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[(1,3,-2),(-3,0,-5),(2,5,0)] =A [(1,0,0)...

`[(1,3,-2),(-3,0,-5),(2,5,0)] =A [(1,0,0),(0,1,0),(0,0,1)]` then, `C_(2) rarr C_(2)-3C_(1) and C_(3) rarr C_(3) +2C_(1)` gives...`[(1,0,0),(3,-9,11),(-2,1,4)] =A [(-1,3,2),(0,-1,0),(0,0,-1)]` `[(1,0,0),(-3,-1,-4),(2,-9,11)] =A [(1,-3,2),(0,1,0),(0,0,1)]` `[(1,0,0),(-3,1,4),(-2,9,-11)] =A [(-1,3,2),(0,-1,0),(0,0,-1)]` `[(1,0,0),(-3,9,-11),(2,-1,4)] =A [(1,-3,2),(0,1,0),(0,0,1)]`

A

`[(1,0,0),(3,-9,11),(-2,1,4)] =A [(-1,3,2),(0,-1,0),(0,0,-1)]`

B

`[(1,0,0),(-3,-1,-4),(2,-9,11)] =A [(1,-3,2),(0,1,0),(0,0,1)]`

C

`[(1,0,0),(-3,1,4),(-2,9,-11)] =A [(-1,3,2),(0,-1,0),(0,0,-1)]`

D

`[(1,0,0),(-3,9,-11),(2,-1,4)] =A [(1,-3,2),(0,1,0),(0,0,1)]`

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The correct Answer is:
D
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