JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
If the system of linear equations \(x+y+3 z=0\) \(x+3 y+k^{2} z=0\) \(3 x+y+3 z=0\) has a non-zero solution \((x, y, z)\) for some \(k \in R ,\) then \(x +\left(\frac{ y }{ z }\right)\) is equal to
- A \(9\)
- B \(-3\)
- C \(-9\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(-3\)
Step-by-step Solution
Detailed explanation
\(x+y+3 z=0\) \(x+3 y+k^{2} z=0\) \(3 x+y+3 z=0\) \(\left|\begin{array}{lll}1 & 1 & 3 \\ 1 & 3 & k^{2} \\ 3 & 1 & 3\end{array}\right|=0\) \(\Rightarrow 9+3+3 k^{2}-27-k^{2}-3=0\) \(\Rightarrow k ^{2}=9\) (i) \(-\) (iii) \(\Rightarrow-2 x =0 \Rightarrow x =0\) Now from (i)…
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