JEE Mathematics 2024 Question with Solution

Given: The system

$$x - 2y + z = -4$$ $$2x + \alpha y + 3z = 5$$ $$3x - y + \beta z = 3$$

has infinitely many solutions.

Find: The value of $12\alpha + 13\beta$.

For infinitely many solutions, we use the condition that the determinant and the relevant minors vanish. From the extracted working:

$$D = \begin{vmatrix} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{vmatrix} = 0$$

and

$$D_2 = \begin{vmatrix} 2 & 1 & 1 \\ 5 & \alpha & 3 \\ 3 & 3 & \beta \end{vmatrix} = 0$$

the solution gives

$$D = 0 \implies \alpha\beta - 3\alpha + 4\beta = 17 \qquad \text{....(1)}$$

and from $D_2 = 0$,

$$1(5\beta - 9) + 4(2\beta - 9) + 1(6 - 15) = 0$$ $$13\beta - 9 - 36 - 9 = 0$$ $$13\beta = 54 \Rightarrow \beta = \frac{54}{13}$$

Substitute this in equation $\text{(1)}$:

$$\frac{54}{13}\alpha - 3\alpha + 4\left(\frac{54}{13}\right) = 17$$ $$\frac{54}{13}\alpha - \frac{39}{13}\alpha + \frac{216}{13} = \frac{221}{13}$$ $$15\alpha = 5 \Rightarrow \alpha = \frac{1}{3}$$

Now compute:

$$12\alpha + 13\beta = 12 \cdot \frac{1}{3} + 13 \cdot \frac{54}{13}$$ $$= 4 + 54 = 58$$

Therefore, the correct option is D.

Given: A system of three linear equations in $x, y, z$.

Find: The value of $12\alpha + 13\beta$ when the system has infinitely many solutions.

Write the coefficient matrix as

$$A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{bmatrix}$$

For infinitely many solutions, the rank condition requires the determinant of the coefficient matrix to be zero, together with consistency conditions from the augmented system.

From the extracted solution:

$$D = 0 \implies \alpha\beta - 3\alpha + 4\beta = 17$$

Also,

$$D_2 = 0$$

leads to

$$13\beta = 54$$

so

$$\beta = \frac{54}{13}$$

Substituting into

$$\alpha\beta - 3\alpha + 4\beta = 17$$

we get

$$\alpha\left(\frac{54}{13}\right) - 3\alpha + 4\left(\frac{54}{13}\right) = 17$$

Multiply through by $13$:

$$54\alpha - 39\alpha + 216 = 221$$ $$15\alpha = 5$$ $$\alpha = \frac{1}{3}$$

Hence,

$$12\alpha + 13\beta = 12\left(\frac{1}{3}\right) + 13\left(\frac{54}{13}\right)$$ $$= 4 + 54 = 58$$

Therefore, the value is $58$, so the correct option is D.

Note: The first extracted approach contains inconsistent determinant expansion, but both the stated correct option and the second approach conclude $58$.