Solve the system of equations `x+2y+3z=1,2x+3y+2z=2 and 3x+3y+4z=1` with the help of matrix inversion.

Explanation

we have
`x+2y+3z=1,2x+3y+2z=2 and 3x+3y+4z=1`
the given system of equation in the matrix form are written as below.
`[(1,2,3),(2,3,2),(3,3,4)]=[(,1),(,2),(,1)]`
` AX=B`
`rArr X=A^(-1)B`
where` " " A=[(1,2,3),(2,3,2),(3,3,4)],X=[(,x),(,y),(,z)]and B= [(,1),(,2),(,1)]`
`|A|=1(12-6)-2(8-6)+3(6-9)`
` =6-4-9=-7!=0`
` therefore A^(-1)` exixts and has unique solution.
Let C be the matrix of confactor of elements in `|A|.` Now, confactor along `R_(1)=6,-2,-3`
confactor along `R_(2)=1,-5,3`
and confacators along `R_(3)=-5,4,-1`
`therefore " " C=[(6,-2,-3),(1,-5,3),(-5,4,-1)]`
`therefore" " A=C^(T)`
`rArr " " adjA=[(6,-2,-3),(1,-5,3),(-5,4,-1)]=[(6,1,-5),(-2,-5,4),(-3,3,-1)]`
`rArr" " A^(-1)=(adjA)/(|A|)=-(1)/(7)[(6,1,-5),(-2,-5,4),(-3,3,-1)]`
`=[(-(6)/(7),-(1)/(7),(5)/(7)),((2)/(7),(5)/(7),-(4)/(7)),((3)/(7),-(3)/(7),(1)/(7))]`
from Eq. (i),`X=A^(-1)B`
`rArr " " [(,x),(,y),(,z)]=[(-(6)/(7),-(1)/(7),(5)/(7)),((2)/(7),(5)/(7),-(4)/(7)),((3)/(7),-(3)/(7),(1)/(7))][(,1),(,2),(,1)]=[(-(3)/(7)),((8)/(7)),(-(2)/(7))]`
Hence `x=-(3)/(7),y=(8)/(7) and z=-(2)/(7)` is the required solution.