Solve system of linear equations, using matrix method, in questions 7 to 14.
x-y+2z=7, 3x+4y-5z=-5, 2x-y+3z=12

Explanation

Given system of equations,
x-y+2z=7,
3x+4y-5z=-5,
2x-y+3z=12
`=[{:(1,-1,2),(3,4,-5),(2,-1,3):}][{:(x),(y),(z):}]=[{:(7),(-5),(12):}]rArrAX=B`
`therefore" "A=[{:(1,-1,2),(3,4,-5),(2,-1,3):}]`
`rArr" "|A|=[{:(1,-1,2),(3,4,-5),(2,-1,3):}]`
=1(12-5)-(-1)(9+10)+2(-3-8)
`=7+19-22=4ne0`
`therefore` A is invertible.
`"Now, "A_(11)=7, A_(12)=-19, A_(13)=-11`
`A_(21)=1, A_(22)=-1, A_(23)=-1`
`A_(31)=-3, A_(32)=11, A_(33)=7`
`therefore" adjA"=[{:(7,-19,-11),(1,-10,-1),(-3,11,7):}]=[{:(7,1,-3),(-19,-1,11),(-11,-1,7):}]`
`"and A"^(-1)=1/|A|"adj A"=1/4[{:(7,1,-3),(-19,-1,11),(-11,-1,7):}]`
`"=1/4[{:(49-5-36),(-133+5+132),(-775+5+84):}]=[{:(2),(1),(3):}]`
`therefore"x=2, y=1, z=3`