Question

The values of $\alpha$, for which $\begin{array}{ccc} 1 & \frac{3}{2} & \alpha +3 \\ 1 & 2 & \alpha + \frac{1}{3} \\ 2\alpha +3 & 3\alpha +1 & 0 \end{array}$ = 0, lie in the interval [JEE (Main)-2024]

• A. (-2, 1)
• B. (0, 3)
• C. (-3, 0)
• D. (-$\frac{3}{2}$,$\frac{3}{2}$)

Answer 1

No real values of $\alpha$ satisfy the equation.

Answer 2

The given determinant is:

$$D = \begin{vmatrix} 1 & \frac{3}{2} & \alpha +3 \\ 1 & 2 & \alpha + \frac{1}{3} \\ 2\alpha +3 & 3\alpha +1 & 0 \end{vmatrix}$$

We need to find the values of $\alpha$ for which $D = 0$. Expanding the determinant along the first row:

$$D = 1 \begin{vmatrix} 2 & \alpha + \frac{1}{3} \\ 3\alpha +1 & 0 \end{vmatrix} - \frac{3}{2} \begin{vmatrix} 1 & \alpha + \frac{1}{3} \\ 2\alpha +3 & 0 \end{vmatrix} + (\alpha +3) \begin{vmatrix} 1 & 2 \\ 2\alpha +3 & 3\alpha +1 \end{vmatrix}$$

Calculate each minor:

1. $1 \times [2(0) - (\alpha + \frac{1}{3})(3\alpha +1)] = -(\alpha + \frac{1}{3})(3\alpha +1) = -\frac{(3\alpha+1)^2}{3}$
2. $-\frac{3}{2} \times [1(0) - (\alpha + \frac{1}{3})(2\alpha +3)] = \frac{3}{2} (\alpha + \frac{1}{3})(2\alpha +3) = \frac{1}{2}(3\alpha+1)(2\alpha+3)$
3. $(\alpha +3) \times [1(3\alpha +1) - 2(2\alpha +3)] = (\alpha +3) [3\alpha +1 - 4\alpha - 6] = (\alpha +3) (-\alpha - 5) = -(\alpha +3)(\alpha + 5)$

Now, sum these terms and set $D=0$:

$$-\frac{(3\alpha+1)^2}{3} + \frac{1}{2}(3\alpha+1)(2\alpha +3) - (\alpha +3)(\alpha + 5) = 0$$

Expanding and simplifying:

$$-\alpha^2 - \frac{9}{2}\alpha - \frac{83}{6} = 0$$

Multiply by -6 to clear denominators:

$$6\alpha^2 + 27\alpha + 83 = 0$$

To find the values of $\alpha$, we calculate the discriminant $\Delta = b^2 - 4ac$:

$$\Delta = (27)^2 - 4(6)(83) = 729 - 1992 = -1263$$

Since the discriminant $\Delta < 0$, the quadratic equation has no real roots.

Therefore, there are no real values of $\alpha$ for which the determinant is zero.