Using matrices, solve the following system of equations for x,y and z.
`3x-2y+3z=8`,`2x+y-z=1`, `4x-3y+2z=4`

Answer

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

Given equations:
1. \( 3x - 2y + 3z = 8 \) (Equation 1)
2. \( 2x + y - z = 1 \) (Equation 2)
3. \( 4x - 3y + 2z = 4 \) (Equation 3)

### Step 1: Write the equations in matrix form \( AX = B \)

We can express the system of equations in the form of matrices:
\[ A = \begin{pmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} \]

### Step 2: Calculate the determinant of matrix \( A \) (denoted as \( \Delta \))

The determinant \( \Delta \) of matrix \( A \) can be calculated using the formula for the determinant of a 3x3 matrix:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Where:
\[ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix \( A \):
\[ \Delta = 3 \cdot (1 \cdot 2 - (-1) \cdot (-3)) - (-2) \cdot (2 \cdot 2 - (-1) \cdot 4) + 3 \cdot (2 \cdot (-3) - 1 \cdot 4) \]
Calculating this step by step:
\[ = 3 \cdot (2 - 3) + 2 \cdot (4 + 4) + 3 \cdot (-6 - 4) \]
\[ = 3 \cdot (-1) + 2 \cdot 8 + 3 \cdot (-10) \]
\[ = -3 + 16 - 30 = -17 \]

### Step 3: Calculate \( \Delta_x \) (replace the first column of \( A \) with \( B \))

To find \( \Delta_x \):
\[ \Delta_x = \begin{vmatrix} 8 & -2 & 3 \\ 1 & 1 & -1 \\ 4 & -3 & 2 \end{vmatrix} \]
Calculating \( \Delta_x \):
\[ = 8 \cdot (1 \cdot 2 - (-1) \cdot (-3)) - (-2) \cdot (1 \cdot 2 - (-1) \cdot 4) + 3 \cdot (1 \cdot (-3) - 1 \cdot 4) \]
\[ = 8 \cdot (2 - 3) + 2 \cdot (2 + 4) + 3 \cdot (-3 - 4) \]
\[ = 8 \cdot (-1) + 2 \cdot 6 + 3 \cdot (-7) \]
\[ = -8 + 12 - 21 = -17 \]

### Step 4: Calculate \( \Delta_y \) (replace the second column of \( A \) with \( B \))

To find \( \Delta_y \):
\[ \Delta_y = \begin{vmatrix} 3 & 8 & 3 \\ 2 & 1 & -1 \\ 4 & 4 & 2 \end{vmatrix} \]
Calculating \( \Delta_y \):
\[ = 3 \cdot (1 \cdot 2 - (-1) \cdot 4) - 8 \cdot (2 \cdot 2 - (-1) \cdot 4) + 3 \cdot (2 \cdot 4 - 1 \cdot 4) \]
\[ = 3 \cdot (2 + 4) - 8 \cdot (4 + 4) + 3 \cdot (8 - 4) \]
\[ = 3 \cdot 6 - 8 \cdot 8 + 3 \cdot 4 \]
\[ = 18 - 64 + 12 = -34 \]

### Step 5: Calculate \( \Delta_z \) (replace the third column of \( A \) with \( B \))

To find \( \Delta_z \):
\[ \Delta_z = \begin{vmatrix} 3 & -2 & 8 \\ 2 & 1 & 1 \\ 4 & -3 & 4 \end{vmatrix} \]
Calculating \( \Delta_z \):
\[ = 3 \cdot (1 \cdot 4 - 1 \cdot (-3)) - (-2) \cdot (2 \cdot 4 - 1 \cdot 4) + 8 \cdot (2 \cdot (-3) - 1 \cdot 4) \]
\[ = 3 \cdot (4 + 3) + 2 \cdot (8 - 4) + 8 \cdot (-6 - 4) \]
\[ = 3 \cdot 7 + 2 \cdot 4 + 8 \cdot (-10) \]
\[ = 21 + 8 - 80 = -51 \]

### Step 6: 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 we found:
\[ x = \frac{-17}{-17} = 1 \]
\[ y = \frac{-34}{-17} = 2 \]
\[ z = \frac{-51}{-17} = 3 \]

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