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If A=1/3[(1, 2, 2),( 2, 1,-2),(x,2,y)] s...

If `A=1/3[(1, 2, 2),( 2, 1,-2),(x,2,y)]` satisfies `A^T A=I` , then `x+y=` (a) 3 (b) 0 (c) -3 (d) 1

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We have given
`A=1/3[(1, 2, 2),( 2, 1,-2),(x,2,y)]`which satisfy `A^T A=I`
Then we have to find `x+y`
As,
`A=1/3[(1, 2, 2),( 2, 1,-2),(x,2,y)]`
`A^T=1/3[(1, 2, x),( 2, 1,2),(2,-2,y)]`
`A^TA=[(9,0,x+2y+4),(0,9,2x+2-2y),(x+2y+4,2x+2-2y,x^2+y^2+4)]xx(1/9)=[(1,0,0),(0,1,0),(0,0,1)]`
Two equation are `x+2y+4=0`
...
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