3 and 4 .Determinants and Matrices
hard

माना $A =\left(\begin{array}{ccc}0 & 2 q & r \\ p & q & - r \\ p & - q & r \end{array}\right)$ । यदि $AA ^{ T }= I _{3}$, तो $| p |$ बराबर है

A

$\frac{1}{{\sqrt 5 }}$

B

$\frac{1}{{\sqrt 3 }}$

C

$\frac{1}{{\sqrt 2 }}$

D

$\frac{1}{{\sqrt 6 }}$

(JEE MAIN-2019)

Solution

$A$ is orthogonal matrix

$\therefore 4{q^2} + {r^2} = {p^2} + {q^2} + {r^2} = 1\,\,\,\,\,\,…….\left( 1 \right)$

         ${p^2} – {q^2} – {r^2} = 0\,\,\,\,\,\,…\left( 2 \right)$

and $2{q^2} – {r^2} = 0\,\,\,\,\,….\left( 3 \right)$

Solving $(1),(2)$ and $(3)$

${p^2} = \frac{1}{2}$

$\left| p \right| = \frac{1}{{\sqrt 2 }}$

Standard 12
Mathematics

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