Given , 2x-y+2z=2, x-2y+z=-4, x+y+`lambda`z=4, then the value of `lambda` such that the given system of equations has no solution is :

Options

A. 3
B. 1
C. 0
D. `-3`

Answer

To find the value of \( \lambda \) such that the given system of equations has no solution, we need to analyze the system of equations:

1. \( 2x - y + 2z = 2 \) (Equation 1)
2. \( x - 2y + z = -4 \) (Equation 2)
3. \( x + y + \lambda z = 4 \) (Equation 3)

### Step 1: Write the system in matrix form
We can represent the system of equations in the form of a matrix equation \( AX = B \), where:

\[ A = \begin{bmatrix} 2 & -1 & 2 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ -4 \\ 4 \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
For the system to have no solution, the determinant of the coefficient matrix \( A \) must be zero. We calculate the determinant \( |A| \):

\[ |A| = \begin{vmatrix} 2 & -1 & 2 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{vmatrix} \]

### Step 3: Calculate the determinant using cofactor expansion
We can expand the determinant along the first row:

\[ |A| = 2 \begin{vmatrix} -2 & 1 \\ 1 & \lambda \end{vmatrix} - (-1) \begin{vmatrix} 1 & 1 \\ 1 & \lambda \end{vmatrix} + 2 \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

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

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

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

### Step 4: Substitute back into the determinant equation
Now substituting these back into the determinant equation:

\[ |A| = 2(-2\lambda - 1) + (\lambda - 1) + 2(3) \]

Expanding this gives:

\[ |A| = -4\lambda - 2 + \lambda - 1 + 6 \]

Combining like terms:

\[ |A| = -4\lambda + \lambda + 3 = -3\lambda + 3 \]

### Step 5: Set the determinant to zero
For the system to have no solution, we set the determinant to zero:

\[ -3\lambda + 3 = 0 \]

### Step 6: Solve for \( \lambda \)
Solving for \( \lambda \):

\[ -3\lambda = -3 \implies \lambda = 1 \]

### Final Answer
Thus, the value of \( \lambda \) such that the given system of equations has no solution is:

\[ \lambda = 1 \]