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

Hint: Write the transpose of a matrix by interchanging the value of entries in each row and column. Multiply the matrix and is transpose using laws of matrix multiplication. Write the $3\times 3$ identity matrix. Substitute the values given in the equation $A{{A}^{T}}=9I$ and compare the terms on both sides to find the value of variables ‘a’ and ‘b’.

Complete step-by-step answer:
We have the matrix $A=\left[ \begin{matrix}
   1 & 2 & 2 \\
   2 & 1 & -2 \\
   a & 2 & b \\
\end{matrix} \right]$ such that $A{{A}^{T}}=9I$. We have to calculate the values of variables ‘a’ and ‘b’.

We will write the transpose of a given matrix by interchanging the value of entries in each row and column.
Thus, the transpose of the matrix $A=\left[ \begin{matrix}
   1 & 2 & 2 \\
   2 & 1 & -2 \\
   a & 2 & b \\
\end{matrix} \right]$ is ${{A}^{T}}=\left[ \begin{matrix}
   1 & 2 & a \\
   2 & 1 & 2 \\
   2 & -2 & b \\
\end{matrix} \right]$.
We will now multiply both the matrices using the law of matrix multiplication.
Let $A={{\left( {{a}_{ij}} \right)}_{n\times n}}$ and $B={{\left( {{b}_{jk}} \right)}_{n\times n}}$ be two matrices. The product of two matrices is $AB={{\left( \sum\limits_{j=1}^{3}{{{a}_{ij}}{{b}_{jk}}} \right)}_{ik}}$.
Thus, we have $A{{A}^{T}}=\left[ \begin{matrix}
   1\left( 1 \right)+2\left( 2 \right)+2\left( 2 \right) & 1\left( 2 \right)+2\left( 1 \right)+2\left( -2 \right) & 1\left( a \right)+2\left( 2 \right)+2\left( b \right) \\
   2\left( 1 \right)+1\left( 2 \right)-2\left( 2 \right) & 2\left( 2 \right)+1\left( 1 \right)-2\left( -2 \right) & 2\left( a \right)+1\left( 2 \right)-2\left( b \right) \\
   a\left( 1 \right)+2\left( 2 \right)+2\left( b \right) & a\left( 2 \right)+2\left( 1 \right)+b\left( -2 \right) & a\left( a \right)+2\left( 2 \right)+b\left( b \right) \\
\end{matrix} \right]$.
Further simplifying the above expression, we have $A{{A}^{T}}=\left[ \begin{matrix}
   9 & 0 & a+4+2b \\
   0 & 9 & 2a+2-2b \\
   a+4+2b & 2a+2-2b & {{a}^{2}}+4+{{b}^{2}} \\
\end{matrix} \right]$.
We know that $A{{A}^{T}}=9I$, where I is the $3\times 3$ matrix. Thus, $I=\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$.
So, we have $A{{A}^{T}}=9I=9\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   9 & 0 & 0 \\
   0 & 9 & 0 \\
   0 & 0 & 9 \\
\end{matrix} \right]$.
We know that $A{{A}^{T}}=\left[ \begin{matrix}
   9 & 0 & a+4+2b \\
   0 & 9 & 2a+2-2b \\
   a+4+2b & 2a+2-2b & {{a}^{2}}+4+{{b}^{2}} \\
\end{matrix} \right]$.
Thus, we have $\left[ \begin{matrix}
   9 & 0 & a+4+2b \\
   0 & 9 & 2a+2-2b \\
   a+4+2b & 2a+2-2b & {{a}^{2}}+4+{{b}^{2}} \\
\end{matrix} \right]=\left[ \begin{matrix}
   9 & 0 & 0 \\
   0 & 9 & 0 \\
   0 & 0 & 9 \\
\end{matrix} \right]$.
Comparing each corresponding term in both the matrices, we have $a+2b+4=0.....\left( 1 \right)$ and $2a-2b+2=0.....\left( 2 \right)$.
We will now solve these two linear equations by the elimination method.
Adding both the equations, we have $\left( a+2b+4 \right)+\left( 2a-2b+2 \right)=0$.
Thus, we have $3a+6=0$.
Rearranging the terms, we have $a=\dfrac{-6}{3}=-2.....\left( 3 \right)$.
Substituting equation (3) in equation (1), we have $-2+2b+4=0$. Simplifying this equation, we have $2b+2=0\Rightarrow b=\dfrac{-2}{2}=-1$.
Hence, we have $\left( a,b \right)=\left( -2,-1 \right)$, which is option (a).

Note: One must know the definition of the transpose of a matrix. We also need to keep in mind that matrix multiplication is not commutative, i.e., for any two matrices A and B, we have $AB\ne BA$. While multiplying two matrices, one must keep in mind that the number of columns of the first matrix must be equal to the number of rows of the second matrix.