The set of all values of `lambda` for which the system of linear equations
`x - 2y - 2z = lambdax`
`x + 2y + z = lambday`
`-x -y = lambdaz`
has a non-trivial solution

Options

A. contains more than two elements
B. is a singleton
C. is an empty set
D. contains exactly two elements

Answer

To solve the given system of linear equations for the values of \( \lambda \) that allow for a non-trivial solution, we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the equations**:
The given equations are:
\[ x - 2y - 2z = \lambda x \quad (1) \]
\[ x + 2y + z = \lambda y \quad (2) \]
\[ -x - y = \lambda z \quad (3) \]

Rearranging these equations, we can express them in standard form:
\[ (1 - \lambda)x - 2y - 2z = 0 \quad (4) \]
\[ x + (2 - \lambda)y + z = 0 \quad (5) \]
\[ -x - y - \lambda z = 0 \quad (6) \]

2. **Set up the coefficient matrix**:
The system can be represented in matrix form \( A\mathbf{v} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \):
\[ A = \begin{bmatrix} 1 - \lambda & -2 & -2 \\ 1 & 2 - \lambda & 1 \\ -1 & -1 & -\lambda \end{bmatrix} \]

3. **Find the determinant of the matrix**:
For the system to have a non-trivial solution, the determinant of the coefficient matrix \( A \) must be zero:
\[ \text{det}(A) = 0 \]

We calculate the determinant:
\[ \text{det}(A) = (1 - \lambda) \begin{vmatrix} 2 - \lambda & 1 \\ -1 & -\lambda \end{vmatrix} - (-2) \begin{vmatrix} 1 & 1 \\ -1 & -\lambda \end{vmatrix} - 2 \begin{vmatrix} 1 & 2 - \lambda \\ -1 & -1 \end{vmatrix} \]

4. **Calculate the 2x2 determinants**:
- For the first determinant:
\[ \begin{vmatrix} 2 - \lambda & 1 \\ -1 & -\lambda \end{vmatrix} = (2 - \lambda)(-\lambda) - (1)(-1) = -\lambda(2 - \lambda) + 1 = \lambda^2 - 2\lambda + 1 \]
- For the second determinant:
\[ \begin{vmatrix} 1 & 1 \\ -1 & -\lambda \end{vmatrix} = (1)(-\lambda) - (1)(-1) = -\lambda + 1 \]
- For the third determinant:
\[ \begin{vmatrix} 1 & 2 - \lambda \\ -1 & -1 \end{vmatrix} = (1)(-1) - (2 - \lambda)(-1) = -1 + (2 - \lambda) = \lambda + 1 \]

5. **Substituting back into the determinant**:
Now substituting back into the determinant expression:
\[ \text{det}(A) = (1 - \lambda)(\lambda^2 - 2\lambda + 1) + 2(-\lambda + 1) - 2(\lambda + 1) \]

Simplifying this:
\[ = (1 - \lambda)(\lambda - 1)^2 + 2 - 2\lambda - 2\lambda - 2 \]
\[ = (1 - \lambda)(\lambda - 1)^2 - 4\lambda \]

6. **Setting the determinant to zero**:
We set the entire expression equal to zero:
\[ (1 - \lambda)(\lambda - 1)^2 = 0 \]

This gives us two cases:
- \( 1 - \lambda = 0 \) which implies \( \lambda = 1 \)
- \( (\lambda - 1)^2 = 0 \) which also implies \( \lambda = 1 \)

7. **Conclusion**:
The only value of \( \lambda \) for which the system has a non-trivial solution is:
\[ \lambda = 1 \]