解下列三元一次方程组(1)x1+x2=3 x2+x3= -6 x3+x1=10 (2) 5x+4y+z=0 3x+y-4

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  • (1)x1+x2=3 (1)

    x2+x3= -6 (2)

    x3+x1=10 (3)

    (1)+(2)+(3)得:

    2x1+2x2+2x3=7

    则:x1+x2+x3=3.5 (4)

    (4)-(1)得:x3=3.5-3=0.5

    (4)-(2)得:x1=3.5-(-6)=9.5

    (4)-(3)得:x2=3.5-10=-6.5

    解得:x1=9.5

    x2=-6.5

    x3=0.5

    (2) 5x+4y+z=0 (1)

    3x+y-4z=11 (2)

    x+y+z= -2 (3)

    (1)×4+(2)得:

    23x+17y=11 (4)

    (1)-(3)得:

    4x+3y=2 (5)

    (5)×23-(4)×4得:

    69y-68y=46-44

    y=2

    代入(5):

    4x+6=2

    4x=-4

    x=-1

    把x=-1 y=2代入(3):

    -1+2+z=-2

    z=-3

    解得:x=-1

    y=2

    z=-

    (3) x+y-z=6 (1)

    x-3y+2z=1 (2)

    3x+2y-z=4 (3)

    (1)+(2)+(3)得:

    5x=11

    x=2.2

    把x=2.2分别代入(1),(3)得:

    2.2+y-z=6

    y-z=3.8 (4)

    6.6+2y-z=4

    2y-z=-2.6 (5)

    (5)-(4)得:

    y=-2.6-3.8=-6.4

    代入(4)

    -6.4-z=3.8

    z=-6.4-3.8=-10.2

    解得:x=2.2

    y=-6.4

    z=-10.2