solve the system of equations `x+y+z=6,x+2y+3z=14` and` x+4y+7z=30` with the help of matrix method.

Explanation

we have, `x+y+z=6,`
` x+2y+3z=14`
and `x+4y+7z=30`
the given system of equations In the matrix form are written as below:
`=[(1,1,1),(1,2,3),(1,4,7)][(,x),(,y),(,z)][(,6),(,14),(,30)]`
`AB=B`
where, `A=[(1,1,1),(1,2,3),(1,4,7)],X=[(,x),(,y),(,z)] and B = [(,6),(,14),(,30)]`
`|A|=1(14-12)-1(7-3)+1(4-2)=2-4=2=0`
`therefore` the equations has on solution or an infinite number of solutions. to decided about this, we proceed to find
`(adj A)B.`
Let c be the matarix of confactors of elements in `|A|`
Now, confactors along `R_(2)=-2,-4,2`
cofactors along `R_(2)=-3,6,-3`
and cofactors along `R_(3) = 1,-2,1`
`therefore" " C=[(2,-4,2),(-3,6,-3),(1,-2,1)]`
`rArr" "adj= C^(T)=[(2,-3,1),(-4,6,-2),(2,-3,1)]`
then `rArr" "(adjA)B=[(2,-3,1),(-4,6,-2),(2,-3,1)][(,6),(,14),(,30)][(,0),(,0),(,0)]`
Hence, both conditions |A|=0 and (adjA)B=O are satisfied, them the system of equations is consistent and has an infinite number of solutions.
proceed as follows:
`[A:B]=[(1,1,1,vdots,6),(1,2,3,vdots,14),(1,4,7,vdots,30)]`
Applying `R_(3) to R_(3)-2R_(2),` then
`[A:B]=[(1,1,1,vdots,6),(0,1,2,vdots,8),(0,2,4,vdots,16)]`
Applying `R_(3) to R_(3)-2R_(2),` then
`[A:B]=[(1,1,1,vdots,6),(0,1,2,vdots,8),(0,0,0,vdots,0)]`
then, Eq. (i) reduces to
`[(1,1,1),(0,1,2),(0,0,0)][(,x),(,y),(,z)]=[(,6),(,8),(,0)] rArr [(x+y+z),(y+2z),(0)]= [(,6),(,8),(,0)]`
on comparing `x+y+z=6 and y+2z=8`
Taking `z=k in R`,then `y=8-2k and x=k-2.`
since, k is arbitaray , henece the number of solutions is infinite.