Let $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$$ and $$|2A|^3 = 2^{21}$$ where $$\alpha, \beta \in Z$$, Then a value of $$\alpha$$ is

Given matrix $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$$ and the condition $$|2A|^3 = 2^{21}$$, with $$\alpha, \beta \in \mathbb{Z}$$.

First, recall that for an $$n \times n$$ matrix, $$|kA| = k^n |A|$$. Here, $$A$$ is a 3x3 matrix, so $$n=3$$. Thus, $$|2A| = 2^3 |A| = 8 |A|$$.

Substitute into the given condition: $$(8 |A|)^3 = 2^{21}$$. Since $$8 = 2^3$$, this becomes: $$(2^3 |A|)^3 = 2^{21}$$. Simplifying the exponents: $$2^{9} |A|^3 = 2^{21}$$.

Divide both sides by $$2^9$$: $$|A|^3 = 2^{12}$$. Taking the cube root: $$|A| = 2^{4} = 16$$. Therefore, the determinant of $$A$$ is 16.

Now, compute the determinant of $$A$$. Expanding along the first row: $$|A| = 1 \cdot \begin{vmatrix} \alpha & \beta \\ \beta & \alpha \end{vmatrix} - 0 \cdot (\ldots) + 0 \cdot (\ldots) = \alpha^2 - \beta^2$$.

So, $$\alpha^2 - \beta^2 = 16$$. Factorizing: $$(\alpha - \beta)(\alpha + \beta) = 16$$. Since $$\alpha$$ and $$\beta$$ are integers, both $$\alpha - \beta$$ and $$\alpha + \beta$$ are integers and must be even (as their product is even and both must have the same parity).

Set $$x = \alpha - \beta$$ and $$y = \alpha + \beta$$, so $$x \cdot y = 16$$, and both $$x$$ and $$y$$ are even integers. The even factor pairs of 16 are: $$(2,8)$$, $$(4,4)$$, $$(8,2)$$, $$(-2,-8)$$, $$(-4,-4)$$, $$(-8,-2)$$.

Solve for $$\alpha$$ and $$\beta$$ in each pair using $$\alpha = \frac{x + y}{2}$$ and $$\beta = \frac{y - x}{2}$$:

• For $$(2,8)$$: $$\alpha = \frac{2+8}{2} = 5$$, $$\beta = \frac{8-2}{2} = 3$$
• For $$(4,4)$$: $$\alpha = \frac{4+4}{2} = 4$$, $$\beta = \frac{4-4}{2} = 0$$
• For $$(8,2)$$: $$\alpha = \frac{8+2}{2} = 5$$, $$\beta = \frac{2-8}{2} = -3$$
• For $$(-2,-8)$$: $$\alpha = \frac{-2-8}{2} = -5$$, $$\beta = \frac{-8-(-2)}{2} = -3$$
• For $$(-4,-4)$$: $$\alpha = \frac{-4-4}{2} = -4$$, $$\beta = \frac{-4-(-4)}{2} = 0$$
• For $$(-8,-2)$$: $$\alpha = \frac{-8-2}{2} = -5$$, $$\beta = \frac{-2-(-8)}{2} = 3$$

The possible values of $$\alpha$$ are $$5, 4, -5, -4$$. Comparing with the options: A. $$3$$, B. $$5$$, C. $$17$$, D. $$9$$, the value $$\alpha = 5$$ is present (option B).

Verification for $$\alpha = 5$$, $$\beta = 3$$: $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 5 & 3 \\ 0 & 3 & 5 \end{bmatrix}$$, $$|A| = 5^2 - 3^2 = 25 - 9 = 16$$. Then $$|2A| = 8 \times 16 = 128$$, and $$|2A|^3 = 128^3 = (2^7)^3 = 2^{21}$$, which satisfies the condition.

Checking other options:

• For $$\alpha = 3$$: $$\alpha^2 - \beta^2 = 9 - \beta^2 = 16 \implies \beta^2 = -7$$, not real.
• For $$\alpha = 17$$: $$289 - \beta^2 = 16 \implies \beta^2 = 273$$, not a perfect square.
• For $$\alpha = 9$$: $$81 - \beta^2 = 16 \implies \beta^2 = 65$$, not a perfect square.

Thus, the only valid value from the options is $$\alpha = 5$$.