A) \[1,2\]
B) \[-1,\text{ }2\]
C) \[-1,-2\]
D) \[2,1\]
E) \[-2,-1\]
Correct Answer: E
Solution :
\[A=\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \\ \end{matrix} \right]\]satisfying \[A{{A}^{T}}=9{{I}_{3}}\] ...(i) \[\Rightarrow \]\[{{A}^{T}}=\left[ \begin{matrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \\ \end{matrix} \right]\]and\[{{I}_{3}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] From Eq. (i), \[\Rightarrow \]\[\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \\ \end{matrix} \right]=\left[ \begin{matrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right]\] \[\Rightarrow \]\[\left[ \begin{matrix} 9 & 0 & a+4+2b \\ 0 & 9 & 2a+2-2b \\ a+2b+4 & 2a+2-2b & {{a}^{2}}+4+{{b}^{2}} \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \\ \end{matrix} \right]\] On comparing, \[a+2b=-4\] ...(ii) \[a-b=-1\] ...(iii) On solving Eqs. (ii) and (iii), \[a=-2,b=-1\]
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