How to find the determinant of a 3×3 matrix

[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 textbooks but it only has examples of how to do it useing a grapgics calculator thanks

 

[Brad Barker]

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

you break it up into three 2x2 determinents!

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

look at the first line of eqt. 27

 

[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':

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

 

[Spunky_Dunky]

form Spunky_Dunkey

thanks very much

 

[Brad Barker]

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':

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

oh, right. that crap.




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


whatever's easiest to you!

 

[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

 

[HallsofIvy]

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

 

[Brad Barker]

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

that's a really good idea. would have really come in handy when i was in 11th grade. (we had the occasional 4x4 determinant! )