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Question

Find the value of k, so that the following system of equations has no solution :
3x+y=1;(2k1)x+(k1)y=(2k1).

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Solution

Given that,
3x+y=1(2k1)x+(k1)y=(2k1)(A:B)=[31:1(2k1)(k1):(2k1)]=[113:13(2k1)(k1):(2k1)]=113:130(k3+23):23(2k1)
It k(A)k(A:B) then it has no solution .
So,
k3+23=0k=2

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