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

1 Answer
Mar 3, 2016

Answer:

#abs((1,1,1,3),(0,3,1,1),(0,0,2,2),(-1,-1,-1,2))=30#

Explanation:

The determinant is unchanged by adding any multiple of one row to another, so first add row #1# to row #4# to find:

#abs((1,1,1,3),(0,3,1,1),(0,0,2,2),(-1,-1,-1,2))=abs((1,1,1,3),(0,3,1,1),(0,0,2,2),(0,0,0,5))#

The determinant of an upper triangular matrix is just the product of the diagonal, so:

#abs((1,1,1,3),(0,3,1,1),(0,0,2,2),(0,0,0,5)) = 1*3*2*5 = 30#