shortcut for finding a inverse  of matrix I need tricks or shortcuts to find the inverse of  $2 \times 2$ and $3 \times 3$ matrices. I have to take a time-based exam, in which I have to find the inverse of square matrices.
 A: So we want to find out a way to compute $2 \times 2 ~\text{ or }~ 3 \times 3$ matrix systems the most efficient way. Well I think the route that we want to go would be to use Cramer's Rule for the $2 \times 2 \text{ or } 3 \times 3$ case. To state the $2 \times 2$ case we will use the following:
For some coefficient matrix A=
$\left[ \begin{array}{rr} 
a & b \\
c & d 
\end{array} \right]$
$A^{-1}=\dfrac{1}{ad-bc} \cdot 

\left[ \begin{array}{rr} 
d & -b \\
-c & a 
\end{array} \right]~ \iff ad-bc \ne 0$  $~~~~~~~~~\Big($i.e., Det(A)$~\ne ~ 0\Big)$
For the $3 \times 3$ case, we will denote that as the folllowing:
$x_{1} = \dfrac{|b~~x_{2}~~x_{3}|}{|\bf{A}|}~~,$
$x_{2} = \dfrac{|x_{1}~~b~~x_{3}|}{|\bf{A}|}~~,$
$x_{3} = \dfrac{|x_{1}~~x_{2}~~b|}{|\bf{A}|}.$
This comes from the matrix equation: ${\bf{A\vec{x}}}={\bf{\vec{b}}},~~~$ where $\vec{x}=[x_{1}~~x_{2}~~x_{3}]^{T}$.
For the elements of matrix $A =  
\left|\begin{array}{rrr} 
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array} \right|,~~$
it can be extended for the solutions $x_{1},~x_{2},~x_{3}$ 
as so knowing that ${\bf|{A}| =} ~ |a_{ij}| ~ \not= ~ 0.$
$x_{1} = \dfrac{1}{|{\bf{A}}|}
\left|\begin{array}{rrr} 
b_1 & a_{12} & a_{13} \\
b_2 & a_{22} & a_{23} \\
b_3 & a_{32} & a_{33}
\end{array} \right|$,
$x_{2} = \dfrac{1}{|{\bf{A}}|}
\left|\begin{array}{rrr} 
a_{11} & b_1 & a_{13} \\
a_{21} & b_2 & a_{23} \\
a_{31} & b_3 & a_{33}
\end{array} \right|$,
$x_{3} = \dfrac{1}{|{\bf{A}}|}
\left|\begin{array}{rrr} 
a_{11} & a_{12} & b_1 \\
a_{21} & a_{22} & b_2 \\
a_{31} & a_{32} & b_3
\end{array} \right|$.
An alternate way of doing this would be using row reducing methods, known as either Gaussian Elimination( ref ) or Gauss-Jordan Elimination( rref ). 
I hope this helped out. Let me know if there if anything you do not understand.
Thanks.
Good Luck.
A: For a 2x2 matrix, the inverse is:
$$
\left(\begin{array}{cc}
a&b\\
c&d
\end{array}\right)^{-1}
=
{1 \over a d - b c}
\left(\begin{array}{rr}
d&-b\\
-c&a
\end{array}\right)~,~~\text{ where } ad-bc \ne 0.
$$
just swap the 'a' and 'd', negate the 'b' and 'c', then divide all by the determinant $a d - b c$.
That's really the most straightforward 'trick', just memorize that pattern.
For 3x3, it's lot more complicated but there is a pattern. As usual compute the determinant first (kind of a pain; but surely you already know the pattern to compute that quickly).
$$
\left(\begin{array}{ccc}
a&b&c\\
d&e&f\\
g&h&i
\end{array}\right)^{-1}
=
{1 \over {\rm{det}}}
\left(\begin{array}{rrr}
e i - f h&-(b i - c h)&b f - c e\\
-(d i - f g)&a i - c g&-(a f -c d)\\
d h - e g&-(a h - b g)&a e - b d
\end{array}\right).
$$
The pattern is that each entry is


*

*the determinant of the submatrix gotten by removing that row and column. I.e. for row 2 column 3 (from $f$'s position), the determinant is $a h - b g$:
$$
\det\left(\begin{array}{cc}
a&b\\
g&h
\end{array}\right)
= a h - b g
$$

*then multiply in the checkerboard pattern. (i.e.1x1 is positive, 1x2 is negative... mathematically it's multiply by $(-1)^r(-1)^c$.

*Then transpose.


See? There's a pattern, but I feel it's about the same symbolic complexity as just doing it brute force Gaussian-elimination style.
A: Your best bet is the Gauss-Jordan method.
