JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
If the system of linear equations \(x - 2y + kz = 1\) ; \(2x + y + z = 2\) ; \(3x - y - kz = 3\) Has a solution \((x, y, z) \ne 0\), then \((x, y)\) lies on the straight line whose equation is
- A \(3x -4y -1 = 0\)
- B \(4x -3y -4 = 0\)
- C \(4x -3y -1 = 0\)
- D \(3x -4y -4 = 0\)
Answer & Solution
Correct Answer
(B) \(4x -3y -4 = 0\)
Step-by-step Solution
Detailed explanation
For infinitly many solution \(\left| {\begin{array}{*{20}{c}} 1&{ - 2}&k\\ 2&1&1\\ 3&{ - 1}&{ - k} \end{array}} \right| - 0\) \( \Rightarrow k = \frac{{ - 1}}{2}\) Also consider \(x - 2y + k = 1\) and \(2x + y + z = 2\) \( \Rightarrow 2x - 4y - z - 2\) \(2x + y + z = 2\)…
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