The system of homogeneous equation `lambdax+(lambda+1)y +(lambda-1)z=0, (lambda+1)x+lambday+(lambda+2)z=0, (lambda-1)x+(lambda+2)y + lambdaz=0` has non-trivial solution for :

Options

A. 1.exactly three real value of `lambda`
B. 2.exactly two real values of `lambda`
C. 3.exactly three real value of `lambda`
D. 4.infinitely many real value of `lambda`

Answer

To solve the problem of finding the values of \(\lambda\) for which the given system of homogeneous equations has non-trivial solutions, we will follow these steps:

### Step 1: Write the system of equations in matrix form
The given system of equations is:
1. \(\lambda x + (\lambda + 1)y + (\lambda - 1)z = 0\)
2. \((\lambda + 1)x + \lambda y + (\lambda + 2)z = 0\)
3. \((\lambda - 1)x + (\lambda + 2)y + \lambda z = 0\)

We can represent this system in matrix form as:
\[ \begin{bmatrix} \lambda & \lambda + 1 & \lambda - 1 \\ \lambda + 1 & \lambda & \lambda + 2 \\ \lambda - 1 & \lambda + 2 & \lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Set up the determinant for the coefficient matrix
For the system to have non-trivial solutions, the determinant of the coefficient matrix must be zero:
\[ \Delta = \begin{vmatrix} \lambda & \lambda + 1 & \lambda - 1 \\ \lambda + 1 & \lambda & \lambda + 2 \\ \lambda - 1 & \lambda + 2 & \lambda \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
We will calculate the determinant \(\Delta\) using cofactor expansion along the first row:
\[ \Delta = \lambda \begin{vmatrix} \lambda & \lambda + 2 \\ \lambda + 2 & \lambda \end{vmatrix} - (\lambda + 1) \begin{vmatrix} \lambda + 1 & \lambda + 2 \\ \lambda - 1 & \lambda \end{vmatrix} + (\lambda - 1) \begin{vmatrix} \lambda + 1 & \lambda \\ \lambda - 1 & \lambda + 2 \end{vmatrix} \]

Calculating the minors:
1. \(\begin{vmatrix} \lambda & \lambda + 2 \\ \lambda + 2 & \lambda \end{vmatrix} = \lambda^2 - (\lambda + 2)(\lambda + 2) = \lambda^2 - (\lambda^2 + 4\lambda + 4) = -4\lambda - 4\)

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

3. \(\begin{vmatrix} \lambda + 1 & \lambda \\ \lambda - 1 & \lambda + 2 \end{vmatrix} = (\lambda + 1)(\lambda + 2) - \lambda(\lambda - 1) = \lambda^2 + 3\lambda + 2 - (\lambda^2 - \lambda) = 4\lambda + 2\)

Putting it all together:
\[ \Delta = \lambda(-4\lambda - 4) - (\lambda + 1)(2\lambda - 2) + (\lambda - 1)(4\lambda + 2) \]

### Step 4: Simplify the determinant expression
Expanding and simplifying:
\[ \Delta = -4\lambda^2 - 4\lambda - (2\lambda^2 - 2 + 2\lambda - 2) + (4\lambda^2 + 2\lambda - 4\lambda - 2) \]
Combine like terms:
\[ \Delta = -4\lambda^2 - 4\lambda - 2\lambda^2 + 2 + 4\lambda^2 - 2 = 0 \]
This simplifies to:
\[ -2\lambda^2 - 4\lambda = 0 \]

### Step 5: Factor the determinant
Factoring out \(-2\lambda\):
\[ -2\lambda(\lambda + 2) = 0 \]

### Step 6: Solve for \(\lambda\)
Setting the factors equal to zero gives:
1. \(-2\lambda = 0 \Rightarrow \lambda = 0\)
2. \(\lambda + 2 = 0 \Rightarrow \lambda = -2\)

### Conclusion
The values of \(\lambda\) for which the system has non-trivial solutions are \(\lambda = 0\) and \(\lambda = -2\). Thus, there are exactly **two real values** of \(\lambda\).