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The question asks to find the value of \(\alpha\) such that the determinant of the given matrix is equal to zero. The matrix provided is:

\[
\begin{pmatrix}
\alpha & 3 & 2 \\
1 & 2\alpha & 6 \\
3 & 3 & 4
\end{pmatrix}
\]

To solve for \(\alpha\), we need to calculate the determinant of the matrix and set it equal to zero.

### Step 1: Calculate the determinant of the matrix.

The determinant of a \(3 \times 3\) matrix \(\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\) is given by:

\[
\text{det} = a(ei - fh) - b(di - fg) + c(dh - eg)
\]

Substituting the elements from the matrix:

\[
\text{det} = \alpha \left(2\alpha \cdot 4 - 6 \cdot 3\right) - 3 \left(1 \cdot 4 - 6 \cdot 3\right) + 2 \left(1 \cdot 3 - 2\alpha \cdot 3\right)
\]

### Step 2: Expand and simplify the determinant expression.

\[
\text{det} = \alpha \left(8\alpha - 18\right) - 3 \left(4 - 18\right) + 2 \left(3 - 6\alpha\right)
\]

\[
\text{det} = \alpha \left(8\alpha - 18\right) - 3 \left(-14\right) + 2 \left(3 - 6\alpha\right)
\]

\[
\text{det} = 8\alpha^2 - 18\alpha + 42 + 6 - 12\alpha
\]

\[
\text{det} = 8\alpha^2 - 30\alpha + 48
\]

### Step 3: Set the determinant equal to zero and solve for \(\alpha\).

\[
8\alpha^2 - 30\alpha + 48 = 0
\]

This is a quadratic equation, which can be solved using the quadratic formula:

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

Here, \(a = 8\), \(b = -30\), and \(c = 48\). Substituting these values into the quadratic formula:

\[
\alpha = \frac{-(-30) \pm \sqrt{(-30)^2 - 4 \cdot 8 \cdot 48}}{2 \cdot 8}
\]

\[
\alpha = \frac{30 \pm \sqrt{900 - 1536}}{16}
\]

\[
\alpha = \frac{30 \pm \sqrt{-636}}{16}
\]

Since the discriminant (\(\sqrt{-636}\)) is negative, the quadratic equation does not have real solutions. Therefore, the matrix does not have a real value for \(\alpha\) that makes the determinant equal to zero.

### Conclusion
There are no real values of \(\alpha\) that satisfy the condition for the determinant of the matrix to be zero.

Would you like more details or have any questions? Here are 5 related questions:

1. How do you calculate the determinant of larger matrices (e.g., 4x4)?
2. Can you explain why a negative discriminant results in no real solutions?
3. How can you determine if a matrix is singular without calculating its determinant?
4. What is the significance of the determinant being zero for a matrix?
5. How would you solve a quadratic equation with a negative discriminant in the context of complex numbers?

**Tip:** When solving quadratic equations, always check the discriminant first to determine the nature of the solutions (real or complex).

Learn how to solve for the value of alpha in a matrix determinant equation. Understand the steps involved in calculating the determinant of a 3x3 matrix and solving the resulting quadratic equation. Suitable for high school students studying matrix algebra and quadratic equations.

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