A) 1
B) 2
C) 3
D) 8
Correct Answer: B
Solution :
We have \[\left| \begin{matrix} a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c \\ \end{matrix} \right|=k\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|\] \[\Rightarrow \] \[\left| \begin{matrix} a & b+c & c+a \\ b & c+a & a+b \\ c & a+b & b+c \\ \end{matrix} \right|+\left| \begin{matrix} b & b+c & c+a \\ c & c+a & a+b \\ a & a+b & b+c \\ \end{matrix} \right|\] \[=k\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|\] \[\Rightarrow \] \[\left| \begin{matrix} a & b & c+a \\ b & c & a+b \\ c & a & b+c \\ \end{matrix} \right|+\left| \begin{matrix} a & c & c+a \\ b & a & a+b \\ c & b & b+c \\ \end{matrix} \right|\] \[+\,\,\left| \begin{matrix} b & c & c+a \\ c & a & a+b \\ a & b & b+c \\ \end{matrix} \right|=k\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|\] \[\Rightarrow \] \[\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|+\left| \begin{matrix} b & c & a \\ c & a & b \\ a & b & c \\ \end{matrix} \right|=k\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|\] \[\Rightarrow \] \[2\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|=k\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|\] \[\Rightarrow \] \[k=2\]
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