Express the following equations in the matrix from and solve them by method of reduction :
`2x-y+z=1, x+2y+3z=8, 3x+y-4z=1`

Explanation

The given equation can be written in the matrix form as :
`[(2,-1,1),(1,2,3),(3,1,-4)][(x),(y),(z)]=[(1),(8),(1)]`
Applying `R_(1)hArr R_(2),` we get
`[(1,2,3),(2,-1,1),(3,1,-4)][(x),(y),(z)]=[(8),(1),(1)]`
Applying `R_(2) rarr R_(2)-2R_(1) and R_(3) rarr R_(3) - 3R_(1),` we get
`[(1,2,3),(0,-5,-5),(0,-5,-13)][(x),(y),(z)]=[(8),(-15),(-23)]`
Applying `R_(3) rarr R_(3) - R_(2)`, we get
`[(1,2,3),(0,-5,-5),(0,0,-8)][(x),(y),(z)]=[(8),(-15),(-8)]`
`therefore" "[(x+2y+3z),(0-5y-5z),(0+0-8z)]=[(8),(-15),(-8)]`
By equality of matrices,
`x+2y+3z=8" ...(i)"`
`-5y-5z=-15" ...(ii)"`
`-8z=-8" ...(iii)"`
From equation (iii), z = 1
Putting z = 8 in equation (ii), we get
`-5y = -15+5`
`-5y=-10 rArr y=2`
Putting the value of y and z in equation (i), we get
`x+2(2)+3(1)=8`
`x=8-7=1`
Hence, x =1, y = 2, and z = 3 is the required solution.