JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let the system of linear equations \(x+y+\alpha z=2\) \(3 x+y+z=4\) \(x+2 z=1\) have a unique solution \(\left(x^{*}, y^{*}, z^{*}\right)\). If \(\left(\alpha, x^{*}\right),\left(y^{*}, \alpha\right)\) and \(\left(x^{*},-y^{*}\right)\) are collinear points, then the sum of absolute values of all possible values of \(\alpha\) is
- A \(4\)
- B \(3\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(\Delta=\left|\begin{array}{lll}1 & 1 & \alpha \\ 3 & 1 & 1 \\ 1 & 0 & 2\end{array}\right|=-(\alpha+3)\) \(\Delta_{1}=\left|\begin{array}{lll}2 & 1 & \alpha \\ 4 & 1 & 1 \\ 1 & 0 & 2\end{array}\right|=-(3+\alpha)\)…
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