Question

If for the matrix, A = \( A = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix} \), and \( A A^T = I_2 \), then the value of \( \alpha^4 + \beta^4 \) is :

• A. 4
• B. 1
• C. 2
• D. 3

Hint:

When a matrix A satisfies AA$^T$ = I, it is called an orthogonal matrix. For an orthogonal matrix, the sum of the squares of the elements in any row or column is equal to 1.

Answer 1

The Correct Option is B

Given the matrix \( A = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix} \).
Its transpose is \( A^T = \begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix} \).
The given condition is \( A A^T = I_2 \).
\( A A^T = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix} \begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix} = \begin{bmatrix} (1)(1) + (-\alpha)(-\alpha) & (1)(\alpha) + (-\alpha)(\beta) \\ (\alpha)(1) + (\beta)(-\alpha) & (\alpha)(\alpha) + (\beta)(\beta) \end{bmatrix} \).
\( A A^T = \begin{bmatrix} 1 + \alpha^2 & \alpha - \alpha\beta \\ \alpha - \alpha\beta & \alpha^2 + \beta^2 \end{bmatrix} \).
We are given \( A A^T = I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Equating the corresponding elements:
1. \( 1 + \alpha^2 = 1 \implies \alpha^2 = 0 \implies \alpha = 0 \).
2. \( \alpha^2 + \beta^2 = 1 \implies 0^2 + \beta^2 = 1 \implies \beta^2 = 1 \).
We need to find the value of \( \alpha^4 + \beta^4 \).
\( \alpha^4 = (\alpha^2)^2 = 0^2 = 0 \).
\( \beta^4 = (\beta^2)^2 = 1^2 = 1 \).
Therefore, \( \alpha^4 + \beta^4 = 0 + 1 = 1 \).