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Q)

The system of equations $ax+y+z=0; x+by+z=0; x+y+cz=0 $ has a non-trivial solution then $\large\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$

\[\begin{array}{1 1}(1)1 & (2)2 \\ (3)-1 & (4)0\end{array}\]

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A)
For given homogenous equation to have non-zero solution $ \begin{bmatrix} a & 1 & 1 \\ 1 & b & 1 \\1 & 1 & c \end{bmatrix}=0$
$R_1 \to R_2 -R_1$
$R_3 \to R_3 -R_1$
$ \begin{bmatrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\1-a & 0 & c-1 \end{bmatrix}=0$
=> $a(b-1)(c-1) -1((1-a)(c-1))+1[0-(b-1)(1-a)]=0$
=> $a(b-1)(c-1) -1((1-a)(c-1))-(b-1)(1-a)=0$
=> $a(b-1)(c-1)+(a-1)(c-1))+(b-1)(a-1)=0$
Divide by $(1-a)(1-b)(1-c)$
$\large\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)}+\frac{(a-1)(c-1)}{(1-a)(1-b)(1-c)}+\frac{(b-1)(a-1)}{(1-a)(1-b)(1-c)}$$=0$
$\large\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$=0$
$\large\frac{-(1-a-1)}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$=0$
$-\large\frac{1-a}{1-a}+\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$=0$
$\large\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$=1$
Hence 1 is the correct answer.
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