Let I be the identity matrix of order 3 × 3 and for the matrix A=[ array ccc λ & 2 & 3 4 & 5 & 6 7 & -1 & 2 array ],|A|=-1 . Let B be the inverse of the matrix adj(A adj(A^2)) . Then |(λ B+1)| is equal to _____

Correct answer: 0 aligned & |A|=| array ccc λ & 2 & 3 4 & 5 & 6 7 & -1 & 2 array |=-1 & λ(16)-2(-34)+3(-39)=-1 & 16 λ=48 λ=3 & B^ -1 =adj(A · adj( A^2)) & Let C=A · adj( A^2) & A C=A^2 adj(A^2)=|A|^2 · I=I C=A^ -1 & Now B^ -1 =adj(A^ -1 )=B=adj(A) & Now λ B+I 3 B+I & Let P=3 B+I & P =3 adj( A)+I & AP=3 Aadj( A)+A & AP=3| A| · I+A & AP=A-3 I & |AP|=|A-3 I| & |A| ·|P|=| array ccc 0 & 2 & 3 4 & 2 & 6 7 & -1 & -1 array |=38 & |P|=-38 aligned

JEE Main 20253 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Matrices

Let I be the identity matrix of order 3 × 3 and for the matrix A=[...

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 _____

Correct answer

Step-by-step 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}

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