Question

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

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