JEE Advance - Mathematics (2023 - Paper 1 Online - No. 14)
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.
The correct option is:
$$ \begin{aligned} & x+2 y+z=7 \\\\ & x+\alpha z=11 \\\\ & 2 x-3 y+\beta z=\gamma \end{aligned} $$
Match each entry in List-I to the correct entries in List-II.
| 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:
$(P) \rightarrow(3) ~~ (Q) \rightarrow(2) ~~(R) \rightarrow(1)~~ (S) \rightarrow(4)$
$(P) \rightarrow(3) ~~(Q) \rightarrow(2) ~~(R) \rightarrow(5)~~ (S) \rightarrow(4)$
$(P) \rightarrow(2)~~ (Q) \rightarrow(1) ~~ (R) \rightarrow(4) ~~ (S) \rightarrow(5)$
$(P) \rightarrow(2) ~~ (Q) \rightarrow(1) ~~ (R) \rightarrow(1) ~~ (S) \rightarrow(3)$
Explanation
Given
$\begin{gathered}x+2 y+z=7 \\\\ x+\alpha z=11 \\\\ 2 x-3 y+\beta z=\gamma\end{gathered}$
Using Cramer's rule
$$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{array}\right| \\\\ & =1(3 \alpha)-2(\beta-2 \alpha)+1(-3) \\\\ & =3 \alpha-2 \beta+4 \alpha-3 \\\\ & =7 \alpha-2 \beta-3 \end{aligned} $$
$\begin{aligned} \Delta_x & =\left|\begin{array}{ccc}7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta\end{array}\right| \\\\ & =7(3 \alpha)-2(11 \beta-\gamma \alpha)+1(-33) \\\\ & =21 \alpha-22 \beta+22 \gamma-33\end{aligned}$
$\begin{aligned} \Delta_y & =\left|\begin{array}{ccc}1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta\end{array}\right| \\\\ & =1(11 \beta-\alpha \gamma)-7(\beta-2 \alpha)+1(\gamma-22) \\\\ & =11 \beta-\alpha \gamma-7 \beta+14 \alpha+\gamma-22 \\\\ & =14 \alpha+4 \beta+\gamma-\alpha \gamma-22\end{aligned}$
$\begin{aligned} \Delta_z & =\left|\begin{array}{ccc}1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma\end{array}\right| \\\\ & =1(33)-2(\gamma-22)+7(-3) \\\\ & =33-2 \gamma+44-21 \\\\ & =-2 \gamma+56\end{aligned}$
For unique solution $\Delta \neq 0$
For infinite solution
$$ \Delta=\Delta x=\Delta y=\Delta z=0 $$
For no solution $\Delta=0$ and atleast one in $\Delta x, \Delta y, \Delta z$ is non zero.
$$ \Delta=0 $$
$$ \Rightarrow \beta=\frac{1}{2}(7 \alpha-3) $$
(P) $ \beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$
then $ \Delta=0, \Delta x=\Delta y=\Delta z=0$
$\therefore$ Infinite solution
(Q) $ \beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$
$\therefore \Delta=0$ and $\Delta_2 \neq 0$
$\Rightarrow$ No solution.
$$ \begin{aligned} & \text { (R) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma \neq 28 \\\\ & \Rightarrow \Delta \neq 0=\text { unique solution } \\\\ & \begin{aligned} & \text { (S) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma=28 \\\\ & \therefore \Delta \neq 0, \Delta=4-2 \beta \\\\ & \Delta x=44-22 \beta \\\\ & \Delta y=4 \beta-8 \\\\ & \Delta z=0 \end{aligned} \end{aligned} $$
$\therefore x=11, y=-2, z=0$ is the solution.
$\begin{gathered}x+2 y+z=7 \\\\ x+\alpha z=11 \\\\ 2 x-3 y+\beta z=\gamma\end{gathered}$
Using Cramer's rule
$$ \begin{aligned} \Delta & =\left|\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{array}\right| \\\\ & =1(3 \alpha)-2(\beta-2 \alpha)+1(-3) \\\\ & =3 \alpha-2 \beta+4 \alpha-3 \\\\ & =7 \alpha-2 \beta-3 \end{aligned} $$
$\begin{aligned} \Delta_x & =\left|\begin{array}{ccc}7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta\end{array}\right| \\\\ & =7(3 \alpha)-2(11 \beta-\gamma \alpha)+1(-33) \\\\ & =21 \alpha-22 \beta+22 \gamma-33\end{aligned}$
$\begin{aligned} \Delta_y & =\left|\begin{array}{ccc}1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta\end{array}\right| \\\\ & =1(11 \beta-\alpha \gamma)-7(\beta-2 \alpha)+1(\gamma-22) \\\\ & =11 \beta-\alpha \gamma-7 \beta+14 \alpha+\gamma-22 \\\\ & =14 \alpha+4 \beta+\gamma-\alpha \gamma-22\end{aligned}$
$\begin{aligned} \Delta_z & =\left|\begin{array}{ccc}1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma\end{array}\right| \\\\ & =1(33)-2(\gamma-22)+7(-3) \\\\ & =33-2 \gamma+44-21 \\\\ & =-2 \gamma+56\end{aligned}$
For unique solution $\Delta \neq 0$
For infinite solution
$$ \Delta=\Delta x=\Delta y=\Delta z=0 $$
For no solution $\Delta=0$ and atleast one in $\Delta x, \Delta y, \Delta z$ is non zero.
$$ \Delta=0 $$
$$ \Rightarrow \beta=\frac{1}{2}(7 \alpha-3) $$
(P) $ \beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$
then $ \Delta=0, \Delta x=\Delta y=\Delta z=0$
$\therefore$ Infinite solution
(Q) $ \beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$
$\therefore \Delta=0$ and $\Delta_2 \neq 0$
$\Rightarrow$ No solution.
$$ \begin{aligned} & \text { (R) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma \neq 28 \\\\ & \Rightarrow \Delta \neq 0=\text { unique solution } \\\\ & \begin{aligned} & \text { (S) } \beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma=28 \\\\ & \therefore \Delta \neq 0, \Delta=4-2 \beta \\\\ & \Delta x=44-22 \beta \\\\ & \Delta y=4 \beta-8 \\\\ & \Delta z=0 \end{aligned} \end{aligned} $$
$\therefore x=11, y=-2, z=0$ is the solution.
Comments (0)
