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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Matrices

Let A=( array ccc 0 & 2 q & r p & q & -r p & -q & r array ) . If...

Let A=\left(02qrpq−rp−qr\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right). If AAT=\mathrm{AA}^{\mathrm{T}}=\mathrm{I}3,_{3}, then ∣p∣|\mathrm{p}| is:

Options

  1. A15\frac{1}{\sqrt{5}}
  2. B13\frac{1}{\sqrt{3}}
  3. C12\frac{1}{\sqrt{2}}
  4. D16\frac{1}{\sqrt{6}}

Correct answer

C. 12\frac{1}{\sqrt{2}}

Step-by-step solution

A=\left[02qrpq−rp−qr\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right]∴A⋅AT=[02qrpq−rp−qr]×[0pp2qq−qr−rr]=[4q2+r22q2−r2−2q2+r22q2−r2p2+q2+r2p2−q2−r2−2q2+r2p2−q2−r2p2+q2+r2]\begin{aligned} \therefore & A \cdot A^{T}=\left[\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right] \times\left[\begin{array}{ccc}0 & p & p \\ 2 q & q & -q \\ r & -r & r\end{array}\right] \\ &=\left[\begin{array}{ccc}4 q^{2}+r^{2} & 2 q^{2}-r^{2} & -2 q^{2}+r^{2} \\ 2 q^{2}-r^{2} & p^{2}+q^{2}+r^{2} & p^{2}-q^{2}-r^{2} \\ -2 q^{2}+r^{2} & p^{2}-q^{2}-r^{2} & p^{2}+q^{2}+r^{2}\end{array}\right] \end{aligned}Given,AAT=I∴4q2+r2=p2+q2+r2=1⇒p2−3q2=0Given, A A^{T}=I \therefore \quad 4 q^{2}+r^{2}=p^{2}+q^{2}+r^{2}=1 \Rightarrow \quad p^{2}-3 q^{2}=0 and r2=1−4q2r^{2}=1-4 q^{2} and 2 q^{2}-r^{2}=0 \Rightarrow r^{2}=2 q^{2} ∴p2=12,q2=16 and r2=13∴∣p∣=12\begin{array}{ll}\therefore \quad & p^{2}=\frac{1}{2}, q^{2}=\frac{1}{6} \text{ and } r^{2}=\frac{1}{3} \\ \therefore \quad & |p|=\frac{1}{\sqrt{2}}\end{array}$
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