Consider the following system of linear equations.
\[
\left\{\begin{array}{c}
\frac{1}{10} x+\frac{2}{5} y-\frac{4}{5} z=8 \\
\frac{3}{5} x+\frac{2}{5} y+\frac{1}{5} z=3 \\
-\frac{2}{5} x-\frac{3}{5} y+\frac{1}{5} z=-4
\end{array}\right.
\]
use the inverse of the coefficient matrix to write the solution matrix $\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ as a product of two matrices.
Then find the solution of the system.
Note that the ALEKS graphing calculator can be used to make computations easier.
Product:
\[
\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=
\]
$\mathrm{x}$
Solution:
\[
x=\square \quad y=\square \quad z=\square
\]

Step 1 ：Write the system of equations in matrix form as A*X = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. The matrix form of the given system is: \[A = \begin{bmatrix} 0.1 & 0.4 & -0.8 \\ 0.6 & 0.4 & 0.2 \\ -0.4 & -0.6 & 0.2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 8 \\ 3 \\ -4 \end{bmatrix}\]

Step 2 ：Find the inverse of the coefficient matrix A. The inverse of A is: \[A_{inv} = \begin{bmatrix} 2 & 4 & 4 \\ -2 & -3 & -5 \\ -2 & -1 & -2 \end{bmatrix}\]

Step 3 ：Multiply both sides of the equation by the inverse of A to find the solution matrix X. The product of the inverse of the coefficient matrix and the constant matrix is: \[X = A_{inv} * B = \begin{bmatrix} 12 \\ -5 \\ -11 \end{bmatrix}\]

Step 4 ：So, the solution of the system is: \[x=\boxed{12}, y=\boxed{-5}, z=\boxed{-11}\]