Page 402 - 35Linear Algebra

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Another way to compute the determinant of a matrix is to use this recursive
formula. Here I take the coefficients of the first row and multiply them by
the determinant of the minors and the cofactor. Then we can use the formula
for a two by two determinant to compute the determinant of the minors

 
1 2 3
1 2 3 2 3 1
det  3 1 2  = 1 − 2 + 3 = 1(1 − 0) − 2(3 − 0) + 3(0 − 0) = −5
0 1 0 1 0 0
0 0 1
Decide which way you prefer and get good at taking determinants, you’ll need
to compute them in a lot of problems.

Hint for Review Problem 5

i
For an arbitrary 3 × 3 matrix A = (a ), we have
j
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
det(A) = a a a + a a a + a a a − a a a − a a a − a a a
2 3 1
3 2 1
3 1 2
1 3 2
1 2 3
2 1 3
and so the complexity is 5a + 12m. Now note that in general, the complexity
c n of the expansion minors formula of an arbitrary n × n matrix should be
c n = (n − 1)a + nc n−1 m
P n i 1 1 1
since det(A) = (−1) a cofactor(a ) and cofactor(a ) is an (n − 1) × (n − 1)
i=1 i i i
matrix. This is one way to prove part (c).

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