Answer:
Option D
Explanation:
Given , $A = \begin{bmatrix}1 & 2 & 2\\2 & 1 &-2 \\ a&2&b\end{bmatrix}$
$A^{T} = \begin{bmatrix}1 & 2 & a\\2 & 1 &2 \\ 2&-2&b\end{bmatrix}$
$A.A^{T} = \begin{bmatrix}1 & 2 & 2\\2 & 1 &-2 \\ a&2&b\end{bmatrix}\begin{bmatrix}1 & 2 & a\\2 & 1 &2 \\ 2&-2&b\end{bmatrix}$
$= \begin{bmatrix}9 & 0 & a+4+2b\\0 & 9 &2a+2-2b \\ a+4+2b&2a+2-2b&a^{2}+4+b^{2}\end{bmatrix}$
It is given that A.AT =9I
= $ \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}$
= 9 $ \begin{bmatrix}1 & 0 & 0\\0 & 1 &0 \\ 0&0&1\end{bmatrix}$
$\Rightarrow$ $\begin{bmatrix}9 & 0 & a+4+2b\\0 & 9 &2a+2-2b \\ a+4+2b&2a+2-2b&a^{2}+4+b^{2}\end{bmatrix}= \begin{bmatrix}9 & 0 & 0\\0 & 9 &0 \\ 0&0&9\end{bmatrix}$
On comparing we get, a+4+2b=0
$\Rightarrow$ a+2b=-4 ......(i)
2a+2-2b=0
$\Rightarrow$ a-b=-1
and $a^{2}+4+b^{2}=9$ .........(iii)
On solving Eqs. (i) and (ii) , we get,
a=-2, b=-1
This satisfies Eq.(iii)
Hence, (a,b)= (-2,-1)
