Question

If $ l,m $ and $ n $ are real numbers such that $ {{l}^{2}}+{{m}^{2}} $ $ +{{n}^{2}}=0, $ then $ \left| \begin{matrix} 1+{{l}^{2}} & lm & ln \\ lm & 1+{{m}^{2}} & mn \\ ln & mn & 1+{{n}^{2}} \\ \end{matrix} \right| $ is equal to

• A. 0
• B. 1
• C. $ l+m+n+2 $
• D. $ 2(Z+m+n)+3 $
• E. $ lmn-1 $

Answer 1

Answer: (B)

$ \left| \begin{matrix} 1+{{l}^{2}} & lm & \ln \\ lm & 1+{{m}^{2}} & mn \\ \ln & mn & 1+{{n}^{2}} \\ \end{matrix} \right| $
$=(1+{{l}^{2}})\left| \begin{matrix} 1+{{m}^{2}} & mn \\ mn & 1+{{n}^{2}} \\ \end{matrix} \right|-lm\left| \begin{matrix} lm & mn \\ \ln & 1+{{n}^{2}} \\ \end{matrix} \right| $
$=ln\left| \begin{matrix} lm & 1+{{m}^{2}} \\ ln & mn \\ \end{matrix} \right| $
$=(1+{{l}^{2}})(1+{{m}^{2}}+{{n}^{2}}+{{m}^{2}}{{n}^{2}}-{{m}^{2}}{{n}^{2}}) $ $ -lm(lm-lm{{n}^{2}}-lm{{n}^{2}}) $ $ +\ln (l{{m}^{2}}n-\ln -l{{m}^{2}}n) $
$=(1+{{l}^{2}})(1+{{m}^{2}}+{{n}^{2}})-{{l}^{2}}{{m}^{2}}-{{l}^{2}}{{n}^{2}} $
$=1+{{m}^{2}}+{{n}^{2}}+{{l}^{2}}+{{l}^{2}}{{m}^{2}}+{{l}^{2}}{{n}^{2}}-{{l}^{2}}{{m}^{2}}-{{l}^{2}}{{n}^{2}} $
$=1+{{m}^{2}}+{{n}^{2}}+{{l}^{2}}=1 $ $ (\because {{l}^{2}}+{{m}^{2}}+{{n}^{2}}=0) $