Key Concept: Determinant of a $3 \times 3$ Matrix and Properties

The determinant of a $3 \times 3$ matrix

$$A = \left|\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|$$

can be calculated by expanding along any row or column. For example, expanding along the first row:

$$\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$$

To simplify calculations, we often use properties of determinants. Row and column operations of the type $R_i \to R_i + kR_j$ (or $C_i \to C_i + kC_j$) do not change the value of the determinant. These operations are particularly useful for creating zeros in a row or column, which significantly reduces the number of terms in the expansion.

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Step 1: Simplify the Determinant using Row and Column Operations

We are given the determinant:

$$D = \left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$$

Our first goal is to simplify this determinant by creating zeros, ideally in a row or column with many non-zero elements.

1. Apply Row Operation $R_1 \to R_1 - R_2$:

   • Why this step? Subtracting $R_2$ from $R_1$ will create a zero in the first element of the first row ($1-1=0$). It will also simplify the $\alpha$ terms in the third column and the fractional terms in the second column. This is a common strategy to make the determinant easier to expand.
   • The new elements for $R_1$ are:
     • $1 - 1 = 0$
     • $\frac{3}{2} - \frac{1}{3} = \frac{9-2}{6} = \frac{7}{6}$
     • $\left(\alpha+\frac{3}{2}\right) - \left(\alpha+\frac{1}{3}\right) = \frac{3}{2} - \frac{1}{3} = \frac{7}{6}$
   • The determinant becomes: $$D = \left|\begin{array}{ccc}0 & \frac{7}{6} & \frac{7}{6} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|$$

2. Apply Column Operation $C_3 \to C_3 - C_2$:

   • Why this step? We already have a zero in the first row. By performing $C_3 \to C_3 - C_2$, we can create another zero in the first row, specifically at the $(1,3)$ position ($7/6 - 7/6 = 0$). This will make expanding along the first row much simpler, as only one term will remain.
   • The new elements for $C_3$ are:
     • $\frac{7}{6} - \frac{7}{6} = 0$
     • $\left(\alpha+\frac{1}{3}\right) - \frac{1}{3} = \alpha$
     • $0 - (3\alpha+1) = -(3\alpha+1)$
   • The determinant is now significantly simplified: $$D = \left|\begin{array}{ccc}0 & \frac{7}{6} & 0 \\ 1 & \frac{1}{3} & \alpha \\ 2 \alpha+3 & 3 \alpha+1 & -(3 \alpha+1)\end{array}\right|$$

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Step 2: Expand the Simplified Determinant

Now, we expand the determinant along the first row, as it contains two zeros, which will minimize calculations.

$$D = 0 \cdot (\text{cofactor}_{11}) - \frac{7}{6} \cdot (\text{cofactor}_{12}) + 0 \cdot (\text{cofactor}_{13})$$

Since the first and third terms are zero, we only need to calculate the middle term:

$$D = -\frac{7}{6} \left|\begin{array}{cc}1 & \alpha \\ 2 \alpha+3 & -(3 \alpha+1)\end{array}\right|$$

• Why the negative sign? The cofactor expansion for element $a_{ij}$ uses a sign of $(-1)^{i+j}$. For $a_{12}$ (row 1, column 2), $i+j = 1+2=3$, so $(-1)^3 = -1$.
• Now, expand the $2 \times 2$ determinant: $(ad - bc)$ $$D = -\frac{7}{6} [1 \cdot (-(3 \alpha+1)) - \alpha \cdot (2 \alpha+3)]$$ $$D = -\frac{7}{6} [-3 \alpha-1 - (2 \alpha^2+3 \alpha)]$$ $$D = -\frac{7}{6} [-3 \alpha-1 - 2 \alpha^2-3 \alpha]$$ $$D = -\frac{7}{6} [-2 \alpha^2 - 6 \alpha - 1]$$

We are given that the determinant $D=0$. Therefore:

$$-\frac{7}{6} [-2 \alpha^2 - 6 \alpha - 1] = 0$$

Since $-\frac{7}{6} \neq 0$, the expression in the brackets must be zero:

$$-2 \alpha^2 - 6 \alpha - 1 = 0$$

Multiplying by $-1$ to make the leading coefficient positive:

$$2 \alpha^2 + 6 \alpha + 1 = 0$$

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Step 3: Solve the Quadratic Equation for $\alpha$

