1. Use the provided inverse matrix to solve the following linear system of equations:
\[
\left\{\begin{array}{c}
3 x+y=1 \\
2 x-y+2 z=-4 \\
x+y+z=3
\end{array}\right.
\]
Inverse 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 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 \(\boxed{z = 0.44444444}\).