1. Key Concept: Properties of Matrices Satisfying $AA^T = kI$

The problem revolves around a matrix $A$ satisfying the condition $AA^T = 9I$, where $I$ is the $3 \times 3$ identity matrix. This condition is a special case of a broader property related to orthogonal matrices.

• Definition of an Orthogonal Matrix: A square matrix $Q$ is orthogonal if $QQ^T = Q^T Q = I$. This implies that its inverse is its transpose, $Q^{-1} = Q^T$.
• Interpretation of $AA^T = kI$: When a matrix $A$ satisfies $AA^T = kI$ (where $k$ is a scalar), it implies two crucial properties about its row vectors:
  1. Orthogonality: The dot product of any two distinct row vectors is zero.
  2. Magnitude: The square of the magnitude (or norm) of each row vector is equal to $k$.

Let $A$ be represented by its row vectors: $A = \begin{bmatrix} \vec{r_1} \\ \vec{r_2} \\ \vec{r_3} \end{bmatrix}$ Then, the product $AA^T$ can be expressed in terms of dot products of these row vectors: $AA^T = \begin{bmatrix} \vec{r_1} \cdot \vec{r_1} & \vec{r_1} \cdot \vec{r_2} & \vec{r_1} \cdot \vec{r_3} \\ \vec{r_2} \cdot \vec{r_1} & \vec{r_2} \cdot \vec{r_2} & \vec{r_2} \cdot \vec{r_3} \\ \vec{r_3} \cdot \vec{r_1} & \vec{r_3} \cdot \vec{r_2} & \vec{r_3} \cdot \vec{r_3} \end{bmatrix}$ Given $AA^T = 9I = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$, we can equate the corresponding elements:

• $\vec{r_i} \cdot \vec{r_i} = ||\vec{r_i}||^2 = 9$ for $i=1, 2, 3$. (Diagonal elements)
• $\vec{r_i} \cdot \vec{r_j} = 0$ for $i \neq j$. (Off-diagonal elements)

This approach significantly simplifies the problem by avoiding full matrix multiplication and directly providing a system of equations for $a$ and $b$.

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2. Step-by-Step Solution

Let the given matrix be $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & { - 2} \\ a & 2 & b \end{bmatrix}$ Let the row vectors of $A$ be $\vec{r_1}$, $\vec{r_2}$, and $\vec{r_3}$:

• $\vec{r_1} = \begin{bmatrix} 1 & 2 & 2 \end{bmatrix}$
• $\vec{r_2} = \begin{bmatrix} 2 & 1 & { - 2} \end{bmatrix}$
• $\vec{r_3} = \begin{bmatrix} a & 2 & b \end{bmatrix}$

Step 1: Check the magnitudes of the known row vectors.

• For $\vec{r_1}$: We calculate the square of its magnitude: $||\vec{r_1}||^2 = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9$ This matches the condition $AA^T = 9I$ (since the diagonal elements are 9). This confirms the first row is consistent with the given property.

• For $\vec{r_2}$: We calculate the square of its magnitude: $||\vec{r_2}||^2 = 2^2 + 1^2 + (-2)^2 = 4 + 1 + 4 = 9$ This also matches the condition.

Step 2: Use the orthogonality condition to form equations for $a$ and $b$.

• Orthogonality of $\vec{r_1}$ and $\vec{r_2}$: Their dot product should be 0: $\vec{r_1} \cdot \vec{r_2} = (1)(2) + (2)(1) + (2)(-2) = 2 + 2 - 4 = 0$ This is consistent with $AA^T = 9I$.

• Orthogonality of $\vec{r_1}$ and $\vec{r_3}$: Their dot product must be 0: $\vec{r_1} \cdot \vec{r_3} = (1)(a) + (2)(2) + (2)(b) = a + 4 + 2b$ Setting this to 0 (as per the $AA^T = 9I$ condition): $a + 4 + 2b = 0 \implies a + 2b = -4 \quad \text{(Equation 1)}$

• Orthogonality of $\vec{r_2}$ and $\vec{r_3}$: Their dot product must be 0: $\vec{r_2} \cdot \vec{r_3} = (2)(a) + (1)(2) + (-2)(b) = 2a + 2 - 2b$ Setting this to 0: $2a + 2 - 2b = 0 \implies 2a - 2b = -2 \implies a - b = -1 \quad \text{(Equation 2)}$

Step 3: Solve the system of linear equations. We have a system of two linear equations with two variables:

1. $a + 2b = -4$
2. $a - b = -1$

To solve, we can subtract Equation 2 from Equation 1: $(a + 2b) - (a - b) = -4 - (-1)$ $a + 2b - a + b = -4 + 1$ $3b = -3$ $b = -1$

Now, substitute the value of $b$ into Equation 2: $a - (-1) = -1$ $a + 1 = -1$ $a = -2$

Thus, the ordered pair $(a, b)$ is $(-2, -1)$.

Step 4: Verify with the magnitude of the third row vector. The square of the magnitude of $\vec{r_3}$ must also be 9: $||\vec{r_3}||^2 = a^2 + 2^2 + b^2 = 9$ Substitute the values $a=-2$ and $b=-1$: $(-2)^2 + 2^2 + (-1)^2 = 4 + 4 + 1 = 9$ This matches the condition, confirming our values for $a$ and $b$ are correct.

Therefore, the ordered pair $(a, b)$ is $(-2, -1)$.

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3. Common Mistakes and Tips

• Full Matrix Multiplication: A common mistake is to perform the entire $AA^T$ multiplication. While correct, it's tedious and prone to calculation errors. Understanding the underlying vector properties of $AA^T = kI$ is much more efficient.
• Sign Errors: Be careful with negative signs when calculating dot products and solving equations.
• Forgetting All Conditions: Ensure you use all relevant conditions. Here, we used orthogonality and magnitude for all rows. The magnitude check for the third row served as a crucial verification.
• Orthogonal vs. Orthonormal: An orthogonal matrix usually implies orthonormal rows/columns (magnitude 1). Here, the magnitude is $\sqrt{9}=3$, so the rows are orthogonal but not orthonormal. This is why it's $9I$ and not just $I$.

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4. Summary and Key Takeaway

The condition $AA^T = kI$ is a powerful indicator that the rows of matrix $A$ (and similarly, its columns) form an orthogonal set of vectors, each with a squared magnitude of $k$. By leveraging this property, we can directly form a system of equations from dot products, avoiding cumbersome matrix multiplication. This method is significantly faster and less error-prone for problems of this type.

The final answer is $\boxed{(-2, -1)}$.