Question

求行列式(0 4 5 -1 2;-5 0 2 0 1;7 2 0 3 -4;-3 1 -1 -5 0;2 -3 0 1 3)

Answer 1

c1+5c5,c3-2c5
10 4 1 -1 2
0 0 0 0 1
-13 2 8 3 -4
-3 1 -1 -5 0
17 -3 -6 1 3
按第2行展开 D=(-1)^(2+5)*
10 4 1 -1
-13 2 8 3
-3 1 -1 -5
17 -3 -6 1
c1+10c4,c2+4c4,c3+c4
0 0 0 -1
17 14 11 3
-53 -19 -6 -5
27 1 -5 1
按第1行展开 D = - (-1)^(1+4)*(-1)*
17 14 11
-53 -19 -6
27 1 -5
r1+r2+r3,r3-r2
-9 -4 0
-53 -19 -6
80 20 1
r2+6r3
-9 -4 0
427 101 0
80 20 1
= (-1)*
-9 -4
427 101
= (-1)*(-909 + 4*427)
= -799