If $\mathrm{A}+\mathrm{B}=\left[\begin{array}{cr}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} &…

If $\mathrm{A}+\mathrm{B}=\left[\begin{array}{cr}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]$ where A is symmetric and B is skew-symmetric matrix, then the matrix $\left(A^{-1} B+A B^{-1}\right)$ at $\theta=\frac{\pi}{6}$ is given by
  1. $\left[\begin{array}{cc}1 & 2 \sqrt{3} \\ 2 \sqrt{3} & 1\end{array}\right]$
  2. $\left[\begin{array}{cc}-1 & -2 \sqrt{3} \\ 2 \sqrt{3} & 1\end{array}\right]$
  3. $\left[\begin{array}{cc}0 & 2 \sqrt{3} \\ 2 \sqrt{3} & 0\end{array}\right]$
  4. $\left[\begin{array}{cc}0 & -2 \sqrt{3} \\ 2 \sqrt{3} & 0\end{array}\right]$

Solution

Given, $A+B=\left[\begin{array}{cc}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]$...(i) $A^T+B^T=\left[\begin{array}{cc} 1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1 \end{array}\right]$ Since $A$ is symmetric and $B$ is skew symmetric $\begin{aligned} & A^T=A, B^T=-B \\ \therefore & A-B=\left[\begin{array}{cc} 1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1 \end{array}\right]...(ii) \end{aligned}$ Adding (i) and (ii), we get $\begin{aligned} 2 A & =\left[\begin{array}{cc} 1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1 \end{array}\right]+\left[\begin{array}{cc} 1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1 \end{array}\right] \\ 2 A & =\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right] \\ \therefore \quad A & =\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}$ Now, subtracting (ii) from (i), we get $\begin{aligned} & 2 B=\left[\begin{array}{cc} 1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1 \end{array}\right]-\left[\begin{array}{cc} 1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1 \end{array}\right] \\ & 2 B=\left[\begin{array}{cc} 0 & 2 \tan \frac{\theta}{2} \\ -2 \tan \frac{\theta}{2} & 0 \end{array}\right] \\ & \therefore B=\left[\begin{array}{cc} 0 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 0 \end{array}\right] \end{aligned}$ $\begin{array}{ll} \therefore & B^{-1}=\frac{1}{\tan ^2 \frac{\theta}{2}}\left[\begin{array}{cc} 0 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 0 \end{array}\right] \\ \therefore & B^{-1}=\left[\begin{array}{cc} 0 & \frac{-1}{\tan \frac{\theta}{2}} \\ \frac{1}{\tan \frac{\theta}{2}} & 0 \end{array}\right] \end{array}$ $\therefore \quad$ Now, $\left(\mathrm{A}^{-1} \mathrm{~B}+\mathrm{AB}^{-1}\right)$ $=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}0 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 0\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}0 & \frac{-1}{\tan \frac{\theta}{2}} \\ \frac{1}{\tan \frac{\theta}{2}} & 0\end{array}\right]$ $=\left[\begin{array}{cc}0 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 0\end{array}\right]+\left[\begin{array}{cc}0 & \frac{-1}{\tan \frac{\theta}{2}} \\ \frac{1}{\tan \frac{\theta}{2}} & 0\end{array}\right]$ $\begin{aligned} & =\left[\begin{array}{cc}0 & \frac{\tan ^2 \frac{\theta}{2}-1}{\tan \frac{\theta}{2}} \\ \frac{1-\tan ^2 \frac{\theta}{2}}{\tan \frac{\theta}{2}} & 0\end{array}\right] \\ & =2\left[\begin{array}{cc}0 & (-1) \frac{1-\tan ^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}} \\ \frac{1-\tan ^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}} & 0\end{array}\right]\end{aligned}$ $\begin{aligned} & \therefore=2\left[\begin{array}{cc}0 & \frac{-1}{\tan \theta} \\ \frac{1}{\tan \theta} & 0\end{array}\right] \\ & \therefore \quad\left(\mathrm{A}^{-1} \mathrm{~B}+\mathrm{AB}^{-1}\right) \text { at } \theta=\frac{\pi}{6}\end{aligned}$ $\begin{aligned} & =2\left[\begin{array}{cc}0 & \frac{-1}{\tan \frac{\pi}{6}} \\ \frac{1}{\tan \frac{\pi}{6}} & 0\end{array}\right] \\ & =2\left[\begin{array}{cc}0 & -\sqrt{3} \\ \sqrt{3} & 0\end{array}\right] \\ & =\left[\begin{array}{cc}0 & -2 \sqrt{3} \\ 2 \sqrt{3} & 0\end{array}\right]\end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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