3 and 4 .Determinants and Matrices
medium

જો સમીકરણ સંહતિ $ax + y + z = 0$, $x + by + z = 0$ અને $x + y + cz = 0 $ ને શૂન્યતર ઉકેલ હોય તો $\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}} = . .. . $ (કે જ્યાં $a,b,c \ne 1$ )

A

$-1$

B

$0$

C

$1$

D

એકપણ નહી.

Solution

(c) As the system of equations has a non-trivial solution

$ \Rightarrow $ $\left| {\,\begin{array}{*{20}{c}}a&1&1\\1&b&1\\1&1&c\end{array}\,} \right|\, = \,0$

$ \Rightarrow $ $\left| {\,\begin{array}{*{20}{c}}a&1&1\\{1 – a}&{b – 1}&0\\{1 – a}&0&{c – 1}\end{array}\,} \right|\, = 0,$

by $\begin{array}{l}{R_2} \to {R_2} – {R_1}\\{R_3} \to {R_3} – {R_1}\end{array}$

$ \Rightarrow $ $a\,(b – 1)\,(c – 1) – 1\,.\,(1 – a)\,(c – 1)$$ – 1\,.\,(1 – a)\,(b – 1) = 0$

$ \Rightarrow $ $\frac{a}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}} = 0$

$ \Rightarrow $ $\frac{1}{{1 – a}} – 1 + \frac{1}{{1 – b}} + \frac{1}{{1 – c}} = 0$

$ \Rightarrow $ $\frac{1}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}} = 1$.

Standard 12
Mathematics

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