| (a - 1) x = y + z |
| (b - 1) y = z + x |
| (c - 1) z = x + y |
A) a + b + c
B) abc
C) 1
D) - 1
Correct Answer: B
Solution :
Given system of equations can be written as \[(a-1)x-y-z=0\] \[-x+(b-1)y-z=0\] \[-x-y+(c-1)z=0\] For non-trivial solution, we have \[\left| \begin{matrix} a-1 & -1 & -1 \\ -1 & b-1 & -1 \\ -1 & -1 & c-1 \\ \end{matrix} \right|=0\] \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\]\[\left| \begin{matrix} a-1 & -1 & -1 \\ 0 & b & -c \\ -1 & -1 & c-1 \\ \end{matrix} \right|\] \[{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\]\[\left| \begin{matrix} a-1 & 0 & -1 \\ 0 & b+c & -c \\ -1 & -c & c-1 \\ \end{matrix} \right|=0\] Apply\[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\left| \begin{matrix} a-1 & 0 & -1 \\ 0 & b+c & -c \\ -a & -c & c \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[(a-1)[bc+{{c}^{2}}-{{c}^{2}}]-1[a(b+c)]=0\] \[\Rightarrow \]\[(a-1)[bc]-ab-ac=0\] \[\Rightarrow \]\[abc-bc-ab-ac=0\] \[\Rightarrow \]\[ab+bc+ca=abc\]
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