જો $\mathrm{a, b, c}$ પૈકી પ્રત્યેક બે અસમાન અને પ્રત્યેક ધન હોય, તો સાબિત કરો કે નિશ્ચાયક $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ નું મૂલ્ય ઋણ છે.

Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$ to the given determinant, we get

$\Delta  = \left| {\begin{array}{*{20}{l}}
  {a + b + c}&b&c \\ 
  {a + b + c}&c&a \\ 
  {a + b + c}&a&b 
\end{array}} \right| = (a + b + c)\left| {\begin{array}{*{20}{c}}
  1&b&c \\ 
  1&c&a \\ 
  1&a&b 
\end{array}} \right|$

${ = (a + b + c)\left| {\begin{array}{*{20}{c}}
  1&b&c \\ 
  0&{c - b}&{a - c} \\ 
  0&{a - b}&{b - c} 
\end{array}} \right|}$ ${{\text{(Applying }}{{\text{R}}_2} \to {{\text{R}}_2} - {{\text{R}}_1},{\text{ and }}{{\text{R}}_3} \to {{\text{R}}_3} - {{\text{R}}_1})}$

$ = (a + b + c)[(c - b)(b - c) - (a - c)(a - b)]$ ${\text{(Expanding along }}{{\text{C}}_1}{\text{ ) }}$

$ = (a + b + c)\left( { - {a^2} - {b^2} - {c^2} + ab + bc + ca} \right)$

which is negative (since $\left.a+b+c>0 \text { and }(a-b)^{2}+(b-c)^{2}+(c-a)^{2}>0\right)$