The value of `x, y, z` for the following system of equations `x + y + z = 6, x - y + 2z = 5, 2x + y - z = 1` are

Options

A. `x = 1, y = 2, z = 3 `
B. `x = 2, y = 1, z = 3`
C. `x = -1, y = 2, z = 3`
D. `x = y = z = 3`

Answer

To solve the system of equations given by:

1. \( x + y + z = 6 \) (Equation 1)
2. \( x - y + 2z = 5 \) (Equation 2)
3. \( 2x + y - z = 1 \) (Equation 3)

we will use Cramer's rule.

### Step 1: Set up the coefficient matrix and calculate the determinant \( D \)

The coefficient matrix \( A \) is:

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

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

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

Using the formula for the determinant of a 3x3 matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the corresponding elements of the remaining rows.

Calculating \( D \):

\[ D = 1 \cdot ((-1) \cdot (-1) - 2 \cdot 1) - 1 \cdot (1 \cdot (-1) - 2 \cdot 2) + 1 \cdot (1 \cdot 1 - (-1) \cdot 2) \]

\[ = 1 \cdot (1 - 2) - 1 \cdot (-1 - 4) + 1 \cdot (1 + 2) \]

\[ = 1 \cdot (-1) - 1 \cdot (-5) + 1 \cdot 3 \]

\[ = -1 + 5 + 3 = 7 \]

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

Replace the first column of the coefficient matrix with the constants from the right-hand side of the equations:

\[ D_x = \begin{vmatrix} 6 & 1 & 1 \\ 5 & -1 & 2 \\ 1 & 1 & -1 \end{vmatrix} \]

Calculating \( D_x \):

\[ D_x = 6 \cdot ((-1) \cdot (-1) - 2 \cdot 1) - 1 \cdot (5 \cdot (-1) - 2 \cdot 1) + 1 \cdot (5 \cdot 1 - (-1) \cdot 6) \]

\[ = 6 \cdot (1 - 2) - 1 \cdot (-5 - 2) + 1 \cdot (5 + 6) \]

\[ = 6 \cdot (-1) - 1 \cdot (-7) + 1 \cdot 11 \]

\[ = -6 + 7 + 11 = 12 \]

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

Replace the second column of the coefficient matrix with the constants:

\[ D_y = \begin{vmatrix} 1 & 6 & 1 \\ 1 & 5 & 2 \\ 2 & 1 & -1 \end{vmatrix} \]

Calculating \( D_y \):

\[ D_y = 1 \cdot (5 \cdot (-1) - 2 \cdot 1) - 6 \cdot (1 \cdot (-1) - 2 \cdot 2) + 1 \cdot (1 \cdot 1 - 5 \cdot 2) \]

\[ = 1 \cdot (-5 - 2) - 6 \cdot (-1 - 4) + 1 \cdot (1 - 10) \]

\[ = 1 \cdot (-7) - 6 \cdot (-5) + 1 \cdot (-9) \]

\[ = -7 + 30 - 9 = 14 \]

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

Replace the third column of the coefficient matrix with the constants:

\[ D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & -1 & 5 \\ 2 & 1 & 1 \end{vmatrix} \]

Calculating \( D_z \):

\[ D_z = 1 \cdot ((-1) \cdot 1 - 5 \cdot 1) - 1 \cdot (1 \cdot 1 - 5 \cdot 2) + 6 \cdot (1 \cdot 1 - (-1) \cdot 2) \]

\[ = 1 \cdot (-1 - 5) - 1 \cdot (1 - 10) + 6 \cdot (1 + 2) \]

\[ = 1 \cdot (-6) - 1 \cdot (-9) + 6 \cdot 3 \]

\[ = -6 + 9 + 18 = 21 \]

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

Using Cramer’s rule:

\[ x = \frac{D_x}{D} = \frac{12}{7}, \quad y = \frac{D_y}{D} = \frac{14}{7} = 2, \quad z = \frac{D_z}{D} = \frac{21}{7} = 3 \]

Thus, the values of \( x, y, z \) are:

\[ x = \frac{12}{7}, \quad y = 2, \quad z = 3 \]

### Final Answer:

The values of \( x, y, z \) are \( \frac{12}{7}, 2, 3 \).

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