A) Zero
B) 3
C) 2
D) 1
Correct Answer: C
Solution :
For non-trivial solution of given system of linear equations \[\left| \begin{matrix} 4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow 8+k(2-k)+2(2k-8)=0\] \[\Rightarrow -{{k}^{2}}+6k-8=0\] \[\Rightarrow {{k}^{2}}-6k+8=0\] \[\Rightarrow k=2,4\] Clearly there exists two values of k.You need to login to perform this action.
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