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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