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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
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
    $\frac{1}{{\sqrt 3 }}$
  • ✓
    $\frac{1}{{\sqrt 2 }}$
  • D
    $\frac{1}{{\sqrt 6 }}$
✓

Answer

Correct option: C.
$\frac{1}{{\sqrt 2 }}$
c
$A$ 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 $(3)$

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

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

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