Q44 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
If $3 A=\left[\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b\end{array}\right]$ and $A A^{\mathrm{T}}=I$, then $\frac{a}{b}+\frac{b}{a}$
- A$-\frac{5}{2}$
- B$\frac{13}{6}$
- C$-\frac{13}{6}$
- Correct answerD$\frac{5}{2}$
Solution
$(3 A)(3 A)^T=\left[\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b\end{array}\right]\left[\begin{array}{ccc}1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b\end{array}\right]$
$\begin{aligned} & 9 A A^T=\left[\begin{array}{ccc}9 & 0 & a+4+2 b \\ 0 & 9 & 2 a+2-2 b \\ a+2 b+4 & 2 a+2-2 b & a^2+4+b^2\end{array}\right] \\ & \Rightarrow a+2 b+4=0 \text { and } 2 a+2-2 b=0 \\ & \Rightarrow a=-2, b=-1 \Rightarrow \frac{a}{b}+\frac{b}{a}=2+\frac{1}{2}=\frac{5}{2} .\end{aligned}$
$\begin{aligned} & 9 A A^T=\left[\begin{array}{ccc}9 & 0 & a+4+2 b \\ 0 & 9 & 2 a+2-2 b \\ a+2 b+4 & 2 a+2-2 b & a^2+4+b^2\end{array}\right] \\ & \Rightarrow a+2 b+4=0 \text { and } 2 a+2-2 b=0 \\ & \Rightarrow a=-2, b=-1 \Rightarrow \frac{a}{b}+\frac{b}{a}=2+\frac{1}{2}=\frac{5}{2} .\end{aligned}$
Hence, the correct answer is option: 'D'.