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Question

Find the value for k which each of the following systems of equations has a unique solution:

kx+3y=(k3),12x+ky=k.


Solution

kx+3y(k3)=0(1)12x+kyk=0(2)a1=k,b1=3,c1=(k3)a2=12,b2=k,c2=k

For a unique solution, we must have, 

a1a2b1b2

Now,

k123kk236k±6

Thus, for all real values of k other than ±6, the given system of equations will have a unique solution.

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