Apply Cramer's rule to solve the simultaneous equations. `(i)x+2y+3z=2` `x-y+z=3` `5x-11y+z=17`

Answer

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

Given equations:
1. \( x + 2y + 3z = 2 \) (Equation 1)
2. \( x - y + z = 3 \) (Equation 2)
3. \( 5x - 11y + z = 17 \) (Equation 3)

### Step 1: Write the equations in matrix form
We can express the system of equations in the form of a matrix equation \( AX = B \), where:
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & -1 & 1 \\ 5 & -11 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ 3 \\ 17 \end{bmatrix} \]

### Step 2: Calculate the determinant \( D \) of matrix \( A \)
The determinant \( D \) is calculated as follows:
\[ D = \begin{vmatrix} 1 & 2 & 3 \\ 1 & -1 & 1 \\ 5 & -11 & 1 \end{vmatrix} \]

Calculating \( D \):
\[ D = 1 \begin{vmatrix} -1 & 1 \\ -11 & 1 \end{vmatrix} - 2 \begin{vmatrix} 1 & 1 \\ 5 & 1 \end{vmatrix} + 3 \begin{vmatrix} 1 & -1 \\ 5 & -11 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 1 \\ -11 & 1 \end{vmatrix} = (-1)(1) - (1)(-11) = -1 + 11 = 10 \)

2. \( \begin{vmatrix} 1 & 1 \\ 5 & 1 \end{vmatrix} = (1)(1) - (1)(5) = 1 - 5 = -4 \)

3. \( \begin{vmatrix} 1 & -1 \\ 5 & -11 \end{vmatrix} = (1)(-11) - (-1)(5) = -11 + 5 = -6 \)

Now substituting back into the determinant calculation:
\[ D = 1(10) - 2(-4) + 3(-6) = 10 + 8 - 18 = 0 \]

### Step 3: Check the determinant
Since \( D = 0 \), the system of equations does not have a unique solution. According to Cramer's Rule, if the determinant is zero, it indicates that the system may either have no solution or infinitely many solutions.

### Conclusion
The determinant \( D \) being zero implies that the values of \( x, y, z \) cannot be uniquely determined from the given equations. Therefore, we conclude that the system does not have a unique solution.

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