How do you find the determinant of #((2, 5, -3, -1), (3, 0, 1, -3), (-6, 0, -4, 9), ( 4, 10, -4, -1))#?

1 Answer
Rui D.
May 3, 2016

#120#

Explanation:

#((color(teal)(2),color(teal)(5),color(teal)(-3),color(teal)(-1)),(3,0,1,-3),(-6,0,-4,9),(color(turquoise)(4),color(turquoise)(10),color(turquoise)(-4),color(turquoise)(-1)))=#
#-># row4 #-2*# row1 #-># row4

#=((color(gold)(2),color(blue)(5),color(gold)(-3),color(gold)(-1)),(3,color(gold)(0),1,-3),(-6,color(gold)(0),-4,9),(0,color(gold)(0),2,1))#

#=5*(-1)^(1+2)*((3,1,-3),(-6,-4,9),(0,2,1))#

#=-5[-12+36-(54-6)]#
#=-5[24-48]=5*24=120#

Related questions
See all questions in Determinant of a Square Matrix
Impact of this question
1529 views around the world
You can reuse this answer
Creative Commons License