Using matrix method, solve the following system of equations: `2/x+3/y+(10)/z=4,4/x-6/y+5/z=1,6/x+9/y-(20)/z=2; x ,\ y ,\ z\ !=0`

Explanation

Putting `1/x=u, 1/y=v and 1/z=w` , the given equation become
`2u+3v+10w=4 " " …(i)`
`4u-6v+5w=2 " " …(ii)`
`6u+9v-20w=2 " " …(iii)`
`Let A=[{:(2 " "3 " "10),(4 " "-6" "5),(6" "9 " "20):}],Y=[{:(u),(v),(w):}] and B=[{:(4),(1),(2):}]`
Then , the given system in matrix form is AY=B
`Now, |A|=[{:(2 " "3 " "10),(4 " "-6" "5),(6" "9 " "20):}]`
=`[{:(2 " "3 " "10),(4 " "-12" "-15),(0" "0 " "-50):}] " "[R_2rarrR_2-2R_1,R_3rarrR_3-3R_1]`
`(-50).(-24-0)=1200 ne 0`
Thus A is invertible
So, the given system has a unique solution ,`Y=A^(-1)B`
The minors of the elements of |A| are
`M_(11)=75,M_(12)=-110,M_(13)=72,`
`M_(21)=-150,M_(22)=-100,M_(23)=0,`
`M_(31)=75,M_(32)=-30,M_(33)=-24,`
So, the cofactors of the elements of |A| are
`A_(11)=75,A_(12)=-110,A_(13)=72,`
`A_(21)=-150,A_(22)=-100,A_(23)=0,`
`A_(31)=75,A_(32)=-30,A_(33)=-24,`
`therefore (adjA)=[{:(75 " "110 " "72),(150 " "-100 " "0),(75 " "30 " "-24):}]=[{:(75 " "150 " "75),(110 " "-100 " "30),(72 " "0 " "-24):}]`
`therefore A^(-1)=1/|A|.(adjA)=1/(1200).[{:(75 " "150 " "75),(110 " "-100 " "30),(72 " "0 " "-24):}]`
`therefore Y=A^(-1)B`
`rArr[{:(u),(v),(w):}]=1/(1200).[{:(75 " "150 " "75),(110 " "-100 " "30),(72 " "0 " "-24):}][{:(4),(1),(2):}]`
`1/(1200).[{:(300+150+150),(440-100+60),(288+0-48):}]=1/(1200).[{:(600),(400),(240):}]`
`=[{:((600)/1200),((420)/(1200)),((240)/(1200)):}]=[{:(1/2),(1/3),(1/5):}]`
`rArru=1/2, v=1/3,w=1/5`
`rArr1/x=1/2,1/y=1/3 and 1/z=1/5`
`rArr x=2, y=3 and z=5.`
Hence x=2, y=3 and z=5