The sum of three numbers is 6. If we multiply the third number 2 and add the first number to the result, we get 7. Be adding second and third numbers to three times the first number we get 12. Use determinants to find the numbers.

Explanation

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.