Solve the following system of equation by Cramer's rule.
x+y+z=9
2x+5y+7z=52
2x+y-z=0

Answer

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

### Given Equations:
1. \( x + y + z = 9 \) (Equation 1)
2. \( 2x + 5y + 7z = 52 \) (Equation 2)
3. \( 2x + y - z = 0 \) (Equation 3)

### Step 1: Formulate the Coefficient Matrix and Calculate the Determinant \( D \)

The coefficient matrix \( A \) is formed from the coefficients of \( x, y, z \):

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{bmatrix} \]

Now, we calculate the determinant \( D \):

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ D = 1 \cdot \begin{vmatrix} 5 & 7 \\ 1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 7 \\ 2 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 5 & 7 \\ 1 & -1 \end{vmatrix} = (5 \cdot -1) - (7 \cdot 1) = -5 - 7 = -12 \)

2. \( \begin{vmatrix} 2 & 7 \\ 2 & -1 \end{vmatrix} = (2 \cdot -1) - (7 \cdot 2) = -2 - 14 = -16 \)

3. \( \begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} = (2 \cdot 1) - (5 \cdot 2) = 2 - 10 = -8 \)

Now substituting back into the determinant formula:

\[ D = 1 \cdot (-12) - 1 \cdot (-16) + 1 \cdot (-8) \]
\[ D = -12 + 16 - 8 = -4 \]

### Step 2: Calculate \( D_1 \)

Replace the first column of \( A \) with the constants from the right-hand side of the equations:

\[ D_1 = \begin{vmatrix} 9 & 1 & 1 \\ 52 & 5 & 7 \\ 0 & 1 & -1 \end{vmatrix} \]

Calculating \( D_1 \):

\[ D_1 = 9 \cdot \begin{vmatrix} 5 & 7 \\ 1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 52 & 7 \\ 0 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 52 & 5 \\ 0 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 5 & 7 \\ 1 & -1 \end{vmatrix} = -12 \) (calculated earlier)

2. \( \begin{vmatrix} 52 & 7 \\ 0 & -1 \end{vmatrix} = (52 \cdot -1) - (7 \cdot 0) = -52 \)

3. \( \begin{vmatrix} 52 & 5 \\ 0 & 1 \end{vmatrix} = (52 \cdot 1) - (5 \cdot 0) = 52 \)

Now substituting back into the determinant formula:

\[ D_1 = 9 \cdot (-12) - 1 \cdot (-52) + 1 \cdot 52 \]
\[ D_1 = -108 + 52 + 52 = -4 \]

### Step 3: Calculate \( D_2 \)

Replace the second column of \( A \) with the constants:

\[ D_2 = \begin{vmatrix} 1 & 9 & 1 \\ 2 & 52 & 7 \\ 2 & 0 & -1 \end{vmatrix} \]

Calculating \( D_2 \):

\[ D_2 = 1 \cdot \begin{vmatrix} 52 & 7 \\ 0 & -1 \end{vmatrix} - 9 \cdot \begin{vmatrix} 2 & 7 \\ 2 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 52 \\ 2 & 0 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 52 & 7 \\ 0 & -1 \end{vmatrix} = -52 \) (calculated earlier)

2. \( \begin{vmatrix} 2 & 7 \\ 2 & -1 \end{vmatrix} = -16 \) (calculated earlier)

3. \( \begin{vmatrix} 2 & 52 \\ 2 & 0 \end{vmatrix} = (2 \cdot 0) - (52 \cdot 2) = -104 \)

Now substituting back into the determinant formula:

\[ D_2 = 1 \cdot (-52) - 9 \cdot (-16) + 1 \cdot (-104) \]
\[ D_2 = -52 + 144 - 104 = -12 \]

### Step 4: Calculate \( D_3 \)

Replace the third column of \( A \) with the constants:

\[ D_3 = \begin{vmatrix} 1 & 1 & 9 \\ 2 & 5 & 52 \\ 2 & 1 & 0 \end{vmatrix} \]

Calculating \( D_3 \):

\[ D_3 = 1 \cdot \begin{vmatrix} 5 & 52 \\ 1 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 52 \\ 2 & 0 \end{vmatrix} + 9 \cdot \begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 5 & 52 \\ 1 & 0 \end{vmatrix} = (5 \cdot 0) - (52 \cdot 1) = -52 \)

2. \( \begin{vmatrix} 2 & 52 \\ 2 & 0 \end{vmatrix} = -104 \) (calculated earlier)

3. \( \begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} = -8 \) (calculated earlier)

Now substituting back into the determinant formula:

\[ D_3 = 1 \cdot (-52) - 1 \cdot (-104) + 9 \cdot (-8) \]
\[ D_3 = -52 + 104 - 72 = -20 \]

### Step 5: Calculate Values of \( x, y, z \)

Using Cramer’s Rule:

\[ x = \frac{D_1}{D} = \frac{-4}{-4} = 1 \]
\[ y = \frac{D_2}{D} = \frac{-12}{-4} = 3 \]
\[ z = \frac{D_3}{D} = \frac{-20}{-4} = 5 \]

### Final Solution:
The solution to the system of equations is:
\[ x = 1, \quad y = 3, \quad z = 5 \]