`x+y+z=6`
`x-y+z=2`
`2x+y-z=1`
then x,y,z are respectively

Options

A. 3,2,1
B. 1,2,3
C. 2,1,3
D. None of these

Answer

To solve the system of equations given by:

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

we can use Cramer's Rule, which involves calculating determinants.

### Step 1: Write the equations in matrix form

The equations can be represented in the form \( AX = B \), where:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 2 \\ 1 \end{bmatrix} \]

### Step 2: Calculate the determinant \( D \) of matrix \( A \)

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

Calculating the determinant:

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

Calculating the 2x2 determinants:

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

2. \( \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (1)(2) = -1 - 2 = -3 \)

3. \( \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = (1)(1) - (-1)(2) = 1 + 2 = 3 \)

Substituting back into the determinant \( D \):

\[ D = 1(0) - 1(-3) + 1(3) = 0 + 3 + 3 = 6 \]

### Step 3: Calculate the determinants \( D_x, D_y, D_z \)

**For \( D_x \)** (replace the first column of \( A \) with \( B \)):

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

Calculating \( D_x \):

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

Calculating the 2x2 determinants:

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

2. \( \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (1)(1) = -2 - 1 = -3 \)

3. \( \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (-1)(1) = 2 + 1 = 3 \)

Substituting back into \( D_x \):

\[ D_x = 6(0) - 1(-3) + 1(3) = 0 + 3 + 3 = 6 \]

**For \( D_y \)** (replace the second column of \( A \) with \( B \)):

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

Calculating \( D_y \):

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

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} = -3 \)

2. \( \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -3 \)

3. \( \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = -3 \)

Substituting back into \( D_y \):

\[ D_y = 1(-3) - 6(-3) + 1(-3) = -3 + 18 - 3 = 12 \]

**For \( D_z \)** (replace the third column of \( A \) with \( B \)):

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

Calculating \( D_z \):

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

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} = -1 - 2 = -3 \)

2. \( \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = -3 \)

3. \( \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1 + 2 = 3 \)

Substituting back into \( D_z \):

\[ D_z = 1(-3) - 1(-3) + 6(3) = -3 + 3 + 18 = 18 \]

### Step 4: Calculate the values of \( x, y, z \)

Using Cramer’s Rule:

\[ x = \frac{D_x}{D} = \frac{6}{6} = 1 \]
\[ y = \frac{D_y}{D} = \frac{12}{6} = 2 \]
\[ z = \frac{D_z}{D} = \frac{18}{6} = 3 \]

### Final Answer

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

\[ x = 1, \quad y = 2, \quad z = 3 \]