3 and 4 .Determinants and Matrices
medium

यदि समीकरण निकाय $ax + y + z = 0$, $x + by + z = 0$ और $x + y + cz = 0$, जहाँ $a,b,c \ne 1$ का एक अशून्य हल है, तो $\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}}$ का मान है

A

$-1$

B

$0$

C

$1$

D

इनमें से कोई नहीं

Solution

(c) चूँकि समीकरण निकाय का अशून्य हल है

$ \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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.