શૂન્યતર a,b,c માટે જો Delta = left| {,begin{a | vedclass

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(d) $\Delta = \left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\1&{1 + b}&1\1&1&{1 + c}\end{array}\,} ight|$

= $abc\,\left|

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(d) $\Delta = \left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\1&{1 + b}&1\1&1&{1 + c}\end{array}\,} ight|$

= $abc\,\left| {\,\begin{array}{*{20}{c}}{\frac{1}{a} + 1}&{\frac{1}{b}}&{\frac{1}{c}}\{\frac{1}{a}}&{\frac{1}{b} + 1}&{\frac{1}{c}}\{\frac{1}{a}}&{\frac{1}{b}}&{\frac{1}{c} + 1}\end{array}\,} ight|$, by $\begin{array}{l}{C_1} 	o \frac{1}{a}{C_1}\{C_2} 	o \frac{1}{b}{C_2}\{C_3} 	o \frac{1}{c}{C_3}\end{array}$

=$abc\,\left| {\,\begin{array}{*{20}{c}}{\left( {1 + \sum \frac{1}{a}} ight)}&{\frac{1}{b}}&{\frac{1}{c}}\{\left( {1 + \sum \frac{1}{a}} ight)}&{\frac{1}{b} + 1}&{\frac{1}{c}}\{\left( {1 + \sum \frac{1}{a}} ight)}&{\frac{1}{b}}&{\frac{1}{c} + 1}\end{array}\,} ight|$ by

${C_1} 	o {C_1} + {C_2} + {C_3}$

=$abc\,\left( {1 + \sum \frac{1}{a}} ight)\,\left| {\,\begin{array}{*{20}{c}}1&{\frac{1}{b}}&{\frac{1}{c}}\1&{\frac{1}{b} + 1}&{\frac{1}{c}}\1&{\frac{1}{b}}&{\frac{1}{c} + 1}\end{array}\,} ight|$

$[{m{By   taking }}\left( {\sum \frac{1}{a} + 1} ight)\,{m{as common]}}$
$\Delta = abc\,\left( {1 + \sum \frac{1}{a}} ight)\,\left| {\,\begin{array}{*{20}{c}}1&{\frac{1}{b}}&{\frac{1}{c}}\0&1&0\0&0&1\end{array}\,} ight|$,

by $\begin{array}{l}{R_2} 	o {R_2} - {R_1}\{R_3} 	o {R_3} - {R_1}\end{array}$

= $abc\,\left( {1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}} ight)\,.\,1,\,$ (by expansion along ${C_1}$)

Therefore,$\Delta = 0$ $ \Rightarrow {m{Either}}\,{m{ }}abc = 0 $or $1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$

But $a,b,c$ are non-zero and hence the product $abc$ cannot be zero.

So the only alternative is that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = - 1$.

શૂન્યતર $a,b,c$ માટે જો $\Delta = \left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = 0$, તો $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = . . $

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