NIMCET 2020 Question Paper with solutions

Question : 15 of 120

Marks: +1, -0

The number of values of k for which the linear equations

4x+ky+z=0

kx+4y+z=0

2x+2y+z=0

posses a non-zero solution is

Solution:

|\table 2,2,1;4,k,1;k,4,1|

=2(k-4)+2(k-4)+(16-k^2)

=-k^2+4k=0

=k(k-4)

K=0,4

No. of value of k = 2

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