If `(x_(0), y_(0), z_(0))`is any solution of the system of equations `2x-y-z=1, -x-y+2z=1 and x-2y+z=2`, then the value of `(x_(0)^(2)-y_(0)^(2)+1)/(z_(0))` (where, `z_(0) ne 0`) is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we will first analyze the given system of equations and then derive the required expression.

### Given System of Equations:
1. \( 2x - y - z = 1 \) (Equation 1)
2. \( -x - y + 2z = 1 \) (Equation 2)
3. \( x - 2y + z = 2 \) (Equation 3)

### Step 1: Form the Coefficient Matrix and Calculate Determinants
We will use determinants to check the nature of the solutions.

#### Determinant \( \Delta_1 \):
\[ \Delta_1 = \begin{vmatrix} 2 & -1 & -1 \\ -1 & -1 & 2 \\ 1 & -2 & 1 \end{vmatrix} \]

Calculating \( \Delta_1 \):
- Expand along the first row:
\[ \Delta_1 = 2 \begin{vmatrix} -1 & 2 \\ -2 & 1 \end{vmatrix} + 1 \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} -1 & -1 \\ 1 & -2 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 2 \\ -2 & 1 \end{vmatrix} = (-1)(1) - (2)(-2) = -1 + 4 = 3 \)
2. \( \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} = (-1)(1) - (2)(1) = -1 - 2 = -3 \)
3. \( \begin{vmatrix} -1 & -1 \\ 1 & -2 \end{vmatrix} = (-1)(-2) - (-1)(1) = 2 + 1 = 3 \)

Putting it all together:
\[ \Delta_1 = 2(3) + 1(-3) - 1(3) = 6 - 3 - 3 = 0 \]

### Step 2: Check for Infinite Solutions
Since \( \Delta_1 = 0 \), we will check \( \Delta_2 \) and \( \Delta_3 \) to confirm if we have infinite solutions.

#### Determinant \( \Delta_2 \):
\[ \Delta_2 = \begin{vmatrix} 1 & -1 & -1 \\ 1 & -1 & 2 \\ 2 & -2 & 1 \end{vmatrix} \]

Calculating \( \Delta_2 \):
- Expand along the first row:
\[ \Delta_2 = 1 \begin{vmatrix} -1 & 2 \\ -2 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 2 & -2 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 2 \\ -2 & 1 \end{vmatrix} = 3 \) (calculated earlier)
2. \( \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = 1 - 4 = -3 \)
3. \( \begin{vmatrix} 1 & -1 \\ 2 & -2 \end{vmatrix} = 0 \)

Putting it all together:
\[ \Delta_2 = 1(3) + 1(-3) - 1(0) = 3 - 3 = 0 \]

#### Determinant \( \Delta_3 \):
\[ \Delta_3 = \begin{vmatrix} 2 & -1 & 1 \\ -1 & -1 & 1 \\ 1 & -2 & 2 \end{vmatrix} \]

Calculating \( \Delta_3 \):
- Expand along the first row:
\[ \Delta_3 = 2 \begin{vmatrix} -1 & 1 \\ -2 & 2 \end{vmatrix} + 1 \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} + 1 \begin{vmatrix} -1 & -1 \\ 1 & -2 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 1 \\ -2 & 2 \end{vmatrix} = -2 + 2 = 0 \)
2. \( \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} = -2 - 1 = -3 \)
3. \( \begin{vmatrix} -1 & -1 \\ 1 & -2 \end{vmatrix} = 2 - 1 = 1 \)

Putting it all together:
\[ \Delta_3 = 2(0) + 1(-3) + 1(1) = -3 + 1 = -2 \]

### Conclusion on Solutions
Since \( \Delta_1 = 0 \), \( \Delta_2 = 0 \), and \( \Delta_3 \neq 0 \), we have infinite solutions.

### Step 3: Express Variables in Terms of a Parameter
Let \( z = \lambda \). Substitute \( z \) into the equations to find \( x \) and \( y \).

From Equation 1:
\[ 2x - y - \lambda = 1 \implies 2x - y = 1 + \lambda \quad (1) \]

From Equation 2:
\[ -x - y + 2\lambda = 1 \implies -x - y = 1 - 2\lambda \quad (2) \]

From Equation 3:
\[ x - 2y + \lambda = 2 \implies x - 2y = 2 - \lambda \quad (3) \]

### Step 4: Solve for \( x \) and \( y \)
From (1) and (2):
Adding (1) and (2):
\[ 2x - y - x - y = 1 + \lambda + 1 - 2\lambda \]
\[ x - 2y = 2 - \lambda \quad (which is same as (3)) \]

From (1):
\[ y = 2x - (1 + \lambda) \]

Substituting \( y \) in terms of \( x \) into (3):
\[ x - 2(2x - (1 + \lambda)) = 2 - \lambda \]
\[ x - 4x + 2 + 2\lambda = 2 - \lambda \]
\[ -3x + 3\lambda = 0 \implies x = \lambda \]

Substituting \( x = \lambda \) back into (1):
\[ 2\lambda - y = 1 + \lambda \implies y = \lambda - 1 \]

### Step 5: Calculate the Required Expression
Now we have:
- \( x_0 = \lambda \)
- \( y_0 = \lambda - 1 \)
- \( z_0 = \lambda \)

We need to find:
\[ \frac{x_0^2 - y_0^2 + 1}{z_0} \]
Substituting the values:
\[ = \frac{\lambda^2 - (\lambda - 1)^2 + 1}{\lambda} \]
\[ = \frac{\lambda^2 - (\lambda^2 - 2\lambda + 1) + 1}{\lambda} \]
\[ = \frac{\lambda^2 - \lambda^2 + 2\lambda - 1 + 1}{\lambda} \]
\[ = \frac{2\lambda}{\lambda} = 2 \]

### Final Answer:
The value of \( \frac{x_0^2 - y_0^2 + 1}{z_0} \) is \( 2 \).