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geerky42
 one year ago
Best ResponseYou've already chosen the best response.0http://lmgtfy.com/?q=How+do+you+find+the+determinant+of+a+3x3+matrix%3F

novagirl114
 one year ago
Best ResponseYou've already chosen the best response.0Do you have a graphing calculator? If so you can input the matrix and the calculator can sovle for the determinant.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0You edited your question... Just Google it or do what @novagirl114 said. http://lmgtfy.com/?q=How+do+you+find+the+determinant+of+a+4x4+matrix%3F

malevolence19
 one year ago
Best ResponseYou've already chosen the best response.0Cofactor expansion. The same way you do 3x3's. If you have: \[A=\left[\begin{matrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31}&a_{32}&a_{33}\end{matrix}\right]\] Then: \[\det(A)=a_{11} \det \left[\begin{matrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{matrix}\right]a_{12} \det \left[\begin{matrix}a_{21} & a_{23} \\ a_{31} & a_{33}\end{matrix}\right]+a_{13} \det \left[\begin{matrix}a_{21} & a_{22} \\ a_{31} & a_{32}\end{matrix}\right]\] Better notation is my opinion for an nxn matrix is: \[\det(A)=\sum_{i_1,i_2,...,i_n} \epsilon_{i_1,i_2,...,i_n}a_{1,i_1}a_{2,i_2}...a_{n,i_n}\] Where epsilon is the Levi Cevita tensor.

malevolence19
 one year ago
Best ResponseYou've already chosen the best response.0But the only difference for a 4x4 is that the "sub determinants" (i.e., my determinants of 2x2's) will be THREE BY THREES! So you'll need to do a cofactor expansion to get 4, 3x3 determinants and then 4 cofactor expansions to get 3,2x2 determinants for EACH sub determinant.
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