$\begin{array}{l}\text { Let } A =\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \quad A ^2=\left[\begin{array}{ll} a ^2+ bc & ab + bd \\ ac + dc & bc + d ^2\end{array}\right] \\ A^3=\left[\begin{array}{cc}a^3+2 a b c+b d c & a^2 b+a b d+b^2 c+b d^2 \\ a^2 c+a d c+b c^2+d^2 c & a b c+2 b c d+d^3\end{array}\right] \\ \text { Given trace(A) }=a+d=3 \\ \text { and trace }\left(A^3\right)=a^3+d^3+3 a b c+3 b c d=-18 \\ \Rightarrow \quad a ^3+ d ^3+3 bc ( a + d )=-18 \\ \Rightarrow \quad a^3+d^3+9 b c=-18 \\ \Rightarrow \quad( a + d )\left(( a + d )^2-3 ad \right)+9 bc =-18 \\ \Rightarrow \quad 3(9-3 ad )+9 bc =-18 \\ \Rightarrow ad – bc =5=\text { determinant of } A \\ \text { M-II } \\ A =\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \quad ; \quad \Delta= ad – bc \\ |A-\lambda I|=(a-\lambda)(d-\lambda)-b c \\\end{array}$

$\begin{array}{l}=\lambda^2-(a+d) \lambda+a d-b c \\ =\lambda^2-3 \lambda+\Delta \\ \Rightarrow \quad O = A ^2-3 A +\Delta I \\ \Rightarrow \quad A ^2=3 A -\Delta I \\ \Rightarrow \quad A ^3=3 A ^2-\Delta A \\ =3(3 A -\Delta I )-\Delta A \\ =(9-\Delta) A-3 \Delta I \\ =(9-\Delta)\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]-3 \Delta\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ \therefore \quad \text { trace } A^3=(9-\Delta)(a+d)-6 \Delta \\ \Rightarrow \quad-18=(9-\Delta)(3)-6 \Delta \\ =27-9 \Delta \\ \Rightarrow \quad 9 \Delta=45 \Rightarrow \Delta=5 \\\end{array}$