`{:A=[(1,2,2),(2,1,-2),(a,2,b)]:}` is a matrix and `A A^T=9I`, then the ordered pair (a,b) is equal to

Options

A. (2,1)
B. (-2,-1)
C. (2,-1)
D. (-2,1)

Explanation

To solve the problem, we need to find the ordered pair (a, b) such that the equation \( A A^T = 9I \) holds true for the matrix \( A \).

### Step-by-Step Solution:

1. **Define the Matrix A**:
\[ A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{pmatrix} \]

2. **Find the Transpose of Matrix A**:
The transpose of matrix \( A \), denoted as \( A^T \), is obtained by swapping rows and columns:
\[ A^T = \begin{pmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{pmatrix} \]

3. **Multiply A and A^T**:
We need to compute the product \( A A^T \):
\[ A A^T = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{pmatrix} \begin{pmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{pmatrix} \]

- **First Row Calculation**:
- First element: \( 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 = 1 + 4 + 4 = 9 \)
- Second element: \( 1 \cdot 2 + 2 \cdot 1 + 2 \cdot -2 = 2 + 2 - 4 = 0 \)
- Third element: \( 1 \cdot a + 2 \cdot 2 + 2 \cdot b = a + 4 + 2b \)

- **Second Row Calculation**:
- First element: \( 2 \cdot 1 + 1 \cdot 2 + -2 \cdot 2 = 2 + 2 - 4 = 0 \)
- Second element: \( 2 \cdot 2 + 1 \cdot 1 + -2 \cdot -2 = 4 + 1 + 4 = 9 \)
- Third element: \( 2 \cdot a + 1 \cdot 2 + -2 \cdot b = 2a + 2 - 2b \)

- **Third Row Calculation**:
- First element: \( a \cdot 1 + 2 \cdot 2 + b \cdot 2 = a + 4 + 2b \)
- Second element: \( a \cdot 2 + 2 \cdot 1 + b \cdot -2 = 2a + 2 - 2b \)
- Third element: \( a \cdot a + 2 \cdot 2 + b \cdot b = a^2 + 4 + b^2 \)

Thus, we have:
\[ A A^T = \begin{pmatrix} 9 & 0 & a + 4 + 2b \\ 0 & 9 & 2a + 2 - 2b \\ a + 4 + 2b & 2a + 2 - 2b & a^2 + 4 + b^2 \end{pmatrix} \]

4. **Set Up the Equation**:
Since \( A A^T = 9I \), we have:
\[ A A^T = \begin{pmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{pmatrix} \]

This gives us the following equations:
- \( a + 4 + 2b = 0 \) (1)
- \( 2a + 2 - 2b = 0 \) (2)
- \( a^2 + 4 + b^2 = 9 \) (3)

5. **Solve the System of Equations**:
From equation (1):
\[ a + 2b = -4 \quad \text{(Rearranging equation (1))} \]

From equation (2):
\[ 2a - 2b = -2 \quad \Rightarrow \quad a - b = -1 \quad \text{(Dividing by 2)} \]

Now we have two equations:
- \( a + 2b = -4 \) (4)
- \( a - b = -1 \) (5)

From equation (5), we can express \( a \) in terms of \( b \):
\[ a = b - 1 \]

Substitute \( a \) in equation (4):
\[ (b - 1) + 2b = -4 \quad \Rightarrow \quad 3b - 1 = -4 \quad \Rightarrow \quad 3b = -3 \quad \Rightarrow \quad b = -1 \]

Substitute \( b \) back into equation (5):
\[ a - (-1) = -1 \quad \Rightarrow \quad a + 1 = -1 \quad \Rightarrow \quad a = -2 \]

6. **Final Ordered Pair**:
The ordered pair \( (a, b) \) is:
\[ (a, b) = (-2, -1) \]