JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
For the system of linear equations : \(x-2 y=1, x-y+k z=-2, k y+4 z=6, k \in R\) consider the following statements : \((A)\) The system has unique solution if \(k \neq 2\), \(k \neq-2\) \((B)\) The system has unique solution if \(k =-2\). \((C)\) The system has unique solution if \(k =2\). \((D)\) The system has no-solution if \(k =2\). \((E)\) The system has infinite number of solutions if \(k \neq-2\) Which of the following statements are correct?
- A \((C)\) and \((D)\) only
- B \((B)\) and \((E)\) only
- C \((A)\) and \((E)\) only
- D \((A)\) and \((D)\) only
Answer & Solution
Correct Answer
(D) \((A)\) and \((D)\) only
Step-by-step Solution
Detailed explanation
\(D =\left|\begin{array}{ccc}1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4\end{array}\right|=4- k ^{2}\) so, \(A\) is correct and \(B, C, E\) are incorrect. If \(k =2\) \(D_{1}=\left|\begin{array}{ccc}1 & -2 & 0 \\ -2 & -1 & 2 \\ 6 & 2 & 4\end{array}\right|=-48 \neq 0\) So no solution…
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