A system of equations `lambdax +y +z =1,x+lambday+z=lambda, x + y + lambdaz = lambda^2` have no solution then value of `lambda` is

Answer

To solve the problem, we need to find the values of \( \lambda \) for which the system of equations has no solution. A system of equations has no solution when the determinant of the coefficient matrix is equal to zero.

Given the system of equations:

1. \( \lambda x + y + z = 1 \)
2. \( x + \lambda y + z = \lambda \)
3. \( x + y + \lambda z = \lambda^2 \)

We can represent this system in matrix form \( A \mathbf{x} = \mathbf{b} \), where:

\[ A = \begin{bmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 1 \\ \lambda \\ \lambda^2 \end{bmatrix} \]

To find the values of \( \lambda \) for which the system has no solution, we need to compute the determinant of matrix \( A \) and set it equal to zero.

### Step 1: Calculate the Determinant of Matrix \( A \)

The determinant of matrix \( A \) is given by:

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

Using the formula for the determinant of a \( 3 \times 3 \) matrix, we expand it as follows:

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

Calculating the \( 2 \times 2 \) determinants:

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

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

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

Substituting these back into the determinant expression:

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

### Step 2: Simplify the Determinant

Simplifying the expression:

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

### Step 3: Set the Determinant Equal to Zero

For the system to have no solution, we set the determinant equal to zero:

\[ \lambda^3 - 3\lambda + 2 = 0 \]

### Step 4: Factor the Polynomial

To factor \( \lambda^3 - 3\lambda + 2 \), we can use the Rational Root Theorem or synthetic division. Testing \( \lambda = 1 \):

\[ 1^3 - 3(1) + 2 = 0 \]

Thus, \( \lambda - 1 \) is a factor. We can perform polynomial long division or synthetic division to factor the cubic polynomial:

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

Now, we can factor \( \lambda^2 + \lambda - 2 \):

\[ \lambda^2 + \lambda - 2 = (\lambda - 1)(\lambda + 2) \]

Thus, we have:

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

### Step 5: Solve for \( \lambda \)

Setting the factors equal to zero:

1. \( \lambda - 1 = 0 \) gives \( \lambda = 1 \) (with multiplicity 2)
2. \( \lambda + 2 = 0 \) gives \( \lambda = -2 \)

### Final Answer

The values of \( \lambda \) for which the system of equations has no solution are:

\[ \lambda = 1 \quad \text{and} \quad \lambda = -2 \]