Step 1 :Given the system of linear equations: \[\left\{\begin{array}{c} 3x+y=1 \\ 2x-y+2z=-4 \\ x+y+z=3 \end{array}\right.\]
Step 2 :We are also given the inverse matrix of the coefficient matrix: \[\left[\begin{array}{ccc}3 & 1 & 0 \\ 2 & -1 & 2 \\ 1 & 1 & 1\end{array}\right]^{-1}=\frac{1}{9}\left[\begin{array}{ccc}3 & 1 & -2 \\ 0 & -3 & 6 \\ -3 & 2 & 5\end{array}\right]\]
Step 3 :We can represent the system of equations in matrix form as AX = B, where A is the matrix of coefficients, X is the matrix of variables, and B is the matrix of constants.
Step 4 :We can solve for X by multiplying both sides of the equation by the inverse of A, which gives us X = A^-1B.
Step 5 :Performing the matrix multiplication, we find X = [-0.77777778, 3.33333333, 0.44444444].
Step 6 :Thus, the solution to the system of equations is \(\boxed{x = -0.77777778}\), \(\boxed{y = 3.33333333}\), and \(\boxed{z = 0.44444444}\).