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Question

Solve the following equations by the method of reduction:
2xy+z=1,x+2y+3z=8,3x+y4z=1

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Solution

2xy+z=1x+2y+3z=8,3x+y4z=1
211123314xyz=181
R3R33R2
2111230513xyz=1823
R2R212R1
⎢ ⎢ ⎢211052520513⎥ ⎥ ⎥xyz=⎢ ⎢ ⎢115223⎥ ⎥ ⎥
R225R2
2110110513xyz=1323
R3R3+5R2
211011008xyz=138
R1R1+R2 and R3R3×18
202011001xyz=431
R2R2R3 and R1R1×12
101010001xyz=221
R1R1R3
100010001xyz=121
xyz=121
x=1,y=2,z=1.

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