What is a determinant?
The determinant of a square matrix A, written det(A) or |A| is a number associated with a matrix. For the 2x2 and 3x3 case, the geometric meaning is the area/volume between the vectors. If they point in about the same (or opposite) direction, then their determinant is smaller than if vectors of the same length pointed more like right angles (See Orthogonal).
See related article Properties of the Determinant
2x2 case
The 2x2 case is easy to do. If one has the following matrix:
$ \begin{bmatrix} a & b\\ c & d\end{bmatrix} $
simply compute $ a*d-c*d $ For example, in the following matrix:
$ \begin{bmatrix} 3 & 1\\ 4 & 5\end{bmatrix} $
the determinant is $ 3*5-4*1=15-4=11 $.
General case
If you have a larger matrix (n by n matrix), then here is the process. (Also see the example, as this is a detailed process)
- 1) Choose a row or column (preferably with many zeroes, or at least with small numbers)
- 2) Then write the matrix n times. In the first copy, circle the first number in the row/column. In the second copy, circle the second number in the row/column, and so on. Write this underneath each matrix.
- 3) In each individual copy, cross out the numbers in the same row or column as the circled number (including the circled number itself).
- 4) The numbers left in each copy should form a n-1 by n-1 matrix. Figure out the determinant in each individual n-1 by n-1 matrix (use this same process), and multiply each one by the number below the matrix. (The circled numbers from step 2)
- 5) Add signs to these numbers by alternating plus/minus/plus/minus. (Start with plus) Then add these up to get the determinant. If you used an even row/column in step 1 (second row/column, fourth row/column...), change the sign on this result to get the right sign.
General case example
For example, here is how to use that method to compute the determinant of the following matrix:
$ \begin{bmatrix} 1 & 2 & 0 & 2\\ 0 & 3 & 2 & 0\\ 1 & 2 & 2 & 3\\ 2 & 3 & 3 & 0\end{bmatrix} $
For step 1, choose the second row, as it has the most zeros. (One could have easily also chose the fourth column.)
$ \begin{bmatrix} 1 & 2 & 0 & 2\\ \color{red}0 & \color{red}3 & \color{red}2 & \color{red}0\\ 1 & 2 & 2 & 3\\ 2 & 3 & 3 & 0\end{bmatrix} $
For step 2, here are 4 copies. For now, all 4 copies are listed, but in practice you don't need to list copies where a 0 will be circled, as they will just be multiplied by 0 in the end. (See step 4 to see this.) The circled number is listed in blue.
$ \begin{bmatrix} 1 & 2 & 0 & 2\\ \color{blue}0 & \color{red}3 & \color{red}2 & \color{red}0\\ 1 & 2 & 2 & 3\\ 2 & 3 & 3 & 0\end{bmatrix} \begin{bmatrix} 1 & 2 & 0 & 2\\ \color{red}0 & \color{blue}3 & \color{red}2 & \color{red}0\\ 1 & 2 & 2 & 3\\ 2 & 3 & 3 & 0\end{bmatrix} \begin{bmatrix} 1 & 2 & 0 & 2\\ \color{red}0 & \color{red}3 & \color{blue}2 & \color{red}0\\ 1 & 2 & 2 & 3\\ 2 & 3 & 3 & 0\end{bmatrix} \begin{bmatrix} 1 & 2 & 0 & 2\\ \color{red}0 & \color{red}3 & \color{red}2 & \color{blue}0\\ 1 & 2 & 2 & 3\\ 2 & 3 & 3 & 0\end{bmatrix} $ $ \begin{bmatrix} \color{white}{==} & 0 & \color{white}{=====} & 3 & \color{white}{====} & 2 & \color{white}{=====} & 0 & \color{white}{==} \end{bmatrix} $
Step 3: Create smaller matrices by crossing out the row and column with the circled (blue) number.
$ \begin{bmatrix} X & 2 & 0 & 2\\ \color{blue}X & \color{red}X & \color{red}X & \color{red}X\\ X & 2 & 2 & 3\\ X & 3 & 3 & 0\end{bmatrix} \begin{bmatrix} 1 & X & 0 & 2\\ \color{red}X & \color{blue}X & \color{red}X & \color{red}X\\ 1 & X & 2 & 3\\ 2 & X & 3 & 0\end{bmatrix} \begin{bmatrix} 1 & 2 & X & 2\\ \color{red}X & \color{red}X & \color{blue}X & \color{red}X\\ 1 & 2 & X & 3\\ 2 & 3 & X & 0\end{bmatrix} \begin{bmatrix} 1 & 2 & 0 & X\\ \color{red}X & \color{red}X & \color{red}X & \color{blue}X\\ 1 & 2 & 2 & X\\ 2 & 3 & 3 & X\end{bmatrix} $ $ \begin{bmatrix} \color{white}{===} & 0 & \color{white}{=======} & 3 & \color{white}{======} & 2 & \color{white}{=======} & 0 & \color{white}{===} \end{bmatrix} $
Now to continue, we need to find the determinant of the 4 smaller 3 by 3 matrices. We can use the special rule for 3 by 3 matrices (Sarrus' Rule) to find these products, or just use the same method on each of these matrices. (In this case, we use Sarrus' Rule since it's easier.)
0+0+12-18-0-12 =-18 and -18*0=0 0+0+6-9-8-0=-11 and -11*3=-33 0+12+6-8-9-0=1 and 1*2=2 6+0+8-6-6-0=2 and 2*0=0
Step 5 Compute the answer by alternating plus and minus (including the zeros) (0)-(-33)+(2)-(0)=35. Since the row used was the second row, the sign changes, to make the final result -35.
