Easiest Method to Evaluate $3\times 3$ Determinants

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.