JEE MAIN - Mathematics (2023 - 13th April Evening Shift - No. 10)
If the system of equations
$$2 x+y-z=5$$
$$2 x-5 y+\lambda z=\mu$$
$$x+2 y-5 z=7$$
has infinitely many solutions, then $$(\lambda+\mu)^{2}+(\lambda-\mu)^{2}$$ is equal to
916
912
920
904
Explanation
$$
\begin{aligned}
& 2 x+y-z=5 \\
& 2 x-5 y+\lambda z=\mu \\
& x+2 y-5 z=7
\end{aligned}
$$
For infinite solution $\Delta=0=\Delta_1=\Delta_2=\Delta_3$
$$ \Delta=\left|\begin{array}{ccc} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{array}\right|=0 $$
$$ \begin{aligned} \Rightarrow& 2(25-2 \lambda)-(-10-\lambda)-(4+5)=0 \\\\ \Rightarrow& 50-4 \lambda+10+\lambda-9=0 \\\\ \Rightarrow& 51=3 \lambda \Rightarrow \lambda=17 \end{aligned} $$
$$ \Delta_3=\left|\begin{array}{ccc} 5 & 2 & 1 \\ \mu & 2 & -5 \\ 7 & 1 & 2 \end{array}\right|=0 $$
$$ \begin{aligned} \Rightarrow & 2(-35-2 \mu)-(14-\mu)+5(4+5)=0 \\\\ \Rightarrow & -70-4 \mu-14+\mu+45=0 \\\\ \Rightarrow & -3 \mu=39 \\\\ \Rightarrow & -\mu=13 \end{aligned} $$
Now $(\lambda+\mu)^2+(\lambda-\mu)^2$
$$ \begin{aligned} & (17+13)^2+(17-13)^2 \\\\ & 900+16 \\\\ & =916 \end{aligned} $$
For infinite solution $\Delta=0=\Delta_1=\Delta_2=\Delta_3$
$$ \Delta=\left|\begin{array}{ccc} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{array}\right|=0 $$
$$ \begin{aligned} \Rightarrow& 2(25-2 \lambda)-(-10-\lambda)-(4+5)=0 \\\\ \Rightarrow& 50-4 \lambda+10+\lambda-9=0 \\\\ \Rightarrow& 51=3 \lambda \Rightarrow \lambda=17 \end{aligned} $$
$$ \Delta_3=\left|\begin{array}{ccc} 5 & 2 & 1 \\ \mu & 2 & -5 \\ 7 & 1 & 2 \end{array}\right|=0 $$
$$ \begin{aligned} \Rightarrow & 2(-35-2 \mu)-(14-\mu)+5(4+5)=0 \\\\ \Rightarrow & -70-4 \mu-14+\mu+45=0 \\\\ \Rightarrow & -3 \mu=39 \\\\ \Rightarrow & -\mu=13 \end{aligned} $$
Now $(\lambda+\mu)^2+(\lambda-\mu)^2$
$$ \begin{aligned} & (17+13)^2+(17-13)^2 \\\\ & 900+16 \\\\ & =916 \end{aligned} $$
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