Let `A=[[0,2q,r] , [p,q,-r] , [p,-q,r]]` If `A A^T=I_3` then `|p|=`

Options

A. `(1)/(sqrt(2))`
B. `(1)/(sqrt(5))`
C. `(1)/(sqrt(6))`
D. `(1)/(sqrt(3))`

Answer

To solve the problem, we need to find the absolute value of \( p \) given the matrix \( A \) and the condition \( A A^T = I_3 \).

### Step-by-Step Solution:

1. **Define the Matrix \( A \)**:
\[ A = \begin{bmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{bmatrix} \]

2. **Compute the Transpose of \( A \)**:
\[ A^T = \begin{bmatrix} 0 & p & p \\ 2q & q & -q \\ r & -r & r \end{bmatrix} \]

3. **Multiply \( A \) and \( A^T \)**:
We need to calculate \( A A^T \):
\[ A A^T = \begin{bmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{bmatrix} \begin{bmatrix} 0 & p & p \\ 2q & q & -q \\ r & -r & r \end{bmatrix} \]

The resulting matrix will be:
\[ A A^T = \begin{bmatrix} (0)(0) + (2q)(2q) + (r)(r) & (0)(p) + (2q)(q) + (r)(-r) & (0)(p) + (2q)(-q) + (r)(r) \\ (p)(0) + (q)(2q) + (-r)(r) & (p)(p) + (q)(q) + (-r)(-r) & (p)(p) + (q)(-q) + (-r)(r) \\ (p)(0) + (-q)(2q) + (r)(r) & (p)(p) + (-q)(q) + (r)(-r) & (p)(p) + (-q)(-q) + (r)(r) \end{bmatrix} \]

Simplifying each entry:
- First row, first column: \( 4q^2 + r^2 \)
- First row, second column: \( 2q^2 - r^2 \)
- First row, third column: \( -2q^2 + r^2 \)
- Second row, first column: \( 2q^2 - r^2 \)
- Second row, second column: \( p^2 + q^2 + r^2 \)
- Second row, third column: \( p^2 - q^2 - r^2 \)
- Third row, first column: \( -2q^2 + r^2 \)
- Third row, second column: \( p^2 - q^2 - r^2 \)
- Third row, third column: \( p^2 + q^2 + r^2 \)

Therefore, we have:
\[ A A^T = \begin{bmatrix} 4q^2 + r^2 & 2q^2 - r^2 & -2q^2 + r^2 \\ 2q^2 - r^2 & p^2 + q^2 + r^2 & p^2 - q^2 - r^2 \\ -2q^2 + r^2 & p^2 - q^2 - r^2 & p^2 + q^2 + r^2 \end{bmatrix} \]

4. **Set \( A A^T \) equal to the Identity Matrix \( I_3 \)**:
\[ A A^T = I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

This gives us the following equations:
- \( 4q^2 + r^2 = 1 \) (1)
- \( 2q^2 - r^2 = 0 \) (2)
- \( p^2 + q^2 + r^2 = 1 \) (3)

5. **Solve the equations**:
From equation (2):
\[ r^2 = 2q^2 \]

Substitute \( r^2 \) in equation (1):
\[ 4q^2 + 2q^2 = 1 \implies 6q^2 = 1 \implies q^2 = \frac{1}{6} \]

Now substitute \( q^2 \) back to find \( r^2 \):
\[ r^2 = 2q^2 = 2 \cdot \frac{1}{6} = \frac{1}{3} \]

Now substitute \( q^2 \) and \( r^2 \) into equation (3):
\[ p^2 + \frac{1}{6} + \frac{1}{3} = 1 \implies p^2 + \frac{1}{6} + \frac{2}{6} = 1 \implies p^2 + \frac{3}{6} = 1 \implies p^2 = 1 - \frac{1}{2} = \frac{1}{2} \]

6. **Find \( |p| \)**:
\[ |p| = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Final Answer:
\[ |p| = \frac{\sqrt{2}}{2} \]