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If A=[(2,-3,3),(2,2,3),(3,-2,2)] then ...

If `A=[(2,-3,3),(2,2,3),(3,-2,2)] ` then
`C_(2)rarrC_(2)+2C_(1)` and then `R_(1)rarrR_(1)+R_(3)` gives

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The correct Answer is:
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`[{:( 2,-3,3),(2,2,3),(3,-2,2):}]`
`Now , A=IA `
` implies [{:( 2,-3,3),( 2,2,3),( 3,-2,2):}]=[{:( 1,0,0),( 0,1,0),( 0,0,1):}]A`
`implies [{:( 1,1,4),( 2,2,3),( 3,-2,2):}]=[{:( 1,1,-1),( 0,1,0),(0,0,1):}]A R_(1) to R_(1) +R_(2) -R_(3)`
`={:( 1,1,4),( 0,0,-5),(0,-5,-10):}]=[{:(1,1,-1),(-2,-1,2),(-3,-3,4):}]A`
` R_(2) to R_(2) -2R_(1)`
` R_(3) to R_(3) - 3R_(1)`
`implies [{:( 1,1,4),( 0,-5,-10),( 0,0,-5):}]-[{:( 1,1,-1),( -3,-3,4),( -2,-1,2):}]AR_(2) harr R_(3)`
`implies [{:(1,1,4),( 0,1,2),(0,0,1):}]=[{:(1,1,-1),( (3)/(5),(3)/(5),-(4)/(5)),( (2)/(5),(1)/(5) ,-(2)/(5)):}]A R_(2) to -(1)/(5) R_(2),`
`R_(3) to -(1)/(5) R_(3)`
`implies [{:( 1,0,2),(0,1,2),(0,0,1):}]=[{:((2)/(5),(2)/(5),-(1)/(5)),((3)/(5),(3)/(5),-(4)/(5)),((2)/(5),(1)/(5),-(2)/(5)):}]AR_(1)to R_(1)- R_(2)`
`implies [{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(-(2)/(5),0,(3)/(5)),(-(1)/(5),(1)/(5),0),((2)/(5),(1)/(5) ,-(2)/(5)):}]A`
`R_(1) to R_(1) - 2R_(3),R_(2) to R_(2) -2R_(3)`
`therefore A^(-1)=[{:(-(2)/(5),0,(3)/(5)),(-(1)/(5),(1)/(5),0),((2)/(5),(1)/(5),-(2)/(5)):}]=-(1)/(5)[{:(2,0,-3),(1,-1,0),(-2,-1,2):}]`
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