If \(\mathrm{A}=\left[\begin{array}{ll}\mathrm{i} & 1 \\ 1 & 0\end{array}\right]\) where \(\mathrm{i}=\sqrt{-1}\) and \(\mathrm{B}=\mathrm{A}^{2029}\), then \(\mathrm{B}^{-1}=\)

\(\begin{aligned} & A=\left[\begin{array}{ll}i & 1 \\ 1 & 0\end{array}\right] \\ & \therefore \quad A^2=\left[\begin{array}{ll}i & 1 \\ 1 & 0\end{array}\right] \times\left[\begin{array}{ll}i & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}0 & i \\ i & 1\end{array}\right] \\ & \therefore \quad A^3=\left[\begin{array}{ll}0 & i \\ i & 1\end{array}\right]\left[\begin{array}{ll}i & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}i & 0 \\ 0 & i\end{array}\right] \\ & \therefore \quad \mathrm{A}^6=\mathrm{A}^3 \times \mathrm{A}^3 \\ & =\left[\begin{array}{ll}\mathrm{i} & 0 \\ 0 & \mathrm{i}\end{array}\right] \times\left[\begin{array}{ll}\mathrm{i} & 0 \\ 0 & \mathrm{i}\end{array}\right] \\ & =\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]=(-1) I_2 \\ & \text { Now, } B=A^{2029}=A^{(6 \times 338+1)} \\ & \therefore \quad \mathrm{B}=\left(\mathrm{A}^6\right)^{338} \times \mathrm{A} \\ & =\left((-1) \mathrm{I}_2\right)^{338} \times \mathrm{A} \\ & =\mathrm{I}_2 \times \mathrm{A} \\ & \therefore \quad \mathrm{B}=\mathrm{A} \\ & \text { Now, }|A|=-1 \\ & \therefore \quad B^{-1}=A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ & \therefore \quad \mathrm{B}^{-1}=-\operatorname{adj} \mathrm{A} \\ & \end{aligned}\)