JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
For the system of linear equations \(x+y+z=6\) ; \(\alpha x+\beta y+7 z=3\) ; \(x+2 y+3 z=14\) which of the following is \(NOT\) true ?
- A If \(\alpha=\beta=7\), then the system has no solution
- B If \(\alpha=\beta\) and \(\alpha \neq 7\) then the system has a unique solution.
- C There is a unique point \((\alpha, \beta)\) on the line \(x +2 y +18=0\) for which the system has infinitely many solutions
- D For every point \((\alpha, \beta) \neq(7,7)\) on the line \(x-2 y+7=0\), the system has infinitely many solutions.
Answer & Solution
Correct Answer
(D) For every point \((\alpha, \beta) \neq(7,7)\) on the line \(x-2 y+7=0\), the system has infinitely many solutions.
Step-by-step Solution
Detailed explanation
By equation \(1\) and \(3\) And \(\begin{array}{c}y+2 z=8 \\ y=8-2 z \\ x=-2+z\end{array}\) Now putting in equation \(2\) \(\alpha(z-2)+\beta(-2 z+8)+7 z=3\) \(\Rightarrow(\alpha-2 \beta+7) z=2 \alpha-8 \beta+3\) So equations have unique solution if \(\alpha-2 \beta+7 \neq 0\)…
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