Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix…

Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $\mathrm{A}=\left[\begin{array}{ccc}\lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2\end{array}\right],|\mathrm{A}|=-1$. Let B be the inverse of the matrix $\operatorname{adj}\left(\mathrm{A} \operatorname{adj}\left(\mathrm{A}^2\right)\right)$. Then $|(\lambda B+1)|$ is equal to _____

Solution

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

Asked in: JEE Main 2025 (03 Apr Shift 2)

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