How do you find the determinant of #|(4,-1,-2), (0, 2, 1), (2,1,3)|#?

Question

1189 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-12T06:52:21

The answer is #=26#

One method is as follows :

# | (a,b,c), (d,e,f), (g,h,i) | #

#=a | (e,f), (h,i) |-b | (d,f), (g,i) |+c | (d,e), (g,h) |#

#=a(ei-fh)-b(di-gf)+c(dh-eg)#

Therefore,

#| (4,-1,-2), (0,2,1), (2,1,3) | #

#=4* | (2,1), (1,3) |+1* | (0,1), (2,3) |-2* | (0,2), (2,1) |#

#=4*(6-1)+1*(0-2)-2*(0-4)#

#=4*5-1*2-2*-4#

#=20-2+8#

#=26#