$\begin{aligned} & \text { M-I } \\ & \text { Let } A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \quad A^2=\left[\begin{array}{cc}a^2+b c & a b+b d \\ a c+d c & b c+d^2\end{array}\right] \\ & A^3=\left[\begin{array}{cc}a^3+2 a b c+b d c & a^2 b+a b d+b^2 c+b d^2 \\ a^2 c+a d c+b c^2+d^2 c & a b c+2 b c d+d^3\end{array}\right] \\ & \text { Given trace }(A)=a+d=3 \\ & \text { and trace }\left(A^3\right)=a^3+d^3+3 a b c+3 b c d=-18 \\ & \Rightarrow \quad a^3+d^3+3 b c(a+d)=-18 \\ & \Rightarrow \quad a^3+d^3+9 b c=-18 \\ & \Rightarrow \quad(a+d)\left((a+d)^2-3 a d\right)+9 b c=-18 \\ & \Rightarrow \quad 3(9-3 a d)+9 b c=-18 \\ & \Rightarrow \quad a d-b c=5=\text { determinant of A } \\ & \text { M-II } \\ & A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \quad ; \quad \Delta=a d-b c \\ & |A-\lambda I|=(a-\lambda)(d-\lambda)-b c\end{aligned}$
$ \begin{aligned} & =\lambda^2-(\mathrm{a}+\mathrm{d}) \lambda+\mathrm{ad}-\mathrm{bc} \\ & =\lambda^2-3 \lambda+\Delta \\ & \Rightarrow \quad \mathrm{O}=\mathrm{A}^2-3 \mathrm{~A}+\Delta \mathrm{I} \\ & \Rightarrow \quad \mathrm{~A}^2=3 \mathrm{~A}-\Delta \mathrm{I} \\ & \Rightarrow \quad \mathrm{~A}^3=3 \mathrm{~A}^2-\triangle \mathrm{A} \\ & =3(3 \mathrm{~A}-\Delta \mathrm{I})-\Delta \mathrm{A} \\ & =(9-\Delta) \mathrm{A}-3 \Delta \mathrm{I} \\ & =(9-\Delta)\left[\begin{array}{ll} \mathrm{a} & \mathrm{~b} \\ \mathrm{c} & \mathrm{~d} \end{array}\right]-3 \Delta\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & \therefore \quad \text { trace } \mathrm{A}^3=(9-\Delta)(\mathrm{a}+\mathrm{d})-6 \Delta \\ & \Rightarrow \quad-18=(9-\Delta)(3)-6 \Delta \\ & =27-9 \Delta \\ & \Rightarrow \quad 9 \Delta=45 \Rightarrow \Delta=5 \end{aligned}$