If the following equations `x+y-3z = 0, (1+lamda)x + (2+lamda) y - 8z = 0, x-(1+lamda)y + (2+lamda)z = 0` are consistent then the value of `lamda` is

Answer

To find the value of \( \lambda \) for which the given system of equations is consistent, we will follow these steps:

### Step 1: Write the equations in matrix form
The given equations are:
1. \( x + y - 3z = 0 \)
2. \( (1 + \lambda)x + (2 + \lambda)y - 8z = 0 \)
3. \( x - (1 + \lambda)y + (2 + \lambda)z = 0 \)

We can express this system in matrix form \( AX = 0 \), where:
\[ A = \begin{bmatrix} 1 & 1 & -3 \\ 1 + \lambda & 2 + \lambda & -8 \\ 1 & -(1 + \lambda) & (2 + \lambda) \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Determine the condition for consistency
For the homogeneous system \( AX = 0 \) to be consistent, the determinant of matrix \( A \) must be zero:
\[ \text{det}(A) = 0 \]

### Step 3: Calculate the determinant of \( A \)
The determinant of matrix \( A \) is given by:
\[ \text{det}(A) = \begin{vmatrix} 1 & 1 & -3 \\ 1 + \lambda & 2 + \lambda & -8 \\ 1 & -(1 + \lambda) & (2 + \lambda) \end{vmatrix} \]

### Step 4: Simplify the determinant
We can simplify the determinant using row operations. Let's perform the operation \( R_1 \rightarrow R_1 - R_2 \):
\[ \text{det}(A) = \begin{vmatrix} 0 & -\lambda & 5 \\ 1 + \lambda & 2 + \lambda & -8 \\ 1 & -(1 + \lambda) & (2 + \lambda) \end{vmatrix} \]

Now we can expand the determinant along the first row:
\[ = 0 \cdot M_{11} - (-\lambda) \cdot M_{12} + 5 \cdot M_{13} \]
Where \( M_{ij} \) is the minor of the element in the \( i^{th} \) row and \( j^{th} \) column.

### Step 5: Calculate the minors
Calculating the minors:
1. \( M_{12} = \begin{vmatrix} 1 + \lambda & -8 \\ 1 & (2 + \lambda) \end{vmatrix} = (1 + \lambda)(2 + \lambda) + 8 \)
2. \( M_{13} = \begin{vmatrix} 1 + \lambda & 2 + \lambda \\ 1 & -(1 + \lambda) \end{vmatrix} = -(1 + \lambda)(1) - (2 + \lambda)(1) = -1 - \lambda - 2 - \lambda = -3 - 2\lambda \)

### Step 6: Substitute back into the determinant
Now substituting back:
\[ \text{det}(A) = \lambda \left( (1 + \lambda)(2 + \lambda) + 8 \right) + 5(-3 - 2\lambda) \]

### Step 7: Set the determinant to zero
Setting the determinant to zero gives us:
\[ \lambda \left( (1 + \lambda)(2 + \lambda) + 8 \right) - 15 - 10\lambda = 0 \]

### Step 8: Solve the equation
Expanding and simplifying:
\[ \lambda (2 + 3\lambda + \lambda^2 + 8) - 15 - 10\lambda = 0 \]
\[ \lambda^3 + 3\lambda^2 - 9\lambda - 15 = 0 \]

### Step 9: Factor the polynomial
Factoring the polynomial gives us:
\[ (\lambda - 1)(3\lambda + 5) = 0 \]

### Step 10: Find the values of \( \lambda \)
From the factors, we find:
1. \( \lambda - 1 = 0 \) → \( \lambda = 1 \)
2. \( 3\lambda + 5 = 0 \) → \( \lambda = -\frac{5}{3} \)

### Final Answer
Thus, the values of \( \lambda \) for which the given system of equations is consistent are:
\[ \lambda = 1 \quad \text{or} \quad \lambda = -\frac{5}{3} \]