Determinants- System of Linear Equations – JEE Main 2 Apr 2025 Shift 1

Solution:

For a system to have infinitely many solutions, the determinant of the coefficient matrix ($\Delta$) must be 0, and the determinants formed by replacing any column with the constants ($\Delta_x, \Delta_y, \Delta_z$) must also be 0.

Step 1: Calculate $\Delta_z$ (replacing the $z$ column with constants 3, -3, 4) and set to 0 to find $\alpha$.
$$ \Delta_z = \begin{vmatrix} 3 & 1 & 3 \\ 2 & \alpha & -3 \\ 1 & 2 & 4 \end{vmatrix} = 0 $$
$$ 3(4\alpha + 6) – 1(8 + 3) + 3(4 – \alpha) = 0 $$
$$ 12\alpha + 18 – 11 + 12 – 3\alpha = 0 $$
$$ 9\alpha + 19 = 0 \Rightarrow \alpha = -\frac{19}{9} $$

Step 2: Calculate $\Delta$ and set to 0 to find $\beta$.
$$ \Delta = \begin{vmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0 $$
$$ 3(\alpha + 2) – 1(2 + 1) + \beta(4 – \alpha) = 0 $$
$$ 3\alpha + 6 – 3 + 4\beta – \alpha\beta = 0 $$
$$ 3\alpha + 3 + \beta(4 – \alpha) = 0 $$

Substitute $\alpha = -19/9$:
$$ 3\left(-\frac{19}{9}\right) + 3 + \beta\left(4 – \left(-\frac{19}{9}\right)\right) = 0 $$
$$ -\frac{19}{3} + 3 + \beta\left(\frac{36+19}{9}\right) = 0 $$
$$ \frac{-19+9}{3} + \beta\left(\frac{55}{9}\right) = 0 $$
$$ -\frac{10}{3} + \frac{55\beta}{9} = 0 \Rightarrow \frac{55\beta}{9} = \frac{30}{9} $$
$$ \beta = \frac{30}{55} = \frac{6}{11} $$

Step 3: Calculate $22\beta – 9\alpha$.
$$ 22\left(\frac{6}{11}\right) – 9\left(-\frac{19}{9}\right) = 2(6) + 19 $$
$$ = 12 + 19 = 31 $$

Ans. (2)