Let `lambda` be a real number for which the system of linear equations
x + y +z =6, 4x + `lambday -lambdaz = lambda -2` and 3x + 2y-4z =-5
has infinitely many solutions. Then `lambda` is a root of the quadratic equation

Options

A. `lambda^2-3lambda-4=0`
B. `lambda^2+3lambda-4=0`
C. `lambda^2-lambda-6=0`
D. `lambda^2+lambda-6=0`

Answer

To solve the problem, we need to find the value of \( \lambda \) for which the given system of linear equations has infinitely many solutions. This occurs when the determinant of the coefficient matrix is zero.

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \( x + y + z = 6 \)
2. \( 4x + \lambda y - \lambda z = \lambda - 2 \)
3. \( 3x + 2y - 4z = -5 \)

The coefficient matrix \( A \) and the constant matrix \( B \) can be represented as follows:

\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{pmatrix}, \quad B = \begin{pmatrix} 6 \\ \lambda - 2 \\ -5 \end{pmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
To find the determinant of matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} \lambda & -\lambda \\ 2 & -4 \end{vmatrix} - 1 \cdot \begin{vmatrix} 4 & -\lambda \\ 3 & -4 \end{vmatrix} + 1 \cdot \begin{vmatrix} 4 & \lambda \\ 3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

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

2. \( \begin{vmatrix} 4 & -\lambda \\ 3 & -4 \end{vmatrix} = 4(-4) - (-\lambda)(3) = -16 + 3\lambda = 3\lambda - 16 \)

3. \( \begin{vmatrix} 4 & \lambda \\ 3 & 2 \end{vmatrix} = 4(2) - \lambda(3) = 8 - 3\lambda \)

Putting it all together:

\[ \text{det}(A) = 1(-2\lambda) - 1(3\lambda - 16) + 1(8 - 3\lambda) \]
\[ = -2\lambda - 3\lambda + 16 + 8 - 3\lambda \]
\[ = -8\lambda + 24 \]

### Step 3: Set the determinant to zero
For the system to have infinitely many solutions, we set the determinant equal to zero:

\[ -8\lambda + 24 = 0 \]

### Step