Let I be the identity matrix of order 3 × 3 and for the matrix A= [ … ],| A|=-1. Let B be the inverse of the matrix adj ( A adj ( A^2 ) ). Then |(λ…

$\begin{aligned}
& |A|=\left|\begin{array}{ccc}
\lambda & 2 & 3 \
4 & 5 & 6 \
7 & -1 & 2
\end{array}\right|=-1 \
& \lambda(16)-2(-34)+3(-39)=-1 \
& 16 \lambda=48 \Rightarrow \lambda=3 \
& \mathrm{B}^{-1}=\operatorname{adj}\left(\mathrm{A} \cdot \operatorname{adj}\left(\mathrm{A}^2\right)\right) \
& \text { Let } \mathrm{C}=\mathrm{A} \cdot \operatorname{adj}\left(\mathrm{A}^2\right) \
& A C=A^2 \operatorname{adj}\left(A^2\right)=|A|^2 \cdot I=I \Rightarrow C=A^{-1} \
& \text { Now } B^{-1}=\operatorname{adj}\left(A^{-1}\right)=B=\operatorname{adj}(A) \
& \text { Now } \lambda \mathrm{B}+\mathrm{I} \Rightarrow 3 \mathrm{B}+\mathrm{I} \
& \text { Let } \mathrm{P}=3 \mathrm{B}+\mathrm{I} \
& \text { P }=3 \operatorname{adj}(\mathrm{A})+\mathrm{I} \
& \mathrm{AP}=3 \operatorname{Aadj}(\mathrm{A})+\mathrm{A} \
& \mathrm{AP}=3|\mathrm{A}| \cdot \mathrm{I}+\mathrm{A} \
& \mathrm{AP}=\mathrm{A}-3 \mathrm{I} \
& |\mathrm{AP}|=|\mathrm{A}-3 \mathrm{I}| \
& |\mathrm{A}| \cdot|\mathrm{P}|=\left|\begin{array}{ccc}
0 & 2 & 3 \
4 & 2 & 6 \
7 & -1 & -1
\end{array}\right|=38 \
& |\mathrm{P}|=-38
\end{aligned}$