A relation \( R = \{(a, b) : a = b - 2, b \geq 6 \} \) is defined on the set \( \mathbb{N} \). Then the correct answer will be:
The given relation is \( R = \{(a, b) : a = b - 2, b \geq 6\} \).
For each pair:
For \( (2, 4) \): \( a = 4 - 2 = 2 \) and \( b = 4 \).
Since \( b \geq 6 \) is not satisfied, \( (2, 4) \notin R \).
For \( (3, 8) \): \( a = 8 - 2 = 6 \) and \( b = 8 \). Since \( a \neq 3 \), \( (3, 8) \notin R \).
For \( (6, 8) \): \( a = 8 - 2 = 6 \) and \( b = 8 \).
Both conditions \( a = b - 2 \) and \( b \geq 6 \) are satisfied, so \( (6, 8) \in R \).
For \( (8, 7) \): \( a = 7 - 2 = 5 \) and \( b = 7 \).
Since \( a \neq 8 \), \( (8, 7) \notin R \).
Thus, the correct pair is \( (6, 8) \).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Find the values of \( x, y, z \) if the matrix \( A \) satisfies the equation \( A^T A = I \), where
\[ A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \]