Solve the following system of equations, using matrix method. `x+2y+z=7,x+3z=11 ,2x-3y=1`

Explanation

The given system of system of equations is
`x+2y+z=7`
`x+0y+3z=11`
`2x-3y+0z=1`
or `[(1,2,1),(1,0,3),(2,-3,0)][(x),(y),(z)]=[(7),(11),(1)]`
or `AX=B`, where
`A=[(1,2,1),(1,0,3),(2,-3,0)], X=[(x),(y),(z)]`, and `B=[(7),(11),(1)]`
Now, `|A|=|(1,2,1),(1,0,3),(2,-3,0)|=18`
So, the given system of equations has a unique solution given by `X=A^(-1) B`. therefore,
adj `A=[(9,6,-3),(-3,-2,7),(6,-2,-2)]^(T)=[(9,-3,6),(6,-2,-2),(-3,7,-2)]`
`implies A^(-1)=1/(|A|)` adj `A=1/18[(9,-3,6),(6,-2,-2),(-3,7,-2)]`
Now, `X=A^(-1) B`
`implies X=1/18[(9,-3,6),(6,-2,-2),(-3,7,-2)][(7),(11),(1)]=1/18 [(63-33+6),(42-22-2),(-21+77-2)]`
`implies [(x),(y),(z)]=1/18 [(36),(18),(54)]=[(2),(1),(3)]`
`implies x=2, y=1, z=3`