InterviewSolution
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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). |
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Answer» `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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