If the system of homogeneous equations tx + (t+1) y + (t-1) z = 0 , (t+1) x + ty (t+1) z = 0, (t+1) x + (t+1)y + tz = 0 in x,y,z has a non-trivial solution, then t is a root of the equation

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3t² - 4t + 1 = 0

2t² + 3t + 1 = 0

2t² - 3t + 1 = 0

3t² + 4t + 1 = 0

answer is C.

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Detailed Solution

$\begin{array}{l} \Rightarrow |\begin{array}{ccccccc} t & t + 1 & t - 1 \\ t + 1 & t & t + 2 \\ t - 1 & t + 2 & t \end{array}| = R_{2} - R_{1} |\begin{array}{ccc} t & t + 1 & t - 1 \\ 1 & - 1 & 3 \\ - 1 & 1 & 1 \end{array}| = 0 \\ \Rightarrow t (- 1 - 3) - (t + 1) (1 + 3) + (t - 1) (1 - 1) = 0 \Rightarrow - 4 t - 4 t = 0 \Rightarrow 8 t - 4 \Rightarrow t = - \frac{1}{2} . \\ 3 t^{2} - 4 t + 1 = 0 \Rightarrow (t - 1) (3 t - 1) =\Rightarrow t = 1 or \frac{1}{3} . \\ 2 t^{2} - 3 t + 1 = 0 \Rightarrow (t - 1) (2 t - 1) = 0 \Rightarrow t = 1 or t = \frac{1}{2} . \end{array}$
is a root of 2t² + 3t + 1 = 0