Q43 / 160 · AP EAMCET 2023 (16 May Shift 2) · Mathematics

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?

  • A
    $\theta_1=\frac{\pi}{2}, \theta_2=\pi$
  • B
    $\theta_1=\frac{\pi}{2}$, such $\theta_2$ does not exist
  • Correct answer
    C
    $\theta_1=\frac{\pi}{4}, \theta_2=\frac{\pi}{2}$
  • D
    such $\theta_1$ does not exist, $\theta_2=\pi$

Solution

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, $\begin{aligned} & A^{-1}=A \Rightarrow\left(\begin{array}{cc} \cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & \cot \theta \end{array}\right)=\left(\begin{array}{cc} -\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & -\cot \theta \end{array}\right) \\ & \Rightarrow \cot \theta=-\cot \theta \Rightarrow 2 \cot \theta=0 \Rightarrow \theta=-=\theta_1 \\ & \text { and } A^{-1}+A=0 \\ & \Rightarrow\left(\begin{array}{cc} \cot \theta-\cot \theta & \operatorname{cosec} \theta+\operatorname{cosec} \theta \\ \operatorname{cosec} \theta+\operatorname{cosec} \theta & \cot \theta-\cot \theta \end{array}\right)=0 \\ & \Rightarrow \operatorname{cosec} \theta=0 \Rightarrow \theta=\theta_2 \operatorname{can} \text { not exist } \end{aligned}$

Hence, the correct answer is option: 'C'.

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