Let α ∈(0, ∞) and A=[ array lll 1 & 2 & α 1 & 0 & 1 0 & 1 & 2 array ] . If det(adj(2 A-A^T) · adj(A-2 A^T))=2^8 , then (det(A))^2 is equal to:

Correct answer: B. 16 aligned & |adj(A-2 ~A^ T )(2 ~A-A^ T )|=28 & |(~A-2 ~A^ T )(2 ~A-A^ T )|=24 & |~A-2 ~A^ T ||2 ~A-A^ T |= ± 16 & (~A-2 ~A^ T )^ T =A^ T -2 ~A & |~A-2 ~A^ T |=|A^ T -2 ~A| & |A-2 ~A^ T |^2=16 & |~A-2 ~A^ T |= ± 4 & [ array ccc 1 & 2 & α 1 & 0 & 1 0 & 1 & 2 array ]-[ array ccc 2 & 2 & 0 4 & 0 & 2 2 α & 2 & 4 array ] & | array lll -1 & 0 & α -3 & 0 & -1 -2 α & -1 & -2 array | & 1+3 α=4 & 3 α=3 & α=1 & |~A|=| array lll 1 & 2 & 1 1 & 0 & 1 0 & 1 & 2 array |=-1-3=-4 & |~A|^2=16 aligned

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Matrices

Let α ∈(0, ∞) and A=[ array lll 1 & 2 & α 1 & 0 & 1 0 & 1 & 2 array...

Let

\alpha \in(0, \infty)

and

A=\left[

\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}

\right]

. If

\operatorname{det}\left(\operatorname{adj}\left(2 A-A^T\right) \cdot \operatorname{adj}\left(A-2 A^T\right)\right)=2^8

, then

(\operatorname{det}(A))^2

is equal to:

Options

Correct answer

B. 16

Step-by-step solution

\begin{aligned} & \left|\operatorname{adj}\left(\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right)\right|=28 \\ & \left|\left(\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right)\right|=24 \\ & \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right|\left|2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right|= \pm 16 \\ & \left(\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}-2 \mathrm{~A} \\ & \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right|=\left|\mathrm{A}^{\mathrm{T}}-2 \mathrm{~A}\right| \\ & \Rightarrow\left|\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right|^2=16 \\ & \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right|= \pm 4 \\ & {\left[\begin{array}{ccc}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]-\left[\begin{array}{ccc}2 & 2 & 0 \\ 4 & 0 & 2 \\ 2 \alpha & 2 & 4\end{array}\right]} \\ & \left|\begin{array}{lll}-1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2\end{array}\right| \\ & 1+3 \alpha=4 \\ & 3 \alpha=3 \\ & \alpha=1 \\ & |\mathrm{~A}|=\left|\begin{array}{lll}1 & 2 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right|=-1-3=-4 \\ & |\mathrm{~A}|^2=16\end{aligned}

Practice Matrices on Quantrex Academy →

More from Matrices