JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 12)
If the system of equations
$$ \begin{aligned} & 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \end{aligned} $$
has infinitely many solutions, then $13 \alpha \beta$ is equal to :
$$ \begin{aligned} & 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \end{aligned} $$
has infinitely many solutions, then $13 \alpha \beta$ is equal to :
1110
1120
1210
1220
Explanation
$\begin{aligned} & \text { Given } 2 x+3 y-z=5 \\\\ & x+\alpha y+3 z=-4 \\\\ & 3 x-y+\beta z=7 \\\\ & \Delta_2=\left|\begin{array}{ccc}2 & -1 & 5 \\ 1 & 3 & -4 \\ 3 & \beta & 7\end{array}\right| \\\\ & \Delta_2=2(21+4 \beta)+1(7+12)+5(\beta-9) \\\\& \Delta_2=42+8 \beta+19+5 \beta-45 \\\\ & \Delta_2=13 \beta+16 \\\\ & \Delta_2=0\end{aligned}$
$\begin{aligned} & \therefore \beta=-\frac{16}{13} \\\\ & \Delta_3=\left|\begin{array}{lll}2 & 3 & 5 \\ 1 & \alpha & -4 \\ 3 & -1 & 7\end{array}\right| \\\\ & \Delta_3=2(7 \alpha-4)-3(7+12)+5(-1-3 \alpha) \\\\ & \Delta_3=14 \alpha-8-57-5-15 \alpha \\\\ & \Delta_3=-\alpha-70\end{aligned}$
$\begin{aligned} & \Delta_3=0 \\\\ & \alpha=-70 \\\\ & 13 \alpha \beta=(13)(-70)\left(-\frac{16}{13}\right) \\\\ & =+1120\end{aligned}$
$\begin{aligned} & \therefore \beta=-\frac{16}{13} \\\\ & \Delta_3=\left|\begin{array}{lll}2 & 3 & 5 \\ 1 & \alpha & -4 \\ 3 & -1 & 7\end{array}\right| \\\\ & \Delta_3=2(7 \alpha-4)-3(7+12)+5(-1-3 \alpha) \\\\ & \Delta_3=14 \alpha-8-57-5-15 \alpha \\\\ & \Delta_3=-\alpha-70\end{aligned}$
$\begin{aligned} & \Delta_3=0 \\\\ & \alpha=-70 \\\\ & 13 \alpha \beta=(13)(-70)\left(-\frac{16}{13}\right) \\\\ & =+1120\end{aligned}$
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