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If $Δ = |\begin{array}{ccc} 1 + y & 1 - y & 1 - y \\ 1 - y & 1 + y & 1 - y \\ 1 - y & 1 - y & 1 + y \end{array}| = 0$ , then value of $y$ are

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Correct option is A

Using C1→C1+C2+C3, we getΔ=(3−y)1 1−y 1−y1 1+y 1−y1 1−y 1+yApplying R3→R3−R2,R2→R2−R1 we getΔ=(3−y)11−y1−y02y00−2y2y=(3−y)4y2∴ Δ=0⇒y=0 or y=3.

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If Δ=1+y1−y1−y1−y1+y1−y1−y1−y1+y=0, then value of y are

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If $Δ = |\begin{array}{ccc} 1 + y & 1 - y & 1 - y \\ 1 - y & 1 + y & 1 - y \\ 1 - y & 1 - y & 1 + y \end{array}| = 0$ , then value of $y$ are