Let S be the set of all ` lamda in R ` for which the system of linear equations
` 2x - y + 2z =2`
` x -2y + lamda x=-4`
` x + lamda y + z=4`
has no solution . Then the set S

Options

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

Answer

To determine the set \( S \) of all \( \lambda \in \mathbb{R} \) for which the given system of linear equations has no solution, we need to analyze the system of equations:

1. \( 2x - y + 2z = 2 \) (Equation 1)
2. \( x - 2y + \lambda x = -4 \) (Equation 2)
3. \( x + \lambda y + z = 4 \) (Equation 3)

### Step 1: Rewrite the equations in standard form
First, we can rewrite Equation 2 as:
\[ (1 + \lambda)x - 2y = -4 \]

Now, our system of equations looks like:
1. \( 2x - y + 2z = 2 \)
2. \( (1 + \lambda)x - 2y = -4 \)
3. \( x + \lambda y + z = 4 \)

### Step 2: Write the augmented matrix
The augmented matrix for this system is:
\[ \begin{bmatrix} 2 & -1 & 2 & | & 2 \\ 1 + \lambda & -2 & 0 & | & -4 \\ 1 & \lambda & 1 & | & 4 \end{bmatrix} \]

### Step 3: Find the determinant of the coefficient matrix
The coefficient matrix is:
\[ \begin{bmatrix} 2 & -1 & 2 \\ 1 + \lambda & -2 & 0 \\ 1 & \lambda & 1 \end{bmatrix} \]

To find the determinant, we can use the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \).

Calculating the determinant:
\[ \text{det}(A) = 2((-2)(1) - (0)(\lambda)) - (-1)((1 + \lambda)(1) - (0)(1)) + 2((1 + \lambda)(\lambda) - (-2)(1)) \]
\[ = 2(-2) - (-1)(1 + \lambda) + 2((1 + \lambda)\lambda + 2) \]
\[ = -4 + 1 + \lambda + 2(1 + \lambda)\lambda + 4 \]
\[ = \lambda + 1 + 2\lambda^2 + 2\lambda + 4 \]
\[ = 2\lambda^2 + 3\lambda + 1 \]

### Step 4: Set the determinant equal to zero
For the system to have no solution, the determinant must be zero:
\[ 2\lambda^2 + 3\lambda + 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = 3, c = 1 \):
\[ D = b^2 - 4ac = 3^2 - 4(2)(1) = 9 - 8 = 1 \]
\[ \lambda = \frac{-3 \pm \sqrt{1}}{2(2)} = \frac{-3 \pm 1}{4} \]
Calculating the two values:
1. \( \lambda_1 = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2} \)
2. \( \lambda_2 = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \)

### Conclusion
The set \( S \) of all \( \lambda \) for which the system has no solution is:
\[ S = \{-1, -\frac{1}{2}\} \]