Question

If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]$ is a matrix satisfying
$A{A}^T=9I_3$, then find the values of $a$ and $b$.

• A. $a=-1,b=-2$
• B. $a=-2,b=-1$
• C. $a=1,b=2$
• D. $a=2,b=1$

Answer 1

The correct option is B $a=-2,b=-1$

We have ,
$A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]\Rightarrow A^T=\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]\therefore A{A}^T=9I_3\Rightarrow \left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ a& 2& b\end{array}\right]\left[\begin{array}{ccc}1& 2& a\\ 2& 1& 2\\ 2& -2& b\end{array}\right]=9\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{ccc}9& 0& a+2b+4\\ 0& 9& 2a+2-2b\\ a+2b+4& 2a+2-2b& a^2+4+b^2\end{array}\right]=\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]\Rightarrow a+2b+4=0,2a+2-2b=0\text{and}{a}^2+4+b^2=9\Rightarrow a+2b+4=0,a-b+1=0\text{and}{a}^2+b^2=5$
Solving $a+2b+4=0$ and $a-b+1=0$, we get
$a=-2,b=-1.$
Clearly, these values satisfy $a^2+b^2=5$.
Hence, $a=-2$ and $b=-1.$