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. $2(Z+m+n)+3$

Answer 1

Answer: (B)

1

Answer 2

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