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JEE Main 2019 · 11 Jan · Shift I

JEE Main 2019 Mathematics question on Matrices and Determinants

Let A=[02qrpq−rp−qr]A = \begin{bmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{bmatrix}. If AAT=I3AA^T = I_3, then ∣p∣|p| is
  1. A.
    15\dfrac{1}{\sqrt{5}}
  2. B.
    12\dfrac{1}{\sqrt{2}}
  3. C.
    13\dfrac{1}{\sqrt{3}}
  4. D.
    16\dfrac{1}{\sqrt{6}}
Show answer and solution

Correct option: B

Solution

From AAT = I, the rows of A must be mutually orthonormal (orthogonal unit vectors).

Unit vector condition on row 1: (0)² + (2q)² + r² = 1 → 4q² + r² = 1

Unit vector condition on row 2: p² + q² + r² = 1

Orthogonality of rows 1 and 2: 0·p + 2q·q + r·(-r) = 0 → 2q² - r² = 0 → r² = 2q²

Substituting r² = 2q² into row 1 equation: 4q² + 2q² = 1 → q² = 1/6, so r² = 1/3.

From row 2: p² = 1 - q² - r² = 1 - 1/6 - 1/3 = 1/2, so |p| = 1/√2.

[JEE Main 2019 · 11 Jan · Shift I]

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