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scalar triple product
cofactor expansion... that's needed to shrink the matrix down to a 2 x 2
A * (B x C)
do you know a 2 by 2 determinant?
yes! i just don't know how to find the determinant for a 3x3 matrix
Determinant of 2 by 2 matrix = a*d - cd [a b] [c d]
ad - bc for determinant of a 2 x 2 matrix for a 3 x 3 matrix we have to use cofactor expansion.
how do i do cofactor expansion?
Cofactor expansion expanding a row ... find the minors and the cofactors of the first row entries ( you can choose second row or third row entries but the sign pattern is different).
Cover first column, 8*det of that 2x2 is the first of 3 terms
look up determinant of 3 x 3 matrix.. lol this is alot of drawing
the signs for the 3 x 3 matrix is like a checkerboard |dw:1438053171537:dw| that's why I use the first drawing a lot.
* first row signs
yeah i just remember the second is minus
cut row one cut column one cut row one cut column two cut row one cut column three
ok I'm going to clean this up a bit.. but where I circled the row and column... you draw a line to cross them out... you should be left with 4 numbers... that's your 2 x 2 matrix. now you can take the determinant of those matrices...
|dw:1438053599579:dw| normally we would have a + sign but that negative -2 on the first row third column entry changes that.
|dw:1438053728550:dw| so we should have this equation after we have done cofactor expansion. Now we can take the determinant of the 2 x 2 matrix and combine like terms. The final answer for the determinant of the 3 x 3 matrix should be just a single number