Understanding Determinants of a 3x3 Matrix: A Beginner's Guide

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To find the determinant of a 3x3 matrix, one effective method is to break it down into three 2x2 determinants. Alternatively, properties can be used to create zeros in the matrix, simplifying the calculation. The determinant can also be calculated directly using a specific formula involving the elements of the matrix. Another approach is to row-reduce the matrix to triangular form, allowing the determinant to be found by multiplying the diagonal elements. Understanding these methods can greatly aid in mastering matrix determinants.
Spunky_Dunky
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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
 
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Spunky_Dunky said:
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
 
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} } &amp; {a_{12} } &amp; {a_{13} } \\<br /> {a_{21} } &amp; {a_{22} } &amp; {a_{23} } \\<br /> {a_{31} } &amp; {a_{32} } &amp; {a_{33} } \\<br /> \end{array} } \right) \Rightarrow \det \left( A \right) = \left| {\begin{array}{*{20}c}<br /> {a_{11} } &amp; {a_{12} } &amp; {a_{13} } \\<br /> {a_{21} } &amp; {a_{22} } &amp; {a_{23} } \\<br /> {a_{31} } &amp; {a_{32} } &amp; {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}
 
form Spunky_Dunkey

thanks very much :smile:
 
TD said:
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} } &amp; {a_{12} } &amp; {a_{13} } \\<br /> {a_{21} } &amp; {a_{22} } &amp; {a_{23} } \\<br /> {a_{31} } &amp; {a_{32} } &amp; {a_{33} } \\<br /> \end{array} } \right) \Rightarrow \det \left( A \right) = \left| {\begin{array}{*{20}c}<br /> {a_{11} } &amp; {a_{12} } &amp; {a_{13} } \\<br /> {a_{21} } &amp; {a_{22} } &amp; {a_{23} } \\<br /> {a_{31} } &amp; {a_{32} } &amp; {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.

:-p


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


whatever's easiest to you!
 
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 :wink:
 
If you row-reduce the matrix to triangular form, finding the determinant is just multiplying the numbers on the diagonal.
 
TD said:
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 :wink:

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