Question

If \( A = \frac{1}{3} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \) is an orthogonal matrix, then

• A. \( a = -2, b = -1 \)
• B. \( a = 2, b = 1 \)
• C. \( a = 2, b = -1 \)
• D. \( a = -2, b = 1 \)

Hint:

Remember that for an orthogonal matrix \( A \), \( AA^T = I \).

Answer 1

The Correct Option is A

Step 1: {Recall the property of orthogonal matrices}
A matrix \( A \) is orthogonal if its transpose \( A^T \) is equal to its inverse \( A^{-1} \), i.e., \( A^T = A^{-1} \). This implies that \( AA^T = I \), where \( I \) is the identity matrix. 
Step 2: {Find 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: {Multiply \( A \) and \( A^T \)}
\[ AA^T = \frac{1}{9} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 & 0 & a+4+2b \\ 0 & 9 & 2a+2-2b \\ a+4+2b & 2a+2-2b & a^2+4+b^2 \end{bmatrix} \] 
Step 4: {Set \( AA^T = I \)}
For \( AA^T = I \), we must have: \[ \frac{1}{9} \begin{bmatrix} 9 & 0 & a+4+2b \\ 0 & 9 & 2a+2-2b \\ a+4+2b & 2a+2-2b & a^2+4+b^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] 
Step 5: {Solve the resulting equations}
From the above equation, we get the following equations:

1. \( a + 4 + 2b = 0 \)
2. \( 2a + 2 - 2b = 0 \)
3. \( a^2 + 4 + b^2 = 9 \)

Step 6: {Solve for \( a \) and \( b \)}
Adding equations (1) and (2), we get \( 3a + 6 = 0 \), which gives \( a = -2 \). Substituting \( a = -2 \) into equation (1), we get \( -2 + 4 + 2b = 0 \), which gives \( 2b = -2 \), so \( b = -1 \). 
Step 7: {Verify the solution}
Substituting \( a = -2 \) and \( b = -1 \) into equation (3), we get \( (-2)^2 + 4 + (-1)^2 = 4 + 4 + 1 = 9 \), which is true. Therefore, \( a = -2 \) and \( b = -1 \).