If `A=[(0,x),(y,0)]` and `A^(3)+A=O` then sum of possible values of xy is

Options

A. 0
B. `-1`
C. 1
D. 2

Answer

B

Explanation

`A^(3)+A=O`
`implies A(A^(2)+I)=O`
`implies |A(A^(2)+I)|=0`
`implies |A|=0` or `|A^(2)+I|=0`
If `|A|=0`, then `xy=0`
`A^(2)+I=[(0,x),(y,0)][(0,x),(y,0)]+[(1,0),(0,1)]`
`=[(xy,0),(0,xy)]+[(1,0),(0,1)]`
`=[(xy+1,0),(0,xy+1)]`
`|A^(2)+I|=0`
`implies (xy+1)^(2)=0`
`implies xy=-1`
So, sum of possible values of `xy` is `-1`.