How do you find the determinant of #((-2, 0, 0, -1), (-2, 0, 1, 0), (0, -2, 0, 1), (0, -2, 1, 0))#?

1 Answer
Feb 15, 2016

Answer:

The determinate of A, |A| = #|A| = 8#

Explanation:

In general you can use "determinant expansion by minors" also or "Laplacian expansion by minors".
- Recipe
#det(A) = |A| = sum_(k=1)^Na_(ij)C_(ij)#
with no implied summation over j and where #C_(ij)# is the cofactor of #a_(ij#) defined by
#C_(ijM_(ij); M_(ij)# is Minor of A formed by eliminating the ith row, jth column of A. So expanding around #a_(1j)#, that is eliminating #a_(1j) and a_(i1)# we write:

#|A| =a_(11)(-1)^(1+1)color(red)(det((a_(22), cdots,a_(2j)), (vdots, vdots, vdots), (a_(i2), cdots, a_(ij)))) + (-1)^(1+2)a_12color(red)(det((a_(21), cdots,a_(2j)), (vdots, vdots, vdots), (a_(i2), cdots, a_(12)))) + cdots +(-1)^(1+j)a_1jcolor(red)(det((a_(21), cdots,a_(2j-1)), (vdots, vdots, vdots), (a_(i2), cdots, a_(ij-1)))) #

The reality is you can expand around any row or column so let's be strategic and expand around the 3rd column, but first your signs:
for a #4xx4=>((+,-,+,-), (-,+,-,+),(+,-,+,-), (-,+,-,+)); 3xx3 =>((+,-,+), (-,+,-),(+,-,+)); # Note 3rd column has only zero and ones, simplifying life.
#|A| = cancel(0*det((-2,0,0),(0,-2,1),(0,-2,0))) - 1*det((-2,0,-1),(0,-2,1),(0,-2,0))+cancel(0*det((-2,0,-1 ),(-2,0,0 ), ( 0,-2,0))) - 1*det((-2,0,-1),(-2,0,0),(0,-2,1))#
#|A|=1*det((2,0,1),(0,2,-1),(0,2,0))+ 1*det((2,0,1),(2,0,0),(0,2,-1))#
Now if you expand the first term around the first columns and the second term around the 2nd column you will end up:

#|A| = 2det((2,-1), (2,0 )) -2det((2,1 ), (2,0)) # cross multiply and subtract
#|A| = 2((2*0)- (2*(-1)) -2((2*0)-(2*1)) = 8#