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Question

Mathematics Question on complex numbers

The values of α\alpha, for which
Problem Figure
lie in the interval

A

(-2, 1)

B

(-3, 0)

C

(−32,32)\left(-\frac{3}{2}, \frac{3}{2}\right)

D

(0, 3)

Answer

(-3, 0)

Explanation

Solution

To find the values of aa, we expand the determinant:

∣113a+32 11a+13 2a+33a+10∣.\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⋅(1⋅0−(a+13)(3a+1))−32⋅(1⋅0−(a+13)(2a+3))+(a+32)⋅(1⋅(3a+1)−1⋅(2a+3)).= 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+13)(3a+1)+32(a+13)(2a+3)+(a+32)(a−2).= -(a + \frac{1}{3})(3a + 1) + \frac{3}{2}(a + \frac{1}{3})(2a + 3) + (a + \frac{3}{2})(a - 2).

Expanding the products:
=−3a2−a−13−3a−13+32(2a2+3a+23).= -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 aa. Solving for aa gives:
a∈(−3,0).a \in (-3, 0).

The Correct answer is: (-3, 0)