Solve the following system of equations by Cramer's Rule :
`3x+y+z=10`, `x+y+z=0`, `5x-9y=1`.

Answer

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

### Given Equations:
1. \( 3x + y + z = 10 \) (Equation 1)
2. \( x + y + z = 0 \) (Equation 2)
3. \( 5x - 9y + 0z = 1 \) (Equation 3)

### Step 1: Write the Coefficient Matrix (A) and the Constant Matrix (B)
The coefficient matrix \( A \) and the constant matrix \( B \) are formed as follows:

\[ A = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 1 & 1 \\ 5 & -9 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 10 \\ 0 \\ 1 \end{bmatrix} \]

### Step 2: Calculate the Determinant of Matrix A (\( \Delta \))
To find \( \Delta \), we calculate the determinant of matrix \( A \):

\[ \Delta = \begin{vmatrix} 3 & 1 & 1 \\ 1 & 1 & 1 \\ 5 & -9 & 0 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ \Delta = 3 \begin{vmatrix} 1 & 1 \\ -9 & 0 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 5 & 0 \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 5 & -9 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 1 \\ -9 & 0 \end{vmatrix} = (1)(0) - (1)(-9) = 9 \)

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

3. \( \begin{vmatrix} 1 & 1 \\ 5 & -9 \end{vmatrix} = (1)(-9) - (1)(5) = -14 \)

Substituting back into the determinant formula:

\[ \Delta = 3(9) - 1(-5) + 1(-14) = 27 + 5 - 14 = 18 \]

### Step 3: Calculate \( \Delta_x \), \( \Delta_y \), and \( \Delta_z \)

#### Finding \( \Delta_x \)
Replace the first column of \( A \) with \( B \):

\[ \Delta_x = \begin{vmatrix} 10 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & -9 & 0 \end{vmatrix} \]

Calculating this determinant:

\[ \Delta_x = 10 \begin{vmatrix} 1 & 1 \\ -9 & 0 \end{vmatrix} - 1 \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} + 1 \begin{vmatrix} 0 & 1 \\ 1 & -9 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 1 \\ -9 & 0 \end{vmatrix} = 9 \) (calculated earlier)

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

3. \( \begin{vmatrix} 0 & 1 \\ 1 & -9 \end{vmatrix} = (0)(-9) - (1)(1) = -1 \)

Substituting back:

\[ \Delta_x = 10(9) - 1(-1) + 1(-1) = 90 + 1 - 1 = 90 \]

#### Finding \( \Delta_y \)
Replace the second column of \( A \) with \( B \):

\[ \Delta_y = \begin{vmatrix} 3 & 10 & 1 \\ 1 & 0 & 1 \\ 5 & 1 & 0 \end{vmatrix} \]

Calculating this determinant:

\[ \Delta_y = 3 \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} - 10 \begin{vmatrix} 1 & 1 \\ 5 & 0 \end{vmatrix} + 1 \begin{vmatrix} 1 & 0 \\ 5 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

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

2. \( \begin{vmatrix} 1 & 1 \\ 5 & 0 \end{vmatrix} = -5 \) (calculated earlier)

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

Substituting back:

\[ \Delta_y = 3(-1) - 10(-5) + 1(1) = -3 + 50 + 1 = 48 \]

#### Finding \( \Delta_z \)
Replace the third column of \( A \) with \( B \):

\[ \Delta_z = \begin{vmatrix} 3 & 1 & 10 \\ 1 & 1 & 0 \\ 5 & -9 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ \Delta_z = 3 \begin{vmatrix} 1 & 0 \\ -9 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 0 \\ 5 & 1 \end{vmatrix} + 10 \begin{vmatrix} 1 & 1 \\ 5 & -9 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 0 \\ -9 & 1 \end{vmatrix} = 1 \)

2. \( \begin{vmatrix} 1 & 0 \\ 5 & 1 \end{vmatrix} = 1 \)

3. \( \begin{vmatrix} 1 & 1 \\ 5 & -9 \end{vmatrix} = -14 \) (calculated earlier)

Substituting back:

\[ \Delta_z = 3(1) - 1(1) + 10(-14) = 3 - 1 - 140 = -138 \]

### Step 4: Apply Cramer’s Rule
Using Cramer's Rule:

\[ x = \frac{\Delta_x}{\Delta} = \frac{90}{18} = 5 \]
\[ y = \frac{\Delta_y}{\Delta} = \frac{48}{18} = \frac{8}{3} \]
\[ z = \frac{\Delta_z}{\Delta} = \frac{-138}{18} = -\frac{23}{3} \]

### Final Solution:
\[ x = 5, \quad y = \frac{8}{3}, \quad z = -\frac{23}{3} \]