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$If the equations x + a y - z = 0, 2 x - y + a z = 0, α x + y + 2 z = 0 have non - trivial solutions,$ $then a can be$

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Correct option is B

1a−12−1aa12=0 Applying R2→R2-2R1 and R3→R3-aR1, we get 1a-10-1-2aa+201-a22+a=0 ⇒-1+2aa+2-a+21-a2=0 ⇒a+2a2-2a-2=0a=-2,1±3

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If the equations x+ay-z=0, 2x-y+az=0, αx+y+2z=0 have non - trivial solutions, then a can be

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$If the equations x + a y - z = 0, 2 x - y + a z = 0, α x + y + 2 z = 0 have non - trivial solutions,$ $then a can be$