The set of the all values of `lamda` for which the system of linear equations
`2x_(1) - 2x_(2) + x_(3) = lamdax_(1)`
`2x_(1) - 3x_(2) + 2x_(3) = lamda x_(2)`
`-x_(1) + 2x_(2) = lamda x_(3)` has a non-trivial solution,

Options

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

Explanation

To find the set of all values of \( \lambda \) for which the given system of linear equations has a non-trivial solution, we need to analyze the determinant of the coefficient matrix. The system of equations is:

1. \( 2x_1 - 2x_2 + x_3 = \lambda x_1 \)
2. \( 2x_1 - 3x_2 + 2x_3 = \lambda x_2 \)
3. \( -x_1 + 2x_2 = \lambda x_3 \)

We can rewrite these equations in standard form:

1. \( (2 - \lambda)x_1 - 2x_2 + x_3 = 0 \)
2. \( 2x_1 + (-3 - \lambda)x_2 + 2x_3 = 0 \)
3. \( -x_1 + 2x_2 - \lambda x_3 = 0 \)

Now, we can express the system in matrix form \( A \mathbf{x} = 0 \), where \( A \) is the coefficient matrix:

\[ A = \begin{bmatrix} 2 - \lambda & -2 & 1 \\ 2 & -3 - \lambda & 2 \\ -1 & 2 & -\lambda \end{bmatrix} \]

For the system to have a non-trivial solution, the determinant of the matrix \( A \) must be zero:

\[ \text{det}(A) = 0 \]

### Step 1: Calculate the Determinant

We will calculate the determinant of matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} 2 - \lambda & -2 & 1 \\ 2 & -3 - \lambda & 2 \\ -1 & 2 & -\lambda \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows respectively.

### Step 2: Expand the Determinant

Expanding the determinant:

\[ = (2 - \lambda) \begin{vmatrix} -3 - \lambda & 2 \\ 2 & -\lambda \end{vmatrix} + 2 \begin{vmatrix} 2 & 2 \\ -1 & -\lambda \end{vmatrix} + 1 \begin{vmatrix} 2 & -3 - \lambda \\ -1 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

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

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

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

### Step 3: Substitute Back

Now substituting back into the determinant expression:

\[ \text{det}(A) = (2 - \lambda)(\lambda^2 + 3\lambda - 4) + 2(-2\lambda + 2) + (1 - \lambda) \]

### Step 4: Set the Determinant to Zero

Setting the determinant equal to zero:

\[ (2 - \lambda)(\lambda^2 + 3\lambda - 4) - 4\lambda + 4 + 1 - \lambda = 0 \]

### Step 5: Solve the Polynomial

This results in a polynomial equation in \( \lambda \). We can simplify and factor this polynomial to find the values of \( \lambda \).

After simplification, we find:

\[ (1 - \lambda)(\lambda - 2)(\lambda + 4) = 0 \]

### Step 6: Find the Values of \( \lambda \)

Thus, the values of \( \lambda \) are:

\[ \lambda = 1, \quad \lambda = 2, \quad \lambda = -4 \]

### Conclusion

The set of all values of \( \lambda \) for which the system of linear equations has a non-trivial solution is:

\[ \{ 1, 2, -4 \} \]