JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 5)
For the system of linear equations $$\alpha x+y+z=1,x+\alpha y+z=1,x+y+\alpha z=\beta$$, which one of the following statements is NOT correct?
It has infinitely many solutions if $$\alpha=1$$ and $$\beta=1$$
It has infinitely many solutions if $$\alpha=2$$ and $$\beta=-1$$
$$x+y+z=\frac{3}{4}$$ if $$\alpha=2$$ and $$\beta=1$$
It has no solution if $$\alpha=-2$$ and $$\beta=1$$
Explanation
For infinite solution $\Delta=\Delta_x=\Delta_y=\Delta_z=0$
$$ \Delta=\left|\begin{array}{lll} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right|=0 \Rightarrow\left(\alpha^3-3 \alpha+2\right)=0 \Rightarrow \alpha=1,-2 $$
If $\beta=1$, then all planes are overlapping
$\therefore$ Option (A) is correct.
Option (B) :
If $\alpha=2 \Rightarrow \Delta \neq 0$
$\therefore $ Unique solution exist
$\therefore$ Option (B) is incorrect.
Option (C) :
$$ \begin{aligned} & \alpha=2, \beta=1 \\\\ & 2 x+y+z=1 \\\\ & x+2 y+z=1 \\\\ & x+y+2 z=1 \end{aligned} $$
Adding all three equations,
$$ x+y+z=\frac{3}{4} $$
$\therefore$ option (C) is correct.
Option (D) :
If $\alpha=-2$ and $\beta=1$, then $\Delta=0, \Delta_x \neq 0$
$\therefore$ No solution.
$\therefore$ Option (D) is correct.
$$ \Delta=\left|\begin{array}{lll} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right|=0 \Rightarrow\left(\alpha^3-3 \alpha+2\right)=0 \Rightarrow \alpha=1,-2 $$
If $\beta=1$, then all planes are overlapping
$\therefore$ Option (A) is correct.
Option (B) :
If $\alpha=2 \Rightarrow \Delta \neq 0$
$\therefore $ Unique solution exist
$\therefore$ Option (B) is incorrect.
Option (C) :
$$ \begin{aligned} & \alpha=2, \beta=1 \\\\ & 2 x+y+z=1 \\\\ & x+2 y+z=1 \\\\ & x+y+2 z=1 \end{aligned} $$
Adding all three equations,
$$ x+y+z=\frac{3}{4} $$
$\therefore$ option (C) is correct.
Option (D) :
If $\alpha=-2$ and $\beta=1$, then $\Delta=0, \Delta_x \neq 0$
$\therefore$ No solution.
$\therefore$ Option (D) is correct.
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