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

$$\begin{gathered} \because A{A}^T=9I \\ \left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]=9\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}9& 0& a+4+2b\\ 0& 9& 2a+2-2b\\ a+4+2b& 2a+2-2b& a^2+4+b^2\end{array}\right]=\left[\begin{array}{l}9\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9\end{array}\right] \end{gathered}$$

On comparing, we get

$a+2b+4=0$                                                               …… (i)

$2a-2b+2=0$                                                            ……. (ii)

From Eqs. (i) and (ii), we get

                                                 $\begin{array}{r}a=-2\\ b=-1\end{array}$

$\therefore$Ordered pair is   $(-2,-1)$