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Question

The system of equation 2x+3yz=0,3x+2y+kz=0 and 4x+y+z=0 have a set of non-zero integral solution, then the least positive value of z is:

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Solution

2x+3yz=03x+2y+kz=04x+y+z=0000=23132k411......[CRAMER'S RULE]Δ=2(2K)3(34K)1(38)=42K9+12K+5=10K
Now, x=1,y=2,z=3
Clearly, 1=2=3=0
K0
x=y=z=0, which is not possible.

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