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If x(1) = 3y(1) + 2y(2) -y(3), " " y(1)...

If `x_(1) = 3y_(1) + 2y_(2) -y_(3), " " y_(1)=z_(1) - z_(2) + z_(3)`
`x_(2) = -y_(1) + 4y_(2) + 5y_(3), y_(2)= z_(2) + 3z_(3)`
`x_( 3)= y_(1) -y_(2) + 3y_(3)," " y_(3) = 2z_(1) + z_(2)`
espress `x_(1), x_(2), x_(3)` in terms of `z_(1) ,z_(2),z_(3)`.

Text Solution

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The correct Answer is:
`x_(1) = z_(1) -2z_(2) + 9 z_(3) , x_(2) = 9z_(1) + 10 z_(2) +11 z_(3) , x_(3) = 7 z_(1) + z_(2) - 2z_(3)`
Since, ` X_(1) = 3y_(1) + 2y_(2) - y_(3)`
`rArr [x_(1)] = [(3,2,-1)] [[y_(1)],[y_(2)],[y_(3)]]`
Putting the values of `y_(1), y_(2), y_(3), ` we get
`[x_(1)] = [(3,2,-1)] [[z_(1)-z_(2)+z_(3)],[0+z_(2)+3z_(3)],[2z_(1)+z_(2)+0]]`
`= [(3,2,-1)] [[1,-1,1],[0,1,3],[2,1,0]][[z_(1)],[z_(2)],[z_(3)]]`
`=[(3+0-2,-3+2-1,3+6+0)][[z_(1)],[z_(2)],[z_(3)]]`
`=[(1,-2,9)][[z_(1)],[z_(2)],[z_(3)]]`
`[x_(1)]= [z_(1)-2z_(2)+9z_(3)]`
`therefore x_(1) = z_(1) - 2x_(2) + 9z_(3) " "...(i)`
Further, `x_(2) = -y_(1) + 4y_(2) + 5y_(3) `
`rArr [x_(2)] = [(-1, 4,5)][[y_(1)],[y_(2)],[y_(3)]]`
Putting the values of `y_(1),y_(2),y_(3)`, we get
`[x_(2)] = [(-1, 4,5)][[z_(1)-z_(2)+z_(3)],[0+z_(2)+3z_(3)],[2z_(1)+z_(2)+0]]`
` = [(-1, 4,5)][[1,-1,1],[0,1,3],[2,1,0]][[z_(1)],[z_(2)],[z_(3)]]`
` = [(-1+0+10, 1+4+5,-1+12+0)][[z_(1)],[z_(2)],[z_(3)]]`
` = [(9,10,11)][[z_(1)],[z_(2)],[z_(3)]]=[9z_(1)+10z_(2)+11z_(3)]`
Hence, `x_(2)=9z_(1)=10z_(2)+11z_(3)` ...(ii)
Further, `X_(3) = y_(1)-y_(2)+2y_(3)`
therefore [x_(3)] = [(1,-1 ,3)][[y_(1)],[y_(2)],[y_(3)]]
Putting the values of `y_(1), y_(2), y_(3)` we get
`rArr [x_(3)] = [(1, -1,3)][[z_(1)-z_(2)+z_(3)],[0+z_(2)+3z_(3)],[2z_(1)+z_(2)+0]]`
`= [(1,-1,3)][[1,-1,1],[0,1,3],[2,1,0]][[z_(1)],[z_(2)],[z_(3)]]`
`= [(1-0+6,-1-1+3,1-3+0)][[z_(1)],[z_(2)],[z_(3)]]`
`= [(7,1,-2)][[z_(1)],[z_(2)],[z_(3)]]=[7z_(1)+z_(2)-2z_(3)]`
`therefore x_(3) = 7z_(1)+z_(2)-2z_(3)` ...(iii)
Hence , from Eqs. (i), (ii) and (iii) we get
`x_(1) =z_(1) - 2z_(2) + 9 z_(3) , x_(2) = 9z_(1)+10z_(2) + 11z_(3), x_(3)=7z_(1)+z_(2)-2z_(3)`
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