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3 and 4 .Determinants and Matrices
normal
If the system of equations $ax + y + z = 0 , x + by + z = 0 \, \& \, x + y + cz = 0$ $(a, b, c \ne 1)$ has a non-trivial solution, then the value of $\frac{1}{{1\, - \,a}}\,\, + \,\,\frac{1}{{1\, - \,b}}\,\, + \,\,\frac{1}{{1\, - \,c}}$ is :
A
$-1$
B
$0$
C
$1$
D
none of these
Solution
$\left| {\,\begin{array}{*{20}{c}}a&1&1\\1&b&1\\1&1&c\end{array}\,} \right|$ Use
${R_1} \to {R_1} – {R_2} \,and\,\, \,{R_2} \to {R_2} – {R_1}$
open by $C_1$ to get $(1 – a) [(1 – b)c + (1 – c)] + (1 – b) (1 – c) = 0$ divide by $(1 – a) (1 – b) (1 – c)$ to get the result
Standard 12
Mathematics
