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Question: Answered & Verified by Expert
Let $A=\left(\begin{array}{ll}-\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & -\cot \theta\end{array}\right)$. If $A^{-1}=A$ at $\theta=\theta_1$ and $A^{-1}$ $+\mathrm{A}=\mathrm{O}$ at $\theta=\theta_2$, then which one of the following is True?
MathematicsMatricesAP EAMCETAP EAMCET 2023 (16 May Shift 2)
Options:
  • A $\theta_1=\frac{\pi}{2}, \theta_2=\pi$
  • B $\theta_1=\frac{\pi}{2}$, such $\theta_2$ does not exist
  • C $\theta_1=\frac{\pi}{4}, \theta_2=\frac{\pi}{2}$
  • D such $\theta_1$ does not exist, $\theta_2=\pi$
Solution:
2671 Upvotes Verified Answer
The correct answer is: $\theta_1=\frac{\pi}{4}, \theta_2=\frac{\pi}{2}$
Given $A=\left(\begin{array}{ll}-\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & -\cot \theta\end{array}\right)$
Now $|A|=\cot ^2 \theta-\operatorname{cosec}^2 \theta=-1$
and $A^{-1}=\left(\begin{array}{cc}\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & \cot \theta\end{array}\right)$
Since,
$\begin{aligned}
& A^{-1}=A \Rightarrow\left(\begin{array}{cc}
\cot \theta & \operatorname{cosec} \theta \\
\operatorname{cosec} \theta & \cot \theta
\end{array}\right)=\left(\begin{array}{cc}
-\cot \theta & \operatorname{cosec} \theta \\
\operatorname{cosec} \theta & -\cot \theta
\end{array}\right) \\
& \Rightarrow \cot \theta=-\cot \theta \Rightarrow 2 \cot \theta=0 \Rightarrow \theta=-=\theta_1 \\
& \text { and } A^{-1}+A=0 \\
& \Rightarrow\left(\begin{array}{cc}
\cot \theta-\cot \theta & \operatorname{cosec} \theta+\operatorname{cosec} \theta \\
\operatorname{cosec} \theta+\operatorname{cosec} \theta & \cot \theta-\cot \theta
\end{array}\right)=0 \\
& \Rightarrow \operatorname{cosec} \theta=0 \Rightarrow \theta=\theta_2 \operatorname{can} \text { not exist }
\end{aligned}$

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