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

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Answer 1 · 2016-03-03T22:01:58

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

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:

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

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