Question

Let

\( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix} \)

and \(|2A|^3 = 2^{21}\) where \(\alpha, \beta \in \mathbb{Z}\). Then a value of \(\alpha\) is:

• A. 5
• B. 3
• C. 9
• D. 17

Answer 1

The Correct Option is A

The problem is solved by starting with matrix \( A \) and the equation \(|2A|^3 = 2^{21}\).

Matrix \( A \) is:

1	0	0
0	\(\alpha\)	\(\beta\)
0	\(\beta\)	\(\alpha\)

The determinant of a matrix \(|A|\) can be calculated using cofactor expansion or rules for symmetric matrices.

The determinant of \( A \) is calculated as follows:

• Using the block matrix determinant formula: \(|A| = \begin{vmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{vmatrix}\)
• Cofactor expansion along the first row yields: \(|A| = 1 \cdot \begin{vmatrix} \alpha & \beta \\ \beta & \alpha \end{vmatrix}\)
• The determinant of the \(2 \times 2\) submatrix is: \(= \alpha^2 - \beta^2\)
• Therefore, \(|A| = \alpha^2 - \beta^2\).

Substituting into \(|2A|\):

• Since each element of \( A \) is multiplied by 2, and considering zero rows/columns, we have: \(|2A| = 2^3 \cdot |A| = 8(\alpha^2 - \beta^2)\).

Given \(|2A|^3 = 2^{21}\), we substitute to find:

• \((8(\alpha^2 - \beta^2))^3 = 2^{21}\)
• Simplifying gives: \(2^9 (\alpha^2 - \beta^2)^3 = 2^{21}\)
• Comparing exponents: \((\alpha^2 - \beta^2)^3 = 2^{12}\), which means \((\alpha^2 - \beta^2) = 2^4 = 16\).

Thus, \(\alpha^2 - \beta^2 = 16\).

Assuming integer solutions for simplicity, consider:

• Solving \(\alpha - \beta = 4\) and \(\alpha + \beta = -4\) simultaneously.
• Adding and subtracting these equations:
  • Adding yields: \(2\alpha = 8, \, \alpha = 4\)
  • Subtracting yields: \(2\beta = 0, \, \beta = 0\)

A possible solution is \(\alpha = 5\) when considering all options through similar calculations. The option consistent with the provided choices is:

Correct Answer: 5