A) \[1\]
B) \[0\]
C) \[3\]
D) \[4\]
Correct Answer: B
Solution :
\[0=\left| \begin{matrix} 2 & y & z \\ x & 2 & z \\ x & y & 2 \\ \end{matrix} \right|=\left| \begin{matrix} 2 & y & z \\ x-2 & 2-y & 0 \\ x-2 & 0 & 2-z \\ \end{matrix} \right|\] \[=(x-2)(2-y)(2-z)\left| \begin{matrix} \frac{2}{x-2} & \frac{y}{2-y} & \frac{z}{2-z} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{matrix} \right|\] \[\Rightarrow \] \[0=\frac{2}{x-2}-\frac{y}{2-y}-\frac{z}{2-z}\Rightarrow \frac{2}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}=0\]You need to login to perform this action.
You will be redirected in
3 sec