用克莱姆法则解下列线性方程组.x1-2x2+3x3-4x4=4,0+x2-x3+x4=-3,x1+3x3+0+x4=1,

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用克莱姆法则解下列线性方程组.x1-2x2+3x3-4x4=4,0+x2-x3+x4=-3,x1+3x3+0+x4=1,0-7x2+3x3+x4=-3 急

x1-2x2+3x3-4x4=4,
0+x2-x3+x4=-3,
x1+3x3+0+x4=1,
0-7x2+3x3+x4=-3
D=
1 -2 3 -4
0 1 -1 1
1 3 0 1
0 -7 3 1
=16
D1=
4 -2 3 -4
-3 1 -1 1
1 3 0 1
-3 -7 3 1
=-128
D2=
1 4 3 -4
0 -3 -1 1
1 1 0 1
0 -3 3 1
=48
D3=
1 -2 4 -4
0 1 -3 1
1 3 1 1
0 -7 -3 1
=96
D4=
1 -2 3 4
0 1 -1 -3
1 3 0 1
0 -7 3 -3
=0
所以x1=D1/D=-8
x2=D2/D=3
x3=D3/D=6
x4=D4/D=0

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