निम्नलिखित प्रश्न 7 से 14 तक प्रत्येक समीकरण निकाय को आव्यूह विधि से हल कीजिये । `x-y+z=4`
`2x+y-3z=0`
`x+y+z=2`

Explanation

Writing the given equations in matrix form
AX=B
where `A=[{:(1,-1,1),(2,1,-3),(1,1,1)]:}X=[{:(x),(y),(z):}],B=[{:(4),(0),(2):}]`
i.e, `A=[{:(1,-1,1),(2,1,-3),(1,1,1)]:}[{:(x),(y),(z):}]=[{:(4),(0),(2):}]`
On applying `R_(2) to R_(2)-2R_(3) to -R_(1)`, we get
`A=[{:(1,-1,1),(0,3,-5),(0,2,0)]:}[{:(x),(y),(z):}]=[{:(4),(-8),(-2):}]`
`:. [{:(x-y+z),(3y-5z),(2y):}]=[{:(4),(-8),(-2):}]`
`:.` By equality of matrices, we get
`x-y+z=4" "...(1)`
`3y-5z=-8" "...(2)`
`2y=-2" "...(3)`
`:.` from equation (3) we, get `y=-1`
Substituting `y=-1` in (2), we get
`3(-1)-5z=-8`
`-3-5z=-8`
`:. -5z=-8`
`:. z=1`
Substituting `y=-1, z=1` in equation, (1), z=1 in equation, (1)
`x+1+1=4:x=2`
`:.` The required solution is `x=2,y=-1,z=1`