The value of `lambda` such that the system x − 2y + z = − 4, 2x − y + 2z = 2 and x + y + `lambda`z = 4
has no solution, is

Options

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

Answer

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

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

### Step 1: Write the system in matrix form
We can express the system of equations in the form of a matrix \( A \) and a vector \( B \):

\[ \begin{bmatrix} 1 & -2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & \lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -4 \\ 2 \\ 4 \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
To find the value of \( \lambda \) for which the system has no solution, we need to set the determinant of the coefficient matrix to zero:

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

### Step 3: Expand the determinant
Using the determinant formula for a 3x3 matrix:

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

Calculating each of the 2x2 determinants:

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

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

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

Substituting back into the determinant:

\[ \text{det}(A) = 1(-\lambda - 2) + 2(2\lambda - 2) + 1(3) \]

Expanding this gives:

\[ -\lambda - 2 + 4\lambda - 4 + 3 = 3\lambda - 3 \]

### Step 4: Set the determinant to zero
For the system to have no solution, we set the determinant to zero:

\[ 3\lambda - 3 = 0 \]

### Step 5: Solve for \( \lambda \)
Solving the equation gives:

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

### Conclusion
The value of \( \lambda \) such that the system has no solution is:

\[ \lambda = 1 \]

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