Determinants Matrices Question 161
Question: If $ l^2_1+m_1^{2}+n_1^{2}=1 $ , etc. and $ l_1l_2+m_1m_2+n_1n_2=0 $ , etc. and $ \Delta = \begin{vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \\ \end{vmatrix} $ , then
Options:
A) $ | \Delta | $ =3
B) $ | \Delta | $ =2
C) $ | \Delta | $ =1
D) $ \Delta $ =0
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] we have $ {{\Delta }^{2}}=\Delta \Delta = \begin{vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \\ \end{vmatrix} $
$ \begin{vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \\ \end{vmatrix} $
$ = \begin{vmatrix} l_1^{2}+m_1^{2}+n_1^{2} & l_1l_2+m_1m_2+n_1n_2 & l_1l_3+m_1m_3+n_1n_3 \\ l_1l_2+m_1m_2+n_1n_2 & l_2^{2}+m_2^{2}+n_2^{2} & l_2l_3+m_2m_3+n_2n_3 \\ l_1l_3+m_1m_3+n_1n_3 & l_2l_3+m_2m_3+n_2n_3 & l_3^{2}+m_3^{2}+n_3^{2} \\ \end{vmatrix} $
$ = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{vmatrix}1 $
$ \Rightarrow \Delta =\pm 1\Rightarrow | \Delta |=1 $
BETA
