If $$A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}$$ is a matrix satisfying the equation $$AA^T = 9I$$, where $$I$$ is $$3 \times 3$$ identity matrix, then the ordered pair $$(a, b)$$ is equal to

Given matrix $$ A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} $$ and the equation $$ AA^T = 9I $$, where $$ I $$ is the $$ 3 \times 3 $$ identity matrix, we need to find the ordered pair $$ (a, b) $$.

First, find the transpose $$ A^T $$. The transpose is obtained by swapping rows and columns. So, the first row of $$ A $$ becomes the first column of $$ A^T $$, the second row becomes the second column, and the third row becomes the third column. Thus, $$ A^T = \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} $$.

Now, compute the product $$ AA^T $$. The element at position $$ (i, j) $$ in $$ AA^T $$ is the dot product of the $$ i $$-th row of $$ A $$ and the $$ j $$-th column of $$ A^T $$.

Calculate each element:

• Element (1,1): Row 1 of $$ A $$ is $$ [1, 2, 2] $$, column 1 of $$ A^T $$ is $$ [1, 2, 2] $$. Dot product: $$ 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 = 1 + 4 + 4 = 9 $$.
• Element (1,2): Row 1 of $$ A $$ is $$ [1, 2, 2] $$, column 2 of $$ A^T $$ is $$ [2, 1, -2] $$. Dot product: $$ 1 \cdot 2 + 2 \cdot 1 + 2 \cdot (-2) = 2 + 2 - 4 = 0 $$.
• Element (1,3): Row 1 of $$ A $$ is $$ [1, 2, 2] $$, column 3 of $$ A^T $$ is $$ [a, 2, b] $$. Dot product: $$ 1 \cdot a + 2 \cdot 2 + 2 \cdot b = a + 4 + 2b $$.
• Element (2,1): Row 2 of $$ A $$ is $$ [2, 1, -2] $$, column 1 of $$ A^T $$ is $$ [1, 2, 2] $$. Dot product: $$ 2 \cdot 1 + 1 \cdot 2 + (-2) \cdot 2 = 2 + 2 - 4 = 0 $$.
• Element (2,2): Row 2 of $$ A $$ is $$ [2, 1, -2] $$, column 2 of $$ A^T $$ is $$ [2, 1, -2] $$. Dot product: $$ 2 \cdot 2 + 1 \cdot 1 + (-2) \cdot (-2) = 4 + 1 + 4 = 9 $$.
• Element (2,3): Row 2 of $$ A $$ is $$ [2, 1, -2] $$, column 3 of $$ A^T $$ is $$ [a, 2, b] $$. Dot product: $$ 2 \cdot a + 1 \cdot 2 + (-2) \cdot b = 2a + 2 - 2b $$.
• Element (3,1): Row 3 of $$ A $$ is $$ [a, 2, b] $$, column 1 of $$ A^T $$ is $$ [1, 2, 2] $$. Dot product: $$ a \cdot 1 + 2 \cdot 2 + b \cdot 2 = a + 4 + 2b $$.
• Element (3,2): Row 3 of $$ A $$ is $$ [a, 2, b] $$, column 2 of $$ A^T $$ is $$ [2, 1, -2] $$. Dot product: $$ a \cdot 2 + 2 \cdot 1 + b \cdot (-2) = 2a + 2 - 2b $$.
• Element (3,3): Row 3 of $$ A $$ is $$ [a, 2, b] $$, column 3 of $$ A^T $$ is $$ [a, 2, b] $$. Dot product: $$ a \cdot a + 2 \cdot 2 + b \cdot b = a^2 + 4 + b^2 $$.

So, $$ AA^T = \begin{bmatrix} 9 & 0 & a + 4 + 2b \\ 0 & 9 & 2a + 2 - 2b \\ a + 4 + 2b & 2a + 2 - 2b & a^2 + 4 + b^2 \end{bmatrix} $$.

Given $$ AA^T = 9I $$, and $$ 9I = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} $$, equate the corresponding elements:

1. From element (1,3): $$ a + 4 + 2b = 0 $$ → $$ a + 2b = -4 $$ ...(i)
2. From element (2,3): $$ 2a + 2 - 2b = 0 $$ → $$ 2a - 2b = -2 $$ → divide by 2: $$ a - b = -1 $$ ...(ii)
3. From element (3,3): $$ a^2 + 4 + b^2 = 9 $$ ...(iii)

Solve equations (i) and (ii) simultaneously. From equation (ii): $$ a = b - 1 $$. Substitute into equation (i):

$$ (b - 1) + 2b = -4 $$ → $$ b - 1 + 2b = -4 $$ → $$ 3b - 1 = -4 $$ → $$ 3b = -3 $$ → $$ b = -1 $$.

Substitute $$ b = -1 $$ into equation (ii): $$ a - (-1) = -1 $$ → $$ a + 1 = -1 $$ → $$ a = -2 $$.

Now verify with equation (iii): $$ a^2 + 4 + b^2 = (-2)^2 + 4 + (-1)^2 = 4 + 4 + 1 = 9 $$, which satisfies the equation.

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

Comparing with the options:

• A: $$ (-2, -1) $$
• B: $$ (2, -1) $$
• C: $$ (-2, 1) $$
• D: $$ (2, 1) $$

Hence, the correct answer is Option A.