Urgent please help! matrices

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solve using matrices
x+3y-3z=12
3x-y+4z=0
-x+2y-z=1

  • Urgent please help! matrices -

    Put in matrix form (the 4th column is the RHS):
    1 3 -3 12
    3 -1 4 0
    -1 2 -1 1
    proceed to transform to echelon form:
    1 3 -3 12
    0 10 -13 36 (3R1-R2)
    0 5 -4 13 (R1+R3)

    1 3 -3 12
    0 10 -13 36
    0 0 5 -10 (2R3-R2)

    1 3 -3 12
    0 10 -13 36
    0 0 1 -2 (R3/5)

    Back substitute
    1 3 0 6 (3R3+R1)
    0 1 0 1 (13R3+R2)/10
    0 0 1 -2

    1 0 0 3 (R1-3R2)
    0 1 0 1
    0 0 1 -2

    So x=3, y=1, z=-2

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