JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 2)
Let the system of linear equations
4x + $$\lambda$$y + 2z = 0
2x $$-$$ y + z = 0
$$\mu$$x + 2y + 3z = 0, $$\lambda$$, $$\mu$$$$\in$$R.
has a non-trivial solution. Then which of the following is true?
4x + $$\lambda$$y + 2z = 0
2x $$-$$ y + z = 0
$$\mu$$x + 2y + 3z = 0, $$\lambda$$, $$\mu$$$$\in$$R.
has a non-trivial solution. Then which of the following is true?
$$\mu$$ = 6, $$\lambda$$$$\in$$R
$$\lambda$$ = 3, $$\mu$$$$\in$$R
$$\mu$$ = $$-$$6, $$\lambda$$$$\in$$R
$$\lambda$$ = 2, $$\mu$$$$\in$$R
Explanation
Given, system of linear equations
4x + $$\lambda$$y + 2z = 0
2x $$-$$ y + z = 0
$$\mu$$x + 2y + 3z = 0
For non-trivial solution, $$\Delta$$ = 0
$$\left| {\matrix{ 4 & \lambda & 2 \cr 2 & { - 1} & 1 \cr \mu & 2 & 3 \cr } } \right| = 0$$
$$ \Rightarrow 4( - 3 - 2) - \lambda (6 - \mu ) + 2(4 + \mu ) = 0$$
$$ \Rightarrow - \lambda (6 - \mu ) - 2(6 - \mu ) = 0$$
$$ \Rightarrow (6 - \mu )(\lambda + 2) = 0$$
$$ \Rightarrow \lambda = - 2$$ and $$\mu \in R$$ or $$\mu$$ = 6 and $$\lambda \in R$$.
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