix is $\begin{bmatrix} a && b \\ c && d \end{bmatrix}$
Why do we swap values $a$ and $d$?
Why do $c$ and $b$ become negative?.
Why do we divide $a$, $b$, $c$ and $d$ by $ad-bc$?
Once again, this truly seems like magic. It's like ''hey, swap these fellers around, twiddle these signs a little, and poof, you're inverse!''.
Thanks in advance for any help :)
 A: The method you are talking about is finding the inverse of a matrix A by computing the adjugate matrix :
https://en.wikipedia.org/wiki/Adjugate_matrix
This technique is very inefficient in practice for "big" matrix (4x4 or bigger), so it's more a theorical tool. In practice, you better use the gaussian elimination :
https://en.wikipedia.org/wiki/Gaussian_elimination#Finding_the_inverse_of_a_matrix
A: First of all, you can check that
\begin{equation}
\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\begin{bmatrix} a & b \\ c & d \end{bmatrix} =
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},
\end{equation}
which confirms that the matrix on the left is the inverse of
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} $ (when $ad - bc \neq 0$).  The only question is how you would discover this formula for the inverse.
You could discover this formula just by guessing.  If you try the simplest way to put zeros in the right positions, you will be on the right track.  Another way to discover it might be to solve the system
\begin{equation}
\begin{bmatrix} a & b \\ c & d \end{bmatrix} 
\begin{bmatrix} x & y \\ z & w \end{bmatrix} 
=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\end{equation}
for the unknowns $x,y,z,w$.  We have four equations and four unknowns and you could solve it by hand.
A: You can derive it by simultaneous transforming $(A|I)\to(I|A^{-1})$. If $a \ne 0$ and $ad-bc\ne 0$ we get: 
$$
\left[
\begin{array}{rr|rr}
a  & b & 1 & 0 \\
c & d & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1  & b/a & 1/a & 0 \\
c & d & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1  & b/a & 1/a & 0 \\
0 & d - cb/a & -c/a & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1  & b/a & 1/a & 0 \\
0 & (ad - bc)/a & -c/a & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1  & b/a & 1/a & 0 \\
0 & 1 & -c/(ad-bc) & a/(ad-bc)
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1  & 0 & 1/a + bc/a(ad-bc) & -b/(ad-bc) \\
0 & 1 & -c/(ad-bc) & a/(ad-bc)
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1  & 0 & ((ad-bc)+bc)/a(ad-bc) & -b/(ad-bc) \\
0 & 1 & -c/(ad-bc) & a/(ad-bc)
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1  & 0 & d/(ad-bc) & -b/(ad-bc) \\
0 & 1 & -c/(ad-bc) & a/(ad-bc)
\end{array}
\right]
$$
If $a = 0$ and $ad - bc \ne 0$ we have $b\ne 0$ and $c\ne 0$ and it goes like this:
$$
\left[
\begin{array}{rr|rr}
0  & b & 1 & 0 \\
c & d & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
c & d & 0 & 1 \\
0 & b & 1 & 0
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & d/c & 0 & 1/c \\
0 & 1 & 1/b & 0
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & 0 & -d/cb & 1/c \\
0 & 1 & 1/b & 0
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & 0 & d/(-bc) & -b/(-bc) \\
0 & 1 & -c/(-bc) & 0/(-bc)
\end{array}
\right]
\to
\left[
\begin{array}{rr|rr}
1 & 0 & d/(ad-bc) & -b/(ad-bc) \\
0 & 1 & -c/(ad-bc) & 0/(ad-bc)
\end{array}
\right]
$$
A: Given that the only thin you knwo about $A$ is that $D:=a d-b c\ne 0$ and you want to find $a',b',c',d'$ auch that (among others)  $aa'+bc'=1$, trying $a'=\frac dD$ and $c'=-\frac cD$ might seem quite natural ...
