The system of equations `lamdax+y +z=1, x+lamday+z=lamda and x+y+lamdaz=lamda^2` has no solution if `lamda` equals

Options

A. 0
B. 1
C. `-1`
D. `-2`

Answer

To determine the values of \( \lambda \) for which the system of equations has no solution, we can analyze the given equations:

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

We can express this system in matrix form as:

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

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

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

### Step 1: Calculate the Determinant

We can calculate the determinant using cofactor expansion along the first row:

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

Calculating the 2x2 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:

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

### Step 2: Simplify the Determinant

Now, simplify the expression:

\[ D = \lambda^3 - \lambda - \lambda + 1 + 1 - \lambda \]
\[ D = \lambda^3 - 3\lambda + 2 \]

### Step 3: Set the Determinant 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 Cubic Equation

We can try to find rational roots using the Rational Root Theorem. Testing possible rational roots (±1, ±2):

1. For \( \lambda = 1 \):
\[ 1^3 - 3(1) + 2 = 0 \quad \text{(This is a root)} \]

2. For \( \lambda = -1 \):
\[ (-1)^3 - 3(-1) + 2 = 0 \quad \text{(Not a root)} \]

3. For \( \lambda = -2 \):
\[ (-2)^3 - 3(-2) + 2 = 0 \quad \text{(This is a root)} \]

### Step 5: Factor the Polynomial

Since \( \lambda = 1 \) and \( \lambda = -2 \) are roots, we can factor the polynomial as:

\[ (\lambda - 1)(\lambda + 2)(\lambda - r) = 0 \]

Where \( r \) is the third root. We can find \( r \) using synthetic division or polynomial long division, but we are primarily interested in the roots that yield no solutions.

### Conclusion

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

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

Since the question asks for the specific value of \( \lambda \) that results in no solution, we conclude:

\[ \text{The value of } \lambda \text{ for which the system has no solution is } \lambda = -2. \]