Using matrices, solve the following system of equations for x,y and z.
`{:(x+2y+z=8),(2x+y-z=1),(x-y+z=2):}`

Answer

To solve the system of equations using matrices, we can follow these steps:

### Step 1: Write the system of equations in matrix form
We have the following system of equations:
1. \( x + 2y + z = 8 \)
2. \( 2x + y - z = 1 \)
3. \( x - y + z = 2 \)

We can express this in the form \( AX = B \), where:
- \( A \) is the coefficient matrix,
- \( X \) is the column matrix of variables,
- \( B \) is the column matrix of constants.

Thus, we have:
\[ A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 8 \\ 1 \\ 2 \end{pmatrix} \]

### Step 2: Calculate the determinant of matrix \( A \) (denoted as \( \Delta \))
To find \( \Delta \), we calculate the determinant of matrix \( A \):
\[ \Delta = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the remaining rows.

Calculating:
\[ \Delta = 1 \cdot (1 \cdot 1 - (-1) \cdot (-1)) - 2 \cdot (2 \cdot 1 - (-1) \cdot 1) + 1 \cdot (2 \cdot (-1) - 1 \cdot 1) \]
\[ = 1 \cdot (1 - 1) - 2 \cdot (2 + 1) + 1 \cdot (-2 - 1) \]
\[ = 0 - 2 \cdot 3 - 3 = -6 - 3 = -9 \]

### Step 3: Calculate \( \Delta_x \), \( \Delta_y \), and \( \Delta_z \)
Now we will calculate \( \Delta_x \), \( \Delta_y \), and \( \Delta_z \) by replacing the respective columns of \( A \) with \( B \).

#### Calculate \( \Delta_x \)
Replace the first column of \( A \) with \( B \):
\[ \Delta_x = \begin{vmatrix} 8 & 2 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 1 \end{vmatrix} \]
Calculating:
\[ = 8 \cdot (1 \cdot 1 - (-1) \cdot (-1)) - 2 \cdot (1 \cdot 1 - (-1) \cdot 2) + 1 \cdot (1 \cdot (-1) - 1 \cdot 2) \]
\[ = 8 \cdot (1 - 1) - 2 \cdot (1 + 2) + 1 \cdot (-1 - 2) \]
\[ = 0 - 2 \cdot 3 - 3 = -6 - 3 = -9 \]

#### Calculate \( \Delta_y \)
Replace the second column of \( A \) with \( B \):
\[ \Delta_y = \begin{vmatrix} 1 & 8 & 1 \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} \]
Calculating:
\[ = 1 \cdot (1 \cdot 1 - (-1) \cdot 2) - 8 \cdot (2 \cdot 1 - (-1) \cdot 1) + 1 \cdot (2 \cdot (-1) - 1 \cdot 1) \]
\[ = 1 \cdot (1 + 2) - 8 \cdot (2 + 1) + 1 \cdot (-2 - 1) \]
\[ = 3 - 8 \cdot 3 - 3 = 3 - 24 - 3 = -24 \]

#### Calculate \( \Delta_z \)
Replace the third column of \( A \) with \( B \):
\[ \Delta_z = \begin{vmatrix} 1 & 2 & 8 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix} \]
Calculating:
\[ = 1 \cdot (1 \cdot 2 - 1 \cdot (-1)) - 2 \cdot (2 \cdot 2 - 1 \cdot 1) + 8 \cdot (2 \cdot (-1) - 1 \cdot 1) \]
\[ = 1 \cdot (2 + 1) - 2 \cdot (4 - 1) + 8 \cdot (-2 - 1) \]
\[ = 3 - 2 \cdot 3 - 8 \cdot 3 = 3 - 6 - 24 = -27 \]

### Step 4: Calculate the values of \( x, y, z \)
Using Cramer’s rule:
\[ x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta} \]
Substituting the values:
\[ x = \frac{-9}{-9} = 1, \quad y = \frac{-24}{-9} = 2, \quad z = \frac{-27}{-9} = 3 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 1, \quad y = 2, \quad z = 3 \]