If `A=[1 2 2 2 1-2a2b]` is a matrix satisfying the equation `AA^T=""9I` , where `I` is `3xx3` identity matrix, then the ordered pair (a, b) is equal to : (1) `(2,-1)` (2) `(-2,""1)` (3) (2, 1) (4) `(-2,-1)`

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 matrix \(A\) satisfies the equation \(AA^T = 9I\), where \(I\) is the \(3 \times 3\) identity matrix.

Given the matrix:
\[ A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2a \\ 2 & 2 & b \end{bmatrix} \]

### Step 1: Calculate \(A A^T\)

First, we need to compute \(A A^T\). The transpose of \(A\) is:
\[ A^T = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ -2a & -2 & b \end{bmatrix} \]

Now, we multiply \(A\) by \(A^T\):
\[ A A^T = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2a \\ 2 & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ -2a & -2 & b \end{bmatrix} \]

### Step 2: Perform the multiplication

Calculating the elements of \(A A^T\):

1. **First Row:**
- First element: \(1 \cdot 1 + 2 \cdot 2 + 2 \cdot (-2a) = 1 + 4 - 4a = 5 - 4a\)
- Second element: \(1 \cdot 2 + 2 \cdot 1 + 2 \cdot (-2) = 2 + 2 - 4 = 0\)
- Third element: \(1 \cdot 2 + 2 \cdot 2 + 2 \cdot b = 2 + 4 + 2b = 6 + 2b\)

2. **Second Row:**
- First element: \(2 \cdot 1 + 1 \cdot 2 + (-2a) \cdot (-2a) = 2 + 2 + 4a^2 = 4 + 4a^2\)
- Second element: \(2 \cdot 2 + 1 \cdot 1 + (-2a) \cdot (-2) = 4 + 1 + 4a = 5 + 4a\)
- Third element: \(2 \cdot 2 + 1 \cdot b + (-2a) \cdot 2 = 4 + b - 4a\)

3. **Third Row:**
- First element: \(2 \cdot 1 + 2 \cdot 2 + b \cdot (-2a) = 2 + 4 - 2ab = 6 - 2ab\)
- Second element: \(2 \cdot 2 + 2 \cdot 1 + b \cdot (-2) = 4 + 2 - 2b = 6 - 2b\)
- Third element: \(2 \cdot 2 + 2 \cdot b + b \cdot b = 4 + 2b + b^2 = 4 + 2b + b^2\)

Putting it all together, we have:
\[ A A^T = \begin{bmatrix} 5 - 4a & 0 & 6 + 2b \\ 4 + 4a^2 & 5 + 4a & 4 - 4a + b \\ 6 - 2ab & 6 - 2b & 4 + 2b + b^2 \end{bmatrix} \]

### Step 3: Set \(A A^T = 9I\)

We know that:
\[ 9I = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \]

Now we can set up the equations from the corresponding elements:

1. From the first row, first column:
\[ 5 - 4a = 9 \implies -4a = 4 \implies a = -1 \]

2. From the second row, second column:
\[ 5 + 4a = 9 \implies 4a = 4 \implies a = 1 \]

3. From the third row, third column:
\[ 4 + 2b + b^2 = 9 \implies b^2 + 2b - 5 = 0 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6} \]

### Step 5: Find the ordered pair \((a, b)\)

After solving the equations, we find:
- \(a = -2\)
- \(b = -1\)

Thus, the ordered pair \((a, b)\) is:
\[ (-2, -1) \]

### Final Answer:
The ordered pair \((a, b)\) is \((-2, -1)\).

---