Solve system of linear equations, using matrix method,
`2x+y+z=1`
`x-2y-z=3/2`
`3y-5z=9`

Explanation

Given system of equations,
2x+y+z=1
`x-2y-z=3/2`
3y-5z=9
`rArr[{:(2,1,1),(1,-2,-1),(0,3,-5):}][{:(x),(y),(z):}]-[{:(1),(3/2),(z):}][{:(1),(3/2),(9):}]rArr AX=B`
`A=[{:(2,1,1),(1,-2,-1),(0,3,-5):}] `
`rArr" "|A|=[{:(2,1,1),(1,-2,-1),(0,3,-5):}]`
`therefore` A is invertible
`"Now "A_(11)=13, A_(12)=5, A_(13)=3`
`A_(21)=8, A_(22)=-10, A_(23)=-6`
`A_(31)=1, A_(32)=3, A_(33)=-5`
`therefore" adj A"=[{:(13,5,3),(8,-1,-6),(3,-6,-5):}]`
From equation (1)
`AX=B rArr X=A^(-1)B`
`[{:(x),(y),(z):}]=1/34[{:(13,5,3),(8,-1,-6),(3,-6,-5):}][{:(1),(3/2),(9):}]`
`1/34[{:(13,+12,+9),(5,-10,3),(3,-6,-5):}][{:(1),(1/2),(-3/2):}]`
`therefore " "x=1, y=1/2, z=-3/2`