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If A=[alphabetagamma-alpha] is such that...

If `A=[alphabetagamma-alpha]` is such that `A^2=I` , then `1+alpha^2+betagamma=0` (b) `1-alpha^2+betagamma=0` (c) `1-alpha^2-betagamma=0` (d) `1+alpha^2-betagamma=0`

A

`1+alpha^(2)+betagamma=0`

B

`1-alpha^(2)-betagamma=0`

C

`1-alpha^(2)+betagamma=0`

D

`1+alpha^(2)-betagamma=0`

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The correct Answer is:
B
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