यदि a, b, c शून्येतर वास्तविक संख्याएँ हैं तथ | vedclass

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Question

Given system of equationa be written as

$\left( {a - 1} \right)x - y - z = 0$

$ - x + \left( {b - 1} \right)y - z = 0$

$ - x - y + \left( {c

Vedclass · Accepted Answer

$abc$

Vedclass · Answer

Given system of equationa be written as

$\left( {a - 1} ight)x - y - z = 0$

$ - x + \left( {b - 1} ight)y - z = 0$

$ - x - y + \left( {c - 1} ight)z = 0$

For non-trivial solution, we have

$\begin{array}{*{20}{c}}
{a - 1}&{ - 1}&{ - 1}\
{ - 1}&{b - 1}&{ - 1}\
{ - 1}&{ - 1}&{c - 1}
\end{array} = 0$

${R_2} 	o {R_2} - {R_3}$

$\begin{array}{*{20}{c}}
{a - 1}&{ - 1}&{ - 1}\
0&b&{ - c}\
{ - 1}&{ - 1}&{c - 1}
\end{array} = 0$

${C_2} 	o {C_2} - {C_3}$

$\begin{array}{*{20}{c}}
{a - 1}&0&{ - 1}\
0&{b + c}&{ - c}\
{ - 1}&{ - c}&{c - 1}
\end{array} = 0$

Apply, ${R_3} 	o {R_3} - {R_1}$

$\begin{array}{*{20}{c}}
{a - 1}&0&{ - 1}\
0&{b + c}&{ - c}\
{ - 1}&{ - c}&c
\end{array} = 0$

$ \Rightarrow \left( {a - 1} ight)\left[ {bc + {c^2} - {c^2}} ight] - 1\left[ {a\left( {b + c} ight)} ight] = 0$

$ \Rightarrow \left( {a - 1} ight)\left[ {bc} ight] - ab - ac = 0$

$ \Rightarrow abc - bc - ab - ac = 0$

$ \Rightarrow ab + bc + ca = abc$

यदि $a, b, c$ शून्येतर वास्तविक संख्याएँ हैं तथा यदि समीकरण निकाय $(a-1) x=y+z$; $(b-1) y=z+x$; $(c-1) z=x+y$ का एक अतुच्छ हल है, तो $a b+b c+c a$ बराबर है

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