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3 and 4 .Determinants and Matrices
hard
Let $A\, = \,\left( {\begin{array}{*{20}{c}}
0&{2q}&r\\
p&q&{ - r}\\
p&{ - q}&r
\end{array}} \right)$. If $A{A^T}\, = \,{I_3},\,\left| p \right|$ then $\left| p \right|$ is
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
