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2/x+3/y-4/z=-3,1/x+2/y+6/z=2,3/x-1/y+2/z...

`2/x+3/y-4/z=-3,1/x+2/y+6/z=2,3/x-1/y+2/z=5`

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The correct Answer is:
x=1,y=-1,z=2
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1/x+1/y+1/z=1,2/x+5/y+3/z=0,1/x+2/y+4/z=3

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2x-y+z=6,x+2y+3z=3,3x+y-z=4

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2x+3y+z=17,x-y+z=3,3x+2y-2z=4

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Find A^(-1) if A = [1,2,-3][2,3,2] [3,-3,-4] Hence solve the equations x+2y-3z =-4, 2x+3y+2z=2, 3x-3y-4z=11.