JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 16)
Let $$\mathrm{S}$$ be the set of values of $$\lambda$$, for which the system of equations
$$6 \lambda x-3 y+3 z=4 \lambda^{2}$$,
$$2 x+6 \lambda y+4 z=1$$,
$$3 x+2 y+3 \lambda z=\lambda$$ has no solution. Then $$12 \sum_\limits{i \in S}|\lambda|$$ is equal to ___________.
$$6 \lambda x-3 y+3 z=4 \lambda^{2}$$,
$$2 x+6 \lambda y+4 z=1$$,
$$3 x+2 y+3 \lambda z=\lambda$$ has no solution. Then $$12 \sum_\limits{i \in S}|\lambda|$$ is equal to ___________.
Answer
24
Explanation
Given that $S$ be the set of values of $\lambda$ for which given system of equations has no solution.
Therefore for the given set of equations
$$ \Delta=\left|\begin{array}{ccc} 6 \lambda & -3 & 3 \\ 2 & 6 \lambda & 4 \\ 3 & 2 & 3 \lambda \end{array}\right|=0 $$
$$ \begin{aligned} &\Rightarrow6 \lambda\left(18 \lambda^2-8\right)+3(6 \lambda-12)+3(4-18 \lambda)=0 \\\\ &\Rightarrow18 \lambda^3-14 \lambda-4=0 \\\\ &\Rightarrow(\lambda-1)(3 \lambda+1)(3 \lambda+2)=0 \\\\ &\Rightarrow \lambda=1,-\frac{1}{3},-\frac{2}{3} \end{aligned} $$
Also for each values of $\lambda=1, \frac{-1}{3}, \frac{-2}{3}$, we have
$$ \left|\begin{array}{ccc} 6 \lambda & -3 & 4 \lambda^2 \\ 2 & 6 \lambda & 1 \\ 3 & 2 & \lambda \end{array}\right| \neq 0 $$
which implies that, for each values of $\lambda$, the given system of equations has no solution.
$$ \begin{aligned} & \text { Therefore } S \in\left\{1, \frac{-1}{3}, \frac{-2}{3}\right\} \text { and } \\\\ &12 \sum_{\lambda \in S}|\lambda| \\\\ & =12\left(|1|+\left|\frac{-1}{3}\right|+\left|\frac{-2}{3}\right|\right) \\\\ & =12\left(1+\frac{1}{3}+\frac{2}{3}\right)=12\left(\frac{6}{3}\right)=24 \end{aligned} $$
Therefore for the given set of equations
$$ \Delta=\left|\begin{array}{ccc} 6 \lambda & -3 & 3 \\ 2 & 6 \lambda & 4 \\ 3 & 2 & 3 \lambda \end{array}\right|=0 $$
$$ \begin{aligned} &\Rightarrow6 \lambda\left(18 \lambda^2-8\right)+3(6 \lambda-12)+3(4-18 \lambda)=0 \\\\ &\Rightarrow18 \lambda^3-14 \lambda-4=0 \\\\ &\Rightarrow(\lambda-1)(3 \lambda+1)(3 \lambda+2)=0 \\\\ &\Rightarrow \lambda=1,-\frac{1}{3},-\frac{2}{3} \end{aligned} $$
Also for each values of $\lambda=1, \frac{-1}{3}, \frac{-2}{3}$, we have
$$ \left|\begin{array}{ccc} 6 \lambda & -3 & 4 \lambda^2 \\ 2 & 6 \lambda & 1 \\ 3 & 2 & \lambda \end{array}\right| \neq 0 $$
which implies that, for each values of $\lambda$, the given system of equations has no solution.
$$ \begin{aligned} & \text { Therefore } S \in\left\{1, \frac{-1}{3}, \frac{-2}{3}\right\} \text { and } \\\\ &12 \sum_{\lambda \in S}|\lambda| \\\\ & =12\left(|1|+\left|\frac{-1}{3}\right|+\left|\frac{-2}{3}\right|\right) \\\\ & =12\left(1+\frac{1}{3}+\frac{2}{3}\right)=12\left(\frac{6}{3}\right)=24 \end{aligned} $$
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