Let A = `[{:(0, 2b , c),(a, b, -c),(a, -b, c):}]` be an orthogonal matrix, then the values of a, b, c, are related with

Options

A. `b = pm (1)/(sqrt(6)) , c = pm (1)/(sqrt(3)) `
B. `a = pm (1)/(sqrt(2)) , c = pm (1)/(sqrt(3)) `
C. `a = pm (1)/(sqrt(2)) , b = pm (1)/(sqrt(6)) `
D. All of these

Answer

To determine the relationship between the values of \( a, b, c \) in the orthogonal matrix \( A = \begin{pmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \end{pmatrix} \), we need to use the property of orthogonal matrices. An orthogonal matrix \( A \) satisfies the condition:

\[ A A^T = I \]

where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.

### Step 1: Find the transpose of matrix \( A \)

The transpose of \( A \) is given by swapping rows with columns:

\[ A^T = \begin{pmatrix} 0 & a & a \\ 2b & b & -b \\ c & -c & c \end{pmatrix} \]

### Step 2: Multiply \( A \) by \( A^T \)

Now, we will calculate the product \( A A^T \):

\[ A A^T = \begin{pmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \end{pmatrix} \begin{pmatrix} 0 & a & a \\ 2b & b & -b \\ c & -c & c \end{pmatrix} \]

Calculating the elements of the resulting matrix:

- First row:
- First element: \( 0 \cdot 0 + 2b \cdot 2b + c \cdot c = 4b^2 + c^2 \)
- Second element: \( 0 \cdot a + 2b \cdot b + c \cdot (-c) = 2b^2 - c^2 \)
- Third element: \( 0 \cdot a + 2b \cdot (-b) + c \cdot c = -2b^2 + c^2 \)

- Second row:
- First element: \( a \cdot 0 + b \cdot 2b + (-c) \cdot c = 2b^2 - c^2 \)
- Second element: \( a \cdot a + b \cdot b + (-c) \cdot (-c) = a^2 + b^2 + c^2 \)
- Third element: \( a \cdot a + b \cdot (-b) + (-c) \cdot c = a^2 - b^2 + c^2 \)

- Third row:
- First element: \( a \cdot 0 + (-b) \cdot 2b + c \cdot c = -2b^2 + c^2 \)
- Second element: \( a \cdot a + (-b) \cdot b + c \cdot (-c) = a^2 - b^2 - c^2 \)
- Third element: \( a \cdot a + (-b) \cdot (-b) + c \cdot c = a^2 + b^2 + c^2 \)

So, we have:

\[ A A^T = \begin{pmatrix} 4b^2 + c^2 & 2b^2 - c^2 & -2b^2 + c^2 \\ 2b^2 - c^2 & a^2 + b^2 + c^2 & a^2 - b^2 + c^2 \\ -2b^2 + c^2 & a^2 - b^2 - c^2 & a^2 + b^2 + c^2 \end{pmatrix} \]

### Step 3: Set \( A A^T \) equal to the identity matrix \( I \)

The identity matrix \( I \) is:

\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Equating the elements of \( A A^T \) with \( I \):

1. From the first row and first column:
\[ 4b^2 + c^2 = 1 \tag{1} \]

2. From the first row and second column:
\[ 2b^2 - c^2 = 0 \tag{2} \]

3. From the second row and second column:
\[ a^2 + b^2 + c^2 = 1 \tag{3} \]

4. From the second row and third column:
\[ a^2 - b^2 + c^2 = 0 \tag{4} \]

### Step 4: Solve the equations

From equation (2):
\[ c^2 = 2b^2 \tag{5} \]

Substituting (5) into (1):
\[ 4b^2 + 2b^2 = 1 \implies 6b^2 = 1 \implies b^2 = \frac{1}{6} \implies b = \pm \frac{1}{\sqrt{6}} \]

Using (5) to find \( c \):
\[ c^2 = 2b^2 = 2 \cdot \frac{1}{6} = \frac{1}{3} \implies c = \pm \frac{1}{\sqrt{3}} \]

Now, substituting \( b^2 \) and \( c^2 \) into (3):
\[ a^2 + \frac{1}{6} + \frac{1}{3} = 1 \implies a^2 + \frac{1}{6} + \frac{2}{6} = 1 \implies a^2 + \frac{3}{6} = 1 \implies a^2 = 1 - \frac{1}{2} = \frac{1}{2} \implies a = \pm \frac{1}{\sqrt{2}} \]

### Final Result

The values of \( a, b, c \) are related as follows:
- \( a = \pm \frac{1}{\sqrt{2}} \)
- \( b = \pm \frac{1}{\sqrt{6}} \)
- \( c = \pm \frac{1}{\sqrt{3}} \)