How do you find the determinant of #((0, 1, 2, 3), (0, 0, 2, 0), (3, 10, 12, 20), (0, 0, 0, 4))#?

1 Answer
Feb 16, 2016

given by

#-2det((0, 1, 3),(3,10,20),(0,0,4))#

and

#det((0, 1, 3),(3,10,20),(0,0,4))#

is given by

#4det((0,1),(3,10))#

and

#det((0,1),(3,10))#

is given by

#0*10-3*1 = -3#

In fact you can develop on what line or column you want, find one which have most #0#

Then remember formula

#det(A) = sum_(n=0)^(n)(-1)^(i+(j+n))a_(i(j+n))det(A_(i(j+n)))#

Complex ? not really

here you develop the determinant in line because #i# is fixed (and #=2#, you just have to read the number which is #a_(ij)# and to have #A_(ij)# just hide with your finger the column and the line which contain #a_(ij)#

here it's simple, #a_(21) = a_(22) = a_(24) = 0# so you just have

#det(A) = (-1)^(2+3)a_(23)det(A_(23))#

repeat that until you have #2xx2# matrix like i did