Kweku 2 years ago How do you find the determinant of a 4x4 matrix?

1. geerky42
2. Kweku

I didnt get you.

3. novagirl114

Do you have a graphing calculator? If so you can input the matrix and the calculator can sovle for the determinant.

4. geerky42

You 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

5. malevolence19

Cofactor 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.

6. malevolence19

But 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.