The system of linear equations:
`x+lambday-z=0`
`lamdax-y-z=0`
`x+y-lambdaz=0`
has non-trivial solutoin for:

Options

A. Exactly one value of `lambda`
B. Exactly two values of `lambda`
C. Exactly three values of `lambda`
D. Infinitely many values of `lambda`

Answer

To determine the values of \(\lambda\) for which the given system of linear equations has a non-trivial solution, we need to analyze the system represented by the following equations:

1. \(x + \lambda y - z = 0\)
2. \(\lambda x - y - z = 0\)
3. \(x + y - \lambda z = 0\)

### Step 1: Write the system in matrix form
We can express the system of equations in matrix form \(A \mathbf{X} = 0\), where \(A\) is the coefficient matrix and \(\mathbf{X} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\).

The coefficient matrix \(A\) is:
\[ A = \begin{pmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{pmatrix} \]

### Step 2: Find the determinant of the matrix
For the system to have a non-trivial solution, the determinant of the matrix \(A\) must be equal to zero. We calculate the determinant \(|A|\):

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

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

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

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} -1 & -1 \\ 1 & -\lambda \end{vmatrix} = (-1)(-\lambda) - (-1)(1) = \lambda + 1\)

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

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

Substituting these back into the determinant calculation:
\[ |A| = 1(\lambda + 1) - \lambda(-\lambda^2 + 1) - 1(\lambda + 1) \]
\[ = \lambda + 1 + \lambda^3 - \lambda - \lambda - 1 \]
\[ = \lambda^3 - \lambda \]

### Step 4: Set the determinant to zero
To find the values of \(\lambda\) that give a non-trivial solution, we set the determinant equal to zero:
\[ \lambda^3 - \lambda = 0 \]

### Step 5: Factor the equation
Factoring out \(\lambda\):
\[ \lambda(\lambda^2 - 1) = 0 \]
\[ \lambda(\lambda - 1)(\lambda + 1) = 0 \]

### Step 6: Solve for \(\lambda\)
Setting each factor to zero gives us:
1. \(\lambda = 0\)
2. \(\lambda - 1 = 0 \Rightarrow \lambda = 1\)
3. \(\lambda + 1 = 0 \Rightarrow \lambda = -1\)

### Conclusion
The values of \(\lambda\) for which the system has a non-trivial solution are:
\[ \lambda = 0, \lambda = 1, \lambda = -1 \]