# Det of a matrix

1. Aug 9, 2005

### Spunky_Dunky

could someone please explain simply how to get the determinate of a 3 * 3 matrix i'm relly stuck i've looked through my text books but it only has examples of how to do it useing a grapgics calculator thanks

2. Aug 9, 2005

you break it up into three 2x2 determinents!

http://mathworld.wolfram.com/Determinant.html

look at the first line of eqt. 27

3. Aug 9, 2005

### TD

You could do that, or use some properties first to create 0's and then develop to a row or column. There's also a direct way, but it's a bit 'long':

$$\begin{gathered} A = \left( {\begin{array}{*{20}c} {a_{11} } & {a_{12} } & {a_{13} } \\ {a_{21} } & {a_{22} } & {a_{23} } \\ {a_{31} } & {a_{32} } & {a_{33} } \\ \end{array} } \right) \Rightarrow \det \left( A \right) = \left| {\begin{array}{*{20}c} {a_{11} } & {a_{12} } & {a_{13} } \\ {a_{21} } & {a_{22} } & {a_{23} } \\ {a_{31} } & {a_{32} } & {a_{33} } \\ \end{array} } \right| \hfill \\ \\ = a_{1,1}\cdot{a}_{2,2}\cdot{a}_{3,3} + a_{1,3}\cdot{a}_{3,2}\cdot{a}_{2,1} + a_{1,2}\cdot{a}_{2,3}\cdot{a}_{3,1} - a_{1,3}\cdot{a}_{2,2}\cdot{a}_{3,1} - a_{1,1}\cdot{a}_{2,3}\cdot{a}_{3,2} - a_{1,2}\cdot{a}_{2,1}\cdot{a}_{3,3} \hfill \\ \end{gathered}$$

4. Aug 9, 2005

### Spunky_Dunky

form Spunky_Dunkey

thanks very much

5. Aug 9, 2005

oh, right. that crap.

:tongue:

my calc III prof went over that, mainly as a curiosity. i've used expansion by minors exclusively.

whatever's easiest to you!

6. Aug 10, 2005

### TD

I usually expand by minors too, but not before I simplified it first using elementary operations. Having to expand it 'in full' is long too hehe

7. Aug 10, 2005

### HallsofIvy

Staff Emeritus
If you row-reduce the matrix to triangular form, finding the determinant is just multiplying the numbers on the diagonal.

8. Aug 10, 2005