The system of equation `-2x + y + z = 1, x – 2y + z = –2, x + y + lambdaz = 4` will have no solution if

Options

A. `lambda = –2`
B. `lambda = –1`
C. `lambda=3`
D. None of these

Answer

To determine the value of \(\lambda\) for which the given system of equations has no solution, we will analyze the system of equations using the determinant of the coefficients.

The given system of equations is:
1. \(-2x + y + z = 1\)
2. \(x - 2y + z = -2\)
3. \(x + y + \lambda z = 4\)

### Step 1: Write the coefficient matrix
The coefficient matrix \(A\) for the system can be written as:
\[ A = \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{bmatrix} \]

### Step 2: Calculate the determinant of the matrix
To find the condition for no solution, we need to calculate the determinant of matrix \(A\) and set it equal to zero. The determinant of a \(3 \times 3\) matrix is calculated as follows:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ \text{det}(A) = -2 \begin{vmatrix} -2 & 1 \\ 1 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \]

### Step 3: Calculate the minors
Calculating the minors:
1. \(\begin{vmatrix} -2 & 1 \\ 1 & \lambda \end{vmatrix} = (-2)(\lambda) - (1)(1) = -2\lambda - 1\)

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

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

### Step 4: Substitute the minors back into the determinant formula
Now substituting these values back into the determinant:
\[ \text{det}(A) = -2(-2\lambda - 1) - 1(\lambda - 1) + 1(3) \]
\[ = 4\lambda + 2 - \lambda + 1 + 3 \]
\[ = 4\lambda - \lambda + 2 + 1 + 3 \]
\[ = 3\lambda + 6 \]

### Step 5: Set the determinant equal to zero
For the system to have no solution, we set the determinant to zero:
\[ 3\lambda + 6 = 0 \]

### Step 6: Solve for \(\lambda\)
Solving for \(\lambda\):
\[ 3\lambda = -6 \]
\[ \lambda = -2 \]

### Conclusion
The value of \(\lambda\) for which the system of equations has no solution is \(\lambda = -2\).