InterviewSolution
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InterviewSolution
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The sum of three numbers is 6. If we multiply the third number 2 andadd the first number to the result, we get 7. Be adding second and thirdnumbers to three times the first number we get 12. Use determinants to findthe numbers. |
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Answer» Let the first , second and third numbers be x,y,z respectively Then `x+y+z " " …(i)` `x+2z=7 " " …(ii)` `3x+y+z=12 " " …(iii)` `Let A=[{:(1 " "1 " "2),(1 " "0 " "2),(3 " "1 " "1):}], X=[{:(x),(y),(z):}] and B=[{:(6),(7),(12):}]` Then , the given system in matrix from is AX=B `Now , |A|=[{:(1 " "1 " "2),(1 " "0 " "2),(3 " "1 " "1):}]=[{:(1 " "1 " "1),(1 " "-1 " "1),(0 " "-2 " "-2):}] " "[{:(R_2rarrR_2-R_1,),(R_3rarrR_3-3R_1):}]` =1,(2+2)=4 `ne` 0 `therefore` A is invertible So, the given system has a unique solution , `X=A^(-1)B` The minors of the elements of |A| are `M_(11)=-2, M_(12)=-5, M_(13)=1,` `M_(21)=0, M_(22)=-2, M_(23)=-2,` `M_(31)=2, M_(32)=1, M_(33)=-1,` The cofactors of the elements of |A| are `A_(11)=-2, A_(12)=-5, A_(13)=1,` `A_(21)=0, A_(22)=-2, A_(23)=-2,` `A_(31)=2, A_(32)=1, A_(33)=-1,` `therefore (adjA)=[{:(-2 " "5 " "1),(0 " " -2 " "2),(2 " "-1 " "-1):}]=[{:(-2 " "0 " "2),(5 " "-2 " "-1),(1 " "2 " "-1):}]` `A^(-1)=1/(|A|).(adjA)=1/4.[{:(-2 " "0 " "2),(5 " "-2 " "-1),(1 " "2 " "-1):}]` `rArrX=A^(-1)B` `rArr[{:(x),(y),(z):}]=1/4..[{:(-2 " "0 " "2),(5 " "-2 " "-1),(1 " "2 " "-1):}][{:(6),(7),(12):}]` `=1/4.[{:(-12+0+24),(30-14-12),(6+14-12):}]=1/4.[{:(12),(4),(8):}]=[{:(3),(1),(2):}]` `rArr` x=3, y=1 z=2 Hence , the required numbers are 3,1,2. |
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