How do you find the determinant of #((9, -4, 3, 5), (2, 7, 6, -5), (4, 1, -2, 0), (7, 3, 4, 10))#?

Question

1544 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-25T19:32:04

The answer is #=-4025#

We reduce the matrix to row echelon form

#A=((9,-4,3,5),(2,7,6,-5),(4,1,-2,0),(7,3,4,10))#

We perform

#R_2larrR_2-2/9R_1#

#=((9,-4,3,5),(0,71/9,16/3,-55/9),(4,1,-2,0),(7,3,4,10))#

#R_3larrR_3-4/9R_1#

#=((9,-4,3,5),(0,71/9,16/3,-55/9),(0,25/9,-10/3,-2/90),(7,3,4,10))#

#R_4larrR_4-7/9r_1#

#=((9,-4,3,5),(0,71/9,16/3,-55/9),(0,25/9,-10/3,-2/90),(0,55/9,5/3,55/9))#

#R_3larrR_3-25/71R-2#

#=((9,-4,3,5),(0,71/9,16/3,-55/9),(0,0,-370/71,-5/71),(0,55/9,5/3,55/9))#

#R_4larrR_4-55/71R_2#

#=((9,-4,3,5),(0,71/9,16/3,-55/9),(0,0,-370/71,-5/71),(0,0,-175/71,770/71))#

#R_4larrR_4-35/74R_3#

#=((9,-4,3,5),(0,71/9,16/3,-55/9),(0,0,-370/71,-5/71),(0,0,0,805/74))#

The determinant of the matrix is equal to the diagonal product of the matrix

#det(A)=9*71/9*(-370/71)*805/74=-5*805=-4025#