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Question

# The number of values of k for which the system of linear equations, (2k+1)x+5ky=k+2 and kx+(k+2)y=2 has no solution, is:

A
Infinitely many
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B
3
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C
1
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D
2
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Solution

## The correct option is C 1A non-homogeneous system of linear equations has a unique non-trivial solution if and only if its determinant is non-zero. If this determinant is zero, then the system has either no nontrivial solutions or an infinite number of solutions.[(k+2)10k(k+3)] [ xy] =[kk−1]Now it is of the Form Ax=BNow to for the system to have no Solution , determinant of A must be 0,as follows⇒|A|=(k+2)(k+3)−k×10=0⇒k2−5k+6=(k−2)(k−3)=0Therefore for k=2,3 system will have no solution.For k=2, we get infinitely many solutions, after substituting the value of k=2 in the equations.Thus k=3Thus, the number of solutions is 1.

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