Solve system of linear equations, using matrix method,
`x-y+z=4`,
`2x+y-3z=0`,
`x+y+z=2`

Explanation

Given system of equations,
x-y+z=4,
2x+y-3z=0, x+y+z=2
`[{:(1,-1,1),(2,1,-3),(1,1,1):}][{:(x),(y),(z):}][{:(4),(0),(2):}]rArr AX=B`
`therefore" "A=[{:(1,-1,1),(2,1,-3),(1,1,1):}]`
`rArr|A|=[{:(1,-1,1),(2,1,-3),(1,1,1):}]`
`=1(1+3)-(-1)(2+3)+1(2-1)`
`=4+5+1=10ne0`
`therefore` A is invertible.
`Now, A_(11)=4, A_(12)=-5, A_(13)=1`
`A_(21)=2, A_(22)=0, A_(23)=-2`
`A_(31)=2, A_(32)=5, A_(33)=3`
`therefore" adj A="[{:(4,-5,1),(2,0,-2),(2,5,3):}][{:(4,2,2),(-5,0,5),(1,-2,3):}]`
`"and A"^(-1)=1/|A|".adj A"1/10"[{:(4,-5,1),(2,0,-2),(2,5,3):}]`
From equation (1),
`rArr" "[{:(x),(y),(z):}]=1/10[{:(4,-5,1),(2,0,-2),(2,5,3):}][{:(4),(0),(2):}]`
`=1/10[{:(16+0+4),(-20+0+10),(4+0+6):}]`
`therefore` x=2, y=-1, z=1