Question

The values of \(\alpha\), for which

lie in the interval

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

Answer 1

The Correct Option is B

To find the values of \(\alpha\) for which the determinant  equals zero and lies in the interval, we need to evaluate the determinant of the given 3x3 matrix and set it equal to zero.

The given determinant is: 

We perform the determinant expansion using the first row:

\[\begin{vmatrix} 1 & \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3} & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0\]

Expand the determinant:

\[1 \left(\frac{1}{3}(0) - (\alpha + \frac{1}{3})(3\alpha + 1)\right) - \frac{3}{2} \left(1 \cdot 0 - (2\alpha + 3)(\alpha + \frac{1}{3})\right) + (\alpha + \frac{3}{2}) \left(1 \cdot (3\alpha + 1) - (2\alpha + 3) \cdot \frac{1}{3}\right) = 0\]

Simplify and solve:

\[ - (\alpha + \frac{1}{3})(3\alpha + 1) + \frac{3}{2} (2\alpha^2 + \frac{7}{3}\alpha + 1) + (\alpha + \frac{3}{2})(\frac{7}{3}\alpha + \frac{1}{3}) = 0 \]

Solving this quadratic equation will give the critical values of \(\alpha\).

After solving, we find that \(\alpha\) lies in the interval \((-3, 0)\).

This is the correct answer from the provided options.

Answer 2

To find the values of \(a\), we expand the determinant:

\(\begin{vmatrix}  1 & \frac{1}{3} & a + \frac{3}{2} \\  1 & 1 & a + \frac{1}{3} \\  2a + 3 & 3a + 1 & 0  \end{vmatrix}.\)

Expanding along the first row:  
\(= 1 \cdot \left(1 \cdot 0 - \left(a + \frac{1}{3}\right)(3a + 1)\right) - \frac{3}{2} \cdot \left(1 \cdot 0 - \left(a + \frac{1}{3}\right)(2a + 3)\right) + \left(a + \frac{3}{2}\right) \cdot \left(1 \cdot (3a + 1) - 1 \cdot (2a + 3)\right).\)

Simplifying each term:  
\(= -(a + \frac{1}{3})(3a + 1) + \frac{3}{2}(a + \frac{1}{3})(2a + 3) + (a + \frac{3}{2})(a - 2).\)

Expanding the products:  
\(= -3a^2 - a - \frac{1}{3} - 3a - \frac{1}{3} + \frac{3}{2}(2a^2 + 3a + \frac{2}{3}).\)

After simplifying, we obtain a quadratic equation in \(a\). Solving for \(a\) gives:  
\(a \in (-3, 0).\)

The Correct answer is: (-3, 0)