If A = begin{bmatrix} 1 & 2 & 2 2 & 1 & -2 | vedclass

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(A) Given $AA^T = 9I$,where $I$ is the $3 	imes 3$ identity matrix.$\begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & b \end{bmatrix} \begin{bmatrix

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$(-2, -1)$

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(A) Given $AA^T = 9I$,where $I$ is the $3 	imes 3$ identity matrix.$\begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a \ 2 & 1 & 2 \ 2 & -2 & b \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 9 \end{bmatrix}$Multiplying the matrices,we look at the elements in the third row:For the element at $(3, 1)$: $a(1) + 2(2) + b(2) = 0 \Rightarrow a + 2b + 4 = 0 \Rightarrow a + 2b = -4$ ... $(i)$For the element at $(3, 2)$: $a(2) + 2(1) + b(-2) = 0 \Rightarrow 2a - 2b + 2 = 0 \Rightarrow a - b = -1$ ... $(ii)$Solving the system of equations:From $(ii)$,$a = b - 1$.Substitute into $(i)$: $(b - 1) + 2b = -4 \Rightarrow 3b = -3 \Rightarrow b = -1$.Then,$a = -1 - 1 = -2$.Thus,the ordered pair $(a, b)$ is $(-2, -1)$.

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 the $3 \times 3$ identity matrix,then the ordered pair $(a, b)$ is equal to:

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