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Matrices And Determinants

JEE Advanced 2023 Paper 1 Online

MCQ+3 / -12023
Let $\alpha, \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations


$$ \begin{aligned} & x+2 y+z=7 \\\\ & x+\alpha z=11 \\\\ & 2 x-3 y+\beta z=\gamma \end{aligned} $$


Match each entry in List-I to the correct entries in List-II.



.tg {border-collapse:collapse;border-spacing:0;width:100%}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-1wig{font-weight:bold;text-align:left;vertical-align:top}
.tg .tg-0lax{text-align:left;vertical-align:top}






























List - I List - II
(P) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has (1) a unique solution
(Q) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has (2) no solution
(R) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has (3) infinitely many solutions
(S) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has (4) $x=11, y=-2$ and $z=0$ as a solution
(5) $x=-15, y=4$ and $z=0$ as a solution


The correct option is:

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