Step 1: Recall the property of orthogonal matrices. A matrix \( A \) is orthogonal if its transpose \( A^T \) equals its inverse \( A^{-1} \), meaning \( A^T = A^{-1} \). This implies \( AA^T = I \), where \( I \) is the identity matrix.
Step 2: Determine the transpose of matrix \( A \).
\[ A^T = \frac{1}{3} \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} \]
Step 3: Compute the product \( AA^T \).
\[ AA^T = \frac{1}{9} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 & 0 & a+4+2b \\ 0 & 9 & 2a+2-2b \\ a+4+2b & 2a+2-2b & a^2+4+b^2 \end{bmatrix} \]
Step 4: Equate \( AA^T \) to \( I \).
For \( AA^T = I \), the following must hold: \[ \frac{1}{9} \begin{bmatrix} 9 & 0 & a+4+2b \\ 0 & 9 & 2a+2-2b \\ a+4+2b & 2a+2-2b & a^2+4+b^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Step 5: Formulate the resulting equations.
The equality from Step 4 yields the following system of equations:
Step 6: Solve for \( a \) and \( b \).
Summing equations (1) and (2) results in \( 3a + 6 = 0 \), yielding \( a = -2 \). Substituting \( a = -2 \) into equation (1) gives \( -2 + 4 + 2b = 0 \), which simplifies to \( 2b = -2 \), hence \( b = -1 \).
Step 7: Verify the calculated values.
Substituting \( a = -2 \) and \( b = -1 \) into equation (3) yields \( (-2)^2 + 4 + (-1)^2 = 4 + 4 + 1 = 9 \). This confirms the correctness of the solution. Therefore, \( a = -2 \) and \( b = -1 \).