If `A=1/3{:[(1,2,2),(2,1,-2),(a,2,b)]:}` is an orthogonal matrix, then

Options

A. `a=2,b=1`
B. `a=-2,b=-1`
C. `a=2,b=-1`
D. `a=-2,b=1`

Explanation

To determine the values of \( a \) and \( b \) for the orthogonal matrix \( A = \frac{1}{3} \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{pmatrix} \), we will use the property that for an orthogonal matrix, \( A A^T = I \), where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.

### Step 1: Calculate the Transpose of Matrix A
The transpose of matrix \( A \) is obtained by swapping rows with columns. Thus,

\[ A^T = \frac{1}{3} \begin{pmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{pmatrix} \]

### Step 2: Multiply A and A^T
Now we will multiply \( A \) and \( A^T \):

\[ A A^T = \left( \frac{1}{3} \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{pmatrix} \right) \left( \frac{1}{3} \begin{pmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{pmatrix} \right) \]

This results in:

\[ A A^T = \frac{1}{9} \begin{pmatrix} 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 & 1 \cdot 2 + 2 \cdot 1 + 2 \cdot -2 & 1 \cdot a + 2 \cdot 2 + 2 \cdot b \\ 2 \cdot 1 + 1 \cdot 2 + -2 \cdot 2 & 2 \cdot 2 + 1 \cdot 1 + -2 \cdot -2 & 2 \cdot a + 1 \cdot 2 + -2 \cdot b \\ a \cdot 1 + 2 \cdot 2 + b \cdot 2 & a \cdot 2 + 2 \cdot 1 + b \cdot -2 & a \cdot a + 2 \cdot 2 + b \cdot b \end{pmatrix} \]

### Step 3: Simplify the Elements
Calculating each element:

1. First row, first column:
\[ 1 + 4 + 4 = 9 \]

2. First row, second column:
\[ 2 + 2 - 4 = 0 \]

3. First row, third column:
\[ a + 4 + 2b \]

4. Second row, first column:
\[ 2 + 2 - 4 = 0 \]

5. Second row, second column:
\[ 4 + 1 + 4 = 9 \]

6. Second row, third column:
\[ 2a + 2 - 2b \]

7. Third row, first column:
\[ a + 4 + 2b \]

8. Third row, second column:
\[ 2a + 2 - 2b \]

9. Third row, third column:
\[ a^2 + 4 + b^2 \]

Thus, we have:

\[ A A^T = \frac{1}{9} \begin{pmatrix} 9 & 0 & a + 4 + 2b \\ 0 & 9 & 2a + 2 - 2b \\ a + 4 + 2b & 2a + 2 - 2b & a^2 + 4 + b^2 \end{pmatrix} \]

### Step 4: Set Equal to Identity Matrix
We set \( A A^T \) equal to the identity matrix \( I \):

\[ \frac{1}{9} \begin{pmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{pmatrix} \]

From this, we can equate the elements:

1. From the first row, third column:
\[ a + 4 + 2b = 0 \quad (1) \]

2. From the second row, third column:
\[ 2a + 2 - 2b = 0 \quad (2) \]

3. From the third row, third column:
\[ a^2 + 4 + b^2 = 9 \quad (3) \]

### Step 5: Solve the Equations
From equation (1):
\[ a + 4 + 2b = 0 \implies a + 2b = -4 \quad (4) \]

From equation (2):
\[ 2a + 2 - 2b = 0 \implies 2a - 2b = -2 \implies a - b = -1 \quad (5) \]

Now we can solve equations (4) and (5) simultaneously.

From (5):
\[ a = b - 1 \quad (6) \]

Substituting (6) into (4):
\[ (b - 1) + 2b = -4 \implies 3b - 1 = -4 \implies 3b = -3 \implies b = -1 \]

Substituting \( b = -1 \) back into (6):
\[ a = -1 - 1 = -2 \]

### Final Values
Thus, the values of \( a \) and \( b \) are:
\[ a = -2, \quad b = -1 \]