JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 5)
For the system of linear equations
$$x+y+z=6$$
$$\alpha x+\beta y+7 z=3$$
$$x+2 y+3 z=14$$
which of the following is NOT true ?
If $$\alpha=\beta=7$$, then the system has no solution
For every point $$(\alpha, \beta) \neq(7,7)$$ on the line $$x-2 y+7=0$$, the system has infinitely many solutions
There is a unique point $$(\alpha, \beta)$$ on the line $$x+2 y+18=0$$ for which the system has infinitely many solutions
If $$\alpha=\beta$$ and $$\alpha \neq 7$$, then the system has a unique solution
Explanation
$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & 7 \\ 1 & 2 & 3\end{array}\right|$
$$ \begin{aligned} & =1(3 \beta-14)-1(3 \alpha-7)+1(2 \alpha-\beta) \\\\ & =3 \beta-14+7-3 \alpha+2 \alpha-\beta \\\\ & =2 \beta-\alpha-7 \end{aligned} $$
So, for $\alpha=\beta \neq 7, \Delta \neq 0$ so unique solution.
$\alpha=\beta=7$, equation (i) and (ii) represent 2 parallel planes so no solution.
If $\alpha-2 \beta+7=0$, but $(\alpha, \beta) \neq(7,7)$, then no solution.
$$ \begin{aligned} & =1(3 \beta-14)-1(3 \alpha-7)+1(2 \alpha-\beta) \\\\ & =3 \beta-14+7-3 \alpha+2 \alpha-\beta \\\\ & =2 \beta-\alpha-7 \end{aligned} $$
So, for $\alpha=\beta \neq 7, \Delta \neq 0$ so unique solution.
$\alpha=\beta=7$, equation (i) and (ii) represent 2 parallel planes so no solution.
If $\alpha-2 \beta+7=0$, but $(\alpha, \beta) \neq(7,7)$, then no solution.
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