Easiest Method to Evaluate $3\times 3$ Determinants After a lot of practice, I developed a method of evaluating $3\times 3$ determinants 
which I call the
Cross - Left Fish - Right Fish.

The method goes like this,
for some $3 \times 3$ determinant $\left| A \right|$,
$$\left| \mathbf A \right| 
= \left| \begin{array}{ccc}
a & b & c\\
d & e & f\\
g & h & i\\
\end{array} \right| 
= \overbrace{\left( \color{blue}{aei} - \color{red}{ceg}\right)}^{\text{Cross}} 
+ \overbrace{\left( \color{blue}{dhc} - \color{red}{dbi} \right)}^{\text{Left Fish}}
+ \overbrace{\left( \color{blue}{fbg} - \color{red}{fha} \right)}^{\text{Right Fish}}  
$$
It's easy to remember,even the fish, because the fish consists of ticks.
In the Left Fish: The normal tick is positive. The upside down tick is negative.
In the Right Fish: The reverse tick is negative. The reverse upside down tick ispositive because two wrongs make a right. 
Now, someone had pointed out that they had heard of a similar method and that my idea isn't anything original. Please tell me. Is he right? Is there such a method? Is that method any easier than what I've made?
What is the easiest method to evaluate determinants?
 A: Rule of Sarrus
The Rule of Sarrus is an easy trick equivalent to the Fish/Cross method.

Write out the first 2 columns of the matrix to the right of the 3rd column, so that you have 5 columns in a row. Then add the products of the diagonals parallel to the leading diagonal (solid) and subtract the products of the other diagonals with positive slope (dashed).
Remember: Add $(-)$ve sloped lines, Subtract $(+)$ve sloped line.
A: You can take any determinant by looking at the matrix like this,
$$ \begin{bmatrix} a & -b & c \\ -d & e & -f \\ g & -h & i \end{bmatrix}.$$
Now moving down any column or row, strike out all the values across and through it. So if we follow the left column, the first step is.
$$ \begin{bmatrix} x & x & x \\ x & e & -f \\ x & -h & i \end{bmatrix}.$$
So we take the determinant of the matrix left over and multiply it by a, which is $ a \cdot \det \begin{bmatrix} e & -f \\ -h & i \end{bmatrix}.$
Next,
$$ \begin{bmatrix} x & -b & c \\ x & x & x \\ x & -h & i \end{bmatrix},$$
so we now get $ -d \cdot \det\begin{bmatrix} -b & c \\ -h & i \end{bmatrix},$
Then the next step would be
$$ \begin{bmatrix} x & -b & c \\ x & e & -f \\ x & x & x \end{bmatrix},$$
producing
$ g\cdot \det\begin{bmatrix} -b & c \\ e & -f \end{bmatrix}.$
Notice that this follows your method exactly, except you are doing it in a different order.
A: I always remember the determinant like this:
$$
\left| \begin{array}{ccc}
a & b & c\\
d & e & f\\
g & h & i\\
\end{array} \right|= a\left| \begin{array}{ccc}
e & f\\
h & i\\
\end{array} \right|-b \left| \begin{array}{ccc}
d & f\\
g & i\\
\end{array} \right|+c \left| \begin{array}{ccc}
d & e\\
g & h\\
\end{array} \right|$$
If you look closely, you will see a pattern, but it's difficult to describe it.
From here on, we only have to evaluate $2\times2$ determinants.
