Determinants Matrices Question 27
Question: If $ \begin{vmatrix} a & \cot A/2 & \lambda \\ b & \cot B/2 & \mu \\ c & cotC/2 & \gamma \\ \end{vmatrix}=0, $ where a, b, c, A, B, and C are elements of a triangle ABC with usual meaning. Then, the value of a $ (\mu -\gamma )+b(\gamma -\lambda )+c(\lambda -\mu )=0 $ is
Options:
A) $ 0 $
B) $ abc $
C) $ ab+bc+ca $
D) $ 2abc $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Given, $ \begin{vmatrix} a & \cot A/2 & \lambda \\ b & \cot B/2 & \mu \\ c & \cot c/2 & \gamma \\ \end{vmatrix}=0 $
$ \Rightarrow \begin{vmatrix} a & \frac{s(s-a)}{\Delta } & \lambda \\ b & \frac{s(s-b)}{\Delta } & \mu \\ c & \frac{s(s-c)}{\Delta } & \gamma \\ \end{vmatrix}=0 $
$ \Rightarrow \frac{1}{r} \begin{vmatrix} a & s-a & \lambda \\ b & s-b & \mu \\ c & s-c & \gamma \\ \end{vmatrix}=0 $ Apply $ C_2\to C_2+C_1 $
$ \Rightarrow \frac{1}{r} \begin{vmatrix} a & s & \lambda \\ b & s & \mu \\ c & s & \gamma \\ \end{vmatrix}=0, $ where $ r=\frac{\Delta }{s} $
$ \Rightarrow \frac{\Delta }{r^{2}} \begin{vmatrix} a & 1 & \lambda \\ b & 1 & \mu \\ c & 1 & \gamma \\ \end{vmatrix}=0 $ Apply $ R_1\to R_1-R_2,R_2\to R_2-R_3 $
$ \Rightarrow \frac{\Delta }{r^{2}} \begin{vmatrix} a-b & 0 & \lambda -\mu \\ b-c & 0 & \mu -\gamma \\ c & 1 & \gamma \\ \end{vmatrix}=0 $
$ \Rightarrow \frac{\Delta }{r^{2}}[(b-c)(\lambda -\mu )-(\mu -\gamma )(a-b)]=0 $
$ \Rightarrow b(\lambda -\mu )-c(\lambda -\mu )-a(\mu -\gamma )+b(\mu -\gamma )=0 $
$ \Rightarrow -a(\mu -\gamma )+b(\lambda -\mu +\mu -\gamma )-c(\lambda -\mu )=0 $
$ \Rightarrow -a(\mu -\gamma )+b(\lambda -\gamma )-c(\lambda -\mu )=0 $
$ \Rightarrow a(\mu -\gamma )+b(\gamma -\lambda )+c(\lambda -\mu )=0 $
BETA
