Ways to calculate the inverse of a matrix, assuming it exists... I'm wondering - Other than by using row reduction on the augmented $[A|I]$ to get $[I|A^{-1}]$, and by reducing a matrix to a product of elementary matrices, is there any other way to determine what the inverse of a matrix is? I seem to recall something about using the adjoint of a matrix but I can't recall the details.
 A: The other way to calculate the inverse of a matrix $A$:
$A^{-1}=\dfrac{1}{detA}adjA$
This assumes $A$ is an invertible square matrix of course.
Another method is Cramer's Rule.
Cramer's Rule says the following:

If $A$ is an invertible $n \times n$ matrix, the solution to the system $Ax=b$ of $n$ equations in the variables $x_1,x_2,\ldots,x_n$ is given by
  $$x_1=\dfrac{detA_1}{detA}, x_2=\dfrac{detA_2}{detA},\ldots,x_n=\dfrac{detA_n}{detA}$$ where, for each $k$, $A_k$ is the matrix obtained from $A$ by replacing column $k$ by $b$.

A: Another way to calculate an inverse is by using the characteristic polynomial  of a matrix. That is, suppose $A$ matrix satisfies the polynomial equation:
$$
A^n + c_{n-1} A^{n-1} + \ldots + c_1 A + c_0 I = 0
$$
The coefficient $c_0$ must be nonzero if and only if $A$ is invertible. Then we can calculate the inverse of $A$ as a linear combination of the powers of $A$:
$$
A^{-1} = -\frac{1}{c_0}(A^{n-1} + c_{n-1}A^{n-2} + \ldots + c_2 A + c_1 I)
$$
A: A "low tech" approach is to simply solve the system of equations determined by $AX=I$.  For example, when $n=2$, we have
$$
\left(
\begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{matrix}
\right)
\left(
\begin{matrix}
x_{11} & x_{12} \\
x_{21} & x_{22} \\
\end{matrix}
\right)
=\left(
\begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix}
\right)$$
which is equivalent to the four equations
\begin{align}
a_{11}x_{11}+a_{12}x_{21} &=1 \\
a_{11}x_{12}+a_{12}x_{22} &=0 \\
a_{21}x_{11}+a_{22}x_{21} &=0 \\
a_{21}x_{12}+a_{22}x_{22} &=1. \\
\end{align}
Solve to find $X=A^{-1}$.
