Solve the following system of equations by Cramer's Rule :
`x+3y=4`, `y+3z=7`, `4x+z=6`

Answer

To solve the system of equations using Cramer's Rule, we will follow these steps:

Given equations:
1. \( x + 3y = 4 \) (Equation 1)
2. \( y + 3z = 7 \) (Equation 2)
3. \( 4x + z = 6 \) (Equation 3)

### Step 1: Write the system in matrix form
We can express the system of equations 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.

From the equations, we can identify:
\[ A = \begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 3 \\ 4 & 0 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ 7 \\ 6 \end{bmatrix} \]

### Step 2: Calculate the determinant of matrix \( A \) (denoted as \( \Delta \))
\[ \Delta = \begin{vmatrix} 1 & 3 & 0 \\ 0 & 1 & 3 \\ 4 & 0 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \Delta = 1 \cdot \begin{vmatrix} 1 & 3 \\ 0 & 1 \end{vmatrix} - 3 \cdot \begin{vmatrix} 0 & 3 \\ 4 & 1 \end{vmatrix} + 0 \cdot \begin{vmatrix} 0 & 1 \\ 4 & 0 \end{vmatrix} \]
\[ = 1 \cdot (1 \cdot 1 - 0 \cdot 3) - 3 \cdot (0 \cdot 1 - 4 \cdot 3) \]
\[ = 1 - 3 \cdot (-12) = 1 + 36 = 37 \]

### Step 3: Calculate \( \Delta_x \) (determinant when replacing the first column with \( B \))
\[ \Delta_x = \begin{vmatrix} 4 & 3 & 0 \\ 7 & 1 & 3 \\ 6 & 0 & 1 \end{vmatrix} \]

Calculating \( \Delta_x \):
\[ \Delta_x = 4 \cdot \begin{vmatrix} 1 & 3 \\ 0 & 1 \end{vmatrix} - 3 \cdot \begin{vmatrix} 7 & 3 \\ 6 & 1 \end{vmatrix} + 0 \cdot \begin{vmatrix} 7 & 1 \\ 6 & 0 \end{vmatrix} \]
\[ = 4 \cdot (1 \cdot 1 - 0 \cdot 3) - 3 \cdot (7 \cdot 1 - 6 \cdot 3) \]
\[ = 4 - 3 \cdot (7 - 18) = 4 - 3 \cdot (-11) = 4 + 33 = 37 \]

### Step 4: Calculate \( \Delta_y \) (determinant when replacing the second column with \( B \))
\[ \Delta_y = \begin{vmatrix} 1 & 4 & 0 \\ 0 & 7 & 3 \\ 4 & 6 & 1 \end{vmatrix} \]

Calculating \( \Delta_y \):
\[ \Delta_y = 1 \cdot \begin{vmatrix} 7 & 3 \\ 6 & 1 \end{vmatrix} - 4 \cdot \begin{vmatrix} 0 & 3 \\ 4 & 1 \end{vmatrix} + 0 \cdot \begin{vmatrix} 0 & 7 \\ 4 & 6 \end{vmatrix} \]
\[ = 1 \cdot (7 \cdot 1 - 6 \cdot 3) - 4 \cdot (0 \cdot 1 - 4 \cdot 3) \]
\[ = 7 - 18 - 4 \cdot (-12) = 7 - 18 + 48 = 37 \]

### Step 5: Calculate \( \Delta_z \) (determinant when replacing the third column with \( B \))
\[ \Delta_z = \begin{vmatrix} 1 & 3 & 4 \\ 0 & 1 & 7 \\ 4 & 0 & 6 \end{vmatrix} \]

Calculating \( \Delta_z \):
\[ \Delta_z = 1 \cdot \begin{vmatrix} 1 & 7 \\ 0 & 6 \end{vmatrix} - 3 \cdot \begin{vmatrix} 0 & 7 \\ 4 & 6 \end{vmatrix} + 4 \cdot \begin{vmatrix} 0 & 1 \\ 4 & 0 \end{vmatrix} \]
\[ = 1 \cdot (1 \cdot 6 - 0 \cdot 7) - 3 \cdot (0 \cdot 6 - 4 \cdot 7) + 4 \cdot (0 \cdot 0 - 4 \cdot 1) \]
\[ = 6 + 3 \cdot 28 - 16 = 6 + 84 - 16 = 74 \]

### Step 6: Solve for \( x, y, z \)
Using Cramer’s Rule:
\[ x = \frac{\Delta_x}{\Delta} = \frac{37}{37} = 1 \]
\[ y = \frac{\Delta_y}{\Delta} = \frac{37}{37} = 1 \]
\[ z = \frac{\Delta_z}{\Delta} = \frac{74}{37} = 2 \]

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