$$\begin{gathered} \mathrm{A} \mathrm{relation} \mathrm{R} \mathrm{in} \mathrm{a} \mathrm{set} \mathrm{A} \mathrm{is} \mathrm{transitive} \mathrm{if} ( {\mathrm{a}}_1, {\mathrm{a}}_2 ) \in \mathrm{R} \mathrm{and} ( {\mathrm{a}}_2, {\mathrm{a}}_3 ) \in \mathrm{implies} \\ ( {\mathrm{a}}_1, {\mathrm{a}}_3 ) \in \mathrm{R}, \mathrm{where} {\mathrm{a}}_1, {\mathrm{a}}_2, {\mathrm{a}}_3 \in \mathrm{A} \\ \mathrm{Now}, ( 1, 2 ), ( 2, 1 ) \in \mathrm{R}, \mathrm{but} ( 1, 1 ) \notin \mathrm{R} \end{gathered}$$

Thus, the given relation R is not transitive.

$$\begin{gathered} \sin \left[ \frac{\mathrm{\pi }}{3} - {\sin }^{-1} \left(\frac{- 1}{2}\right) \right] \\ \mathrm{Let} {\sin }^{-1} \left(\frac{- 1}{2}\right) = \mathrm{x} \\ \Rightarrow \left(\frac{- 1}{2}\right) = \mathrm{sinx} \\ \Rightarrow \mathrm{sinx} = - \sin \frac{\mathrm{\pi }}{6} = \sin \left(-\frac{\mathrm{\pi }}{6} \right) = \sin \left( 2\mathrm{\pi } - \frac{\mathrm{\pi }}{6} \right) \\ \Rightarrow \mathrm{x} = 2\mathrm{\pi } - \frac{\mathrm{\pi }}{6} \end{gathered}$$

$$\begin{gathered} \therefore \sin \left[ \frac{\mathrm{\pi }}{3} - {\sin }^{-1} \left( \frac{- 1}{2}\right) \right] = \sin \left[ \frac{\mathrm{\pi }}{3} - \left( 2\mathrm{\pi } - \frac{\mathrm{\pi }}{6} \right) \right] \\ = \sin \left[ - \frac{9\mathrm{\pi }}{6} \right] \\ = - \sin \left[ \frac{3\mathrm{\pi }}{2} \right] \\ = - \sin \left[ \mathrm{\pi } + \frac{\mathrm{\pi }}{2} \right] \\ = - \left[ - \sin \frac{\mathrm{\pi }}{2} \right] \\ = + \sin \frac{\mathrm{\pi }}{2} \\ = 1 \\ \mathrm{Thus}, \sin \left[ \frac{\mathrm{\pi }}{3} - {\sin }^{-1} \left( \frac{- 1}{2}\right) \right] = 1 \end{gathered}$$

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{elements} \mathrm{of} \mathrm{the} \mathrm{matrix} \mathrm{A} = \left[ {\mathrm{a}}_{\mathrm{ij}} \right] \mathrm{are} \mathrm{given} \mathrm{by} \\ {\mathrm{a}}_{\mathrm{ij}} = \frac{\mathrm{i}}{\mathrm{j}} \end{gathered}$$

For a₁₂, the value of  i = 1  and  j = 2.

${\mathrm{a}}_{12} = \frac{1}{2}$

$$\begin{gathered} \mathrm{Let} \mathrm{A} = \left[ \begin{array}{cc}5 - \mathrm{x} & \mathrm{x} + 1\\ 2 & 4\end{array} \right] \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{martix} \mathrm{A} \mathrm{is} \mathrm{singular}, \mathrm{therefore} \left| \mathrm{A} \right| = 0 \\ \Rightarrow \left|\begin{array}{cc} 5 - \mathrm{x} & \mathrm{x} + 1 \\ 2 & 4 \end{array}\right| = 0 \\ \Rightarrow 4 \left(5 - \mathrm{x} \right) - 2 \left( \mathrm{x} + 1 \right) = 0 \\ \Rightarrow 20 - 4 \mathrm{x} - 2 \mathrm{x} - 2 = 0 \\ \Rightarrow - 6 \mathrm{x} + 18 = 0 \\ \Rightarrow \mathrm{x} = \frac{- 18}{-6} = 3 \end{gathered}$$

Thus, when  x = 3,  the given matrix A is singular.

$$\begin{gathered} \mathrm{A} = \left[\begin{array}{cc}2& 5 \\ 1 & 3 \end{array}\right] \\ {\mathrm{A}}^{-1} = \frac{1}{\left| \mathrm{A} \right|} \left[ \begin{array}{cc} 3& - 5\\ - 1& 2\end{array} \right] \\ = \frac{1}{\left[ \left( 2 \mathrm{x} 3 \right) - \left( 1 \mathrm{x} 5 \right) \right]} \left[ \begin{array}{cc} 3& - 5\\ - 1& 2\end{array} \right] \\ = \frac{1}{6 - 5} \left[ \begin{array}{cc} 3& - 5\\ - 1& 2\end{array} \right] \\ {\mathrm{A}}^{-1} = \left[ \begin{array}{cc} 3& - 5\\ - 1& 2\end{array} \right] \end{gathered}$$