Where does the method for finding the inverse of a 3x3 matrix come from? The method seems quite arbitrary, creating a matrix of minors, swapping signs, transposing, etc. I would be very grateful if someone could explain how this process was derived. Thanks!
 A: Suppose we have a $n\times m$ matrix $A$ with nonzero determinant $det(A) \neq 0$. $B$ is an inverse of $A$ if the following relation is verified:
$$ AB = I$$
By the formula for matrix multiplication we have:
$$ \sum_{k=1}^{m} a_{ik} b_{kj} = \delta_{i,j}$$
Where $\delta_{i,j}$ is Kronecker's delta ($\delta_{i,j} = 1$ when $i=j$ and $\delta_{i,j} = 0$ whenever $i\neq j$).
To get an explicit formula for $B$, let's fix $\;i=1\;$ and write the following system of equations:
$$ a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} + ... +a_{1m}b_{m1} = 1$$
$$ a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} + ... +a_{2m}b_{m1} = 0$$
$$...$$
$$ a_{n1}b_{11} + a_{n2}b_{21} + a_{n3}b_{31} + ... +a_{nm}b_{m1} = 0$$
Note that this is a system with variables $b_{11}, b_{21}, ..., b_{m1}$. We can then use Cramer's rule to get explicit formulas for the $b$'s. We'll calculate $b_{11}$ explicitly to get a feel of what's going on:
$$ b_{11} = \frac{det\left(
\begin{matrix}
1 & a_{12} & a_{13} & ... & a_{1m}\\
0 & a_{22} & a_{23} & ... & a_{2m}\\
...\\
0 & a_{n2} & a_{n3} & ... & a_{nm}
\end{matrix}
\right)}{det\left(
\begin{matrix}
a_{11} & a_{12} & a_{13} & ... & a_{1m}\\
a_{21} & a_{22} & a_{23} & ... & a_{2m}\\
...\\
a_{n1} & a_{n2} & a_{n3} & ... & a_{nm}
\end{matrix}
\right)}$$
If we expand the determinant of the matrix in the numerator with respect to the first column, we get the equivalent expression:
$$ b_{11}= \frac{det(M_{11})}{det(A)}$$
Where $M_{11}$ is the minor obtained deleting the first row and first column. Any other case can be brought back to this form by exchanging rows and columns. To see this, note that we can take any other matrix:
$$ \left(\begin{matrix}
a_{11} & a_{12} & a_{13} & ...& 0 &... & a_{1m}\\
a_{21} & a_{22} & a_{23} & ...& 0 &... & a_{2m}\\
...\\
a_{21} & a_{22} & a_{23} & ...& 1 &... & a_{2m}\\
...\\
a_{n1} & a_{n2} & a_{n3} & ...& 0 &... & a_{nm}
\end{matrix}\right)$$
To the form above by exchanging $i-1$ rows and $j-1$ columns. The cofactor of the $M_{ij}$ minor that is obtained by deleting the $i-th$ row and the $j-th$ colum will then just be $m_{ij} = (-1)^{(i-1) + (j-1)}$. We then get the general expression for the coefficients of the inverse matrix:
$$ b_{ij} = \frac{m_{ij} M_{ij}}{det(A)} = \frac{(-1)^{(i-1) + (j-1)} M_{ij}}{det(A)} $$
