The number of distinct real values of ` lamda ` for which the system of linear equations ` x + y + z = lamda x , x + y + z = lamday, x + y + z + lamda z ` has non - trival solution.

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to find the number of distinct real values of \( \lambda \) for which the given system of linear equations has a non-trivial solution. The equations are:

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

### Step 1: Rewrite the equations
We can rewrite the equations in a standard form:

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

### Step 2: Form the coefficient matrix
The system can be represented in matrix form \( A \mathbf{v} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \):

\[ A = \begin{pmatrix} 1 - \lambda & 1 & 1 \\ 1 & 1 - \lambda & 1 \\ 1 & 1 & 1 + \lambda \end{pmatrix} \]

### Step 3: Find the determinant of the matrix
For the system to have a non-trivial solution, the determinant of the matrix \( A \) must be zero:

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

Calculating the determinant:

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

Calculating the 2x2 determinants:

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

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

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

Putting it all together:

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

This simplifies to:

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

### Step 4: Set the determinant to zero
Setting the determinant to zero gives us:

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

This leads to two cases:

1. \( 1 - \lambda = 0 \) → \( \lambda = 1 \)
2. \( 1 - \lambda^2 = 0 \) → \( \lambda^2 = 1 \) → \( \lambda = 1 \) or \( \lambda = -1 \)

### Step 5: Identify distinct values of \( \lambda \)
The distinct real values of \( \lambda \) from the above equations are:

- \( \lambda = 1 \)
- \( \lambda = -1 \)

### Conclusion
Thus, the distinct real values of \( \lambda \) for which the system has a non-trivial solution are \( \lambda = 1 \) and \( \lambda = -1 \).

Therefore, the number of distinct real values of \( \lambda \) is **2**.

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