CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$\triangle =\left| \begin{matrix} arg{ z }_{ 1 } & arg{ z }_{ 2 } & arg{ z }_{ 3 } \\ arg{ z }_{ 2 } & arg{ z }_{ 3 } & arg{ z }_{ 1 } \\ arg{ z }_{ 3 } & arg{ z }_{ 1 } & arg{ z }_{ 2 } \end{matrix} \right| $$, the, $$\triangle$$ is divided by: 


A
arg(z1+z2+z3)
loader
B
arg(z1.z2.z3)
loader
C
(argz1+argz2+argz3)
loader
D
N.O.T
loader

Solution

The correct option is B $$arg(z_{1}.z_{2}.z_{3})$$

$$\triangle =\left |\begin{matrix} arg{z}_{1}&arg{z}_{2}&arg{z}_{3}\\arg{z}_{2}&arg{z}_{3}&arg{z}_{1} \\ arg{z}_{3}&arg{z}_{1}&arg{z}_{2}\end{matrix}\right |$$


$$\Rightarrow \triangle =(arg{z}_{1}+arg{z}_{2}+arg{z}_{3}) \left |\begin{matrix} 1&arg{z}_{2}&arg{z}_{3}\\1&arg{z}_{3}&arg{z}_{1} \\ 1&arg{z}_{1}&arg{z}_{2}\end{matrix}\right |$$


Using $$C_1\rightarrow C_1+C_2+C_3$$


$$\Rightarrow \triangle =arg({z}_{1}{z}_{2}{z}_{3}) \left |\begin{matrix} 1&arg{z}_{2}&arg{z}_{3}\\1&arg{z}_{3}&arg{z}_{1} \\ 1&arg{z}_{1}&arg{z}_{2}\end{matrix}\right |$$


Hence, $$\triangle $$ is divisible by $$arg(z_1z_2z_3)$$ 


Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More



footer-image