A) (2,1)
B) -(-2,-1)
C) (2,-1)
D) (-2,1)
Correct Answer: B
Solution :
\[A{{A}^{T}}=9I\]\[\left[ \begin{matrix} 9 & 0 & a+4+2b \\ 0 & a & 2a+2-2b \\ a+4+2b & 2a+2-2b & {{a}^{2}}+4+{{b}^{2}} \\ \end{matrix} \right]=\left[ \begin{matrix} 9 & 0 & 0 \\ 0 & 0 & 9 \\ 0 & 0 & 9 \\ \end{matrix} \right]\] \[\Rightarrow \]\[\left\{ \begin{align} & a+4+2b=0 \\ & 2a+2-2b=9 \\ \end{align} \right.\Rightarrow (a,b)=(-2,-1)\]44RAYou need to login to perform this action.
You will be redirected in
3 sec