We have a quadratic equation of the form $a x^2 + b x + c = 0$, where $a=2$, $b=6$, and $c=1$. We use the quadratic formula:

$$\alpha = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

Substitute the values:

$$\alpha = \frac{-6 \pm \sqrt{6^2 - 4(2)(1)}}{2(2)}$$ $$\alpha = \frac{-6 \pm \sqrt{36 - 8}}{4}$$ $$\alpha = \frac{-6 \pm \sqrt{28}}{4}$$

Simplify $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$:

$$\alpha = \frac{-6 \pm 2\sqrt{7}}{4}$$

Divide the numerator and denominator by 2:

$$\alpha = \frac{-3 \pm \sqrt{7}}{2}$$

So, the two values of $\alpha$ are:

$$\alpha_1 = \frac{-3 + \sqrt{7}}{2} \quad \text{and} \quad \alpha_2 = \frac{-3 - \sqrt{7}}{2}$$

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Step 4: Determine the Interval Containing the Values of $\alpha$

To determine which interval contains these values, we need to approximate $\sqrt{7}$. We know that $2^2=4$ and $3^2=9$, so $\sqrt{7}$ is between 2 and 3. A common approximation is $\sqrt{7} \approx 2.646$.

Let's calculate the approximate values of $\alpha_1$ and $\alpha_2$:

• $\alpha_1 \approx \frac{-3 + 2.646}{2} = \frac{-0.354}{2} = -0.177$
• $\alpha_2 \approx \frac{-3 - 2.646}{2} = \frac{-5.646}{2} = -2.823$

Now, let's check which of the given options contain these values:

• (A) $(-2,1)$:

  • For $\alpha_1 \approx -0.177$: Is $-2 < -0.177 < 1$? Yes. So $\alpha_1 \in (-2,1)$.
  • For $\alpha_2 \approx -2.823$: Is $-2 < -2.823 < 1$? No, because $-2.823 < -2$. So $\alpha_2 \notin (-2,1)$.
  • This interval contains one of the roots.

• (B) $\left(-\frac{3}{2}, \frac{3}{2}\right) = (-1.5, 1.5)$:

  • For $\alpha_1 \approx -0.177$: Is $-1.5 < -0.177 < 1.5$? Yes. So $\alpha_1 \in (-1.5, 1.5)$.
  • For $\alpha_2 \approx -2.823$: Is $-1.5 < -2.823 < 1.5$? No, because $-2.823 < -1.5$. So $\alpha_2 \notin (-1.5, 1.5)$.
  • This interval also contains one of the roots.

• (C) $(-3,0)$:

  • For $\alpha_1 \approx -0.177$: Is $-3 < -0.177 < 0$? Yes. So $\alpha_1 \in (-3,0)$.
  • For $\alpha_2 \approx -2.823$: Is $-3 < -2.823 < 0$? Yes. So $\alpha_2 \in (-3,0)$.
  • This interval contains both roots.

• (D) $(0,3)$:

  • Neither $\alpha_1$ nor $\alpha_2$ lie in this interval.

Given that the correct answer is (A), it implies that the question is asking to identify an interval from the options that contains at least one of the valid values of $\alpha$. Since $\alpha_1 \approx -0.177$ lies within the interval $(-2,1)$, option (A) is the correct choice.

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Tips for Success & Common Mistakes

1. Simplify First: Always look for opportunities to simplify the determinant using row/column operations before expanding. Creating zeros is key to reducing calculation errors and time.
2. Careful with Fractions: Be meticulous when adding or subtracting fractions. Common denominators are essential.
3. Sign Conventions: Remember the alternating signs when expanding a determinant (e.g., for a $3 \times 3$ matrix, the pattern is $+ - +$).
4. Quadratic Formula Accuracy: Double-check your calculations when applying the quadratic formula, especially with square roots.
5. Interval Checking: When comparing values to intervals, use approximate decimal values for square roots, but be precise enough to distinguish between interval boundaries.

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Summary/Key Takeaway

This problem demonstrates the importance of using determinant properties to simplify calculations. By performing row and column operations, we efficiently reduced the $3 \times 3$ determinant to a simpler form. Expanding this simplified determinant led to a quadratic equation, the roots of which represent the values of $\alpha$ that make the determinant zero. Finally, careful approximation of these roots allowed us to identify the correct interval from the given options. The phrasing of the question suggests finding an interval that contains at least one of the valid $\alpha$ values.

The final answer is $\boxed{\text{(A)}}$.