Question

算出D，D1，D2，D3，D4，要有详细过程，急!急！急！

Answer 1

这只能一步步来解，系数行列式D=
3 -2 0 -4
2 -3 -1 -5
1 2 -1 4
3 1 2 11 r3+r1
=
3 -2 0 -4
2 -3 -1 -5
4 0 -1 0
3 1 2 11 c1+4c3
=
3 -2 0 -4
-2 -3 -1 -5
0 0 -1 0
11 1 2 11 按第3行展开
=
3 -2 -4
-2 -3 -5
11 1 11 *(-1) c1-11c2,c3-11c2
=
25 -2 18
31 -3 28
0 1 0 *(-1)
=25*28 -31*18=142
D1=
3 -2 0 -4
-2 -3 -1 -5
-2 2 -1 4
0 1 2 11 r1+1.5r3,r2-r3
=
0 1 -1.5 2
0 -5 0 -9
-2 2 -1 4
0 1 2 11 按c1展开
=
1 -1.5 2
-5 0 -9
1 2 11 *(-2) c3-c1*(9/5)
=
1 -1.5 1/5
-5 0 0
1 2 46/5 *(-2)
= -10 *(-0.3*46-0.4)= 142
D2=
3 3 0 -4
2 -2 -1 -5
1 -2 -1 4
3 0 2 11 r3-r2, r4+2r2
=
3 3 0 -4
2 -2 -1 -5
-1 0 0 9
7 -4 0 1 第2列展开
=
3 3 -4
-1 0 9
7 -4 1 c3+9c1
=
3 3 23
-1 0 0
7 -4 64= 284
D3=
3 -2 3 -4
2 -3 -2 -5
1 2 -2 4
3 1 0 11 r1+r3,r2-r3
=
4 0 1 0
1 -5 0 -9
1 2 -2 4
3 1 0 11 c1-4c2
=
0 0 1 0
1 -5 0 -9
9 2 -2 4
3 1 0 11 按第1行展开
=
1 -5 -9
9 2 4
3 1 11 d2+5d1,d3+9d1
=
1 0 0
9 47 85
3 16 38
=47*38 -85*16=426
D4=
3 -2 0 3
2 -3 -1 -2
1 2 -1 -2
3 1 2 0 r4+2r3, r2-r3
=
3 -2 0 3
1 -5 0 0
1 2 -1 -2
7 -5 0 -4 按第3列展开
=
3 -2 3
1 -5 0
7 -5 -4 *(-1) c2+5c1
=
3 13 3
1 0 0
7 30 -4 *(-1) 按r2展开
= -52 -90= -142
于是解得
x1=D1/D= 142/142=1
x2=D2/D=284/142=2
x3=D3/D=426/142=3
x4=D4/D= -142/142= -1