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If the system of equations

$\begin{array} \\ax +y+z= 0 \\ x+by +z = 0 \\ x+y + cz = 0 \end{array}$

with $a,b,c \neq 1$ has a non trivial solutions, the value of $$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}$$ is

1. $1$
2. $-1$
3. $3$
4. $-3$

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@Sayan Bose

pls edit the question

first equation will be ax+y+z=0

For any Ax = 0, the non-trivial solution of x implies that |A| = 0

i.e $\begin{vmatrix} a &1 &1 \\ 1 &b &1 \\ 1&1 &c \end{vmatrix}$ = 0

Now looking into the qsn. we need some kind of (1-a), (1-b) & (1-c)

So, applying c1 = c1 - c2  &  c3= c2 - c3 on the determinant we get,

$\begin{vmatrix} -(1-a) &1 &0 \\ 1-b &b &-(1-b) \\ 0&1 &1-c \end{vmatrix}$

so, -b(1-a)(1-c) - (1-a)(1-b) - (1-b)(1-c) = 0

or, b(1-a)(1-c) + (1-a)(1-b) + (1-b)(1-c) = 0

or, $\frac{b}{1-b} + \frac{1}{1-c} + \frac{1}{1-a}$ = 0   [ Divide both side by (1-a)(1-b)(1-c) ]

or, $\frac{1+b-1}{1-b} + \frac{1+c-c}{1-c} + \frac{1+a-a}{1-a}$ = 0

or, $\frac{b-1}{1-b} + \frac{1}{1-b} + \frac{1}{1-c} + \frac{1}{1-a}$ = 0

or, $\frac{1}{1-b} + \frac{1}{1-c} + \frac{1}{1-a}$ = 1
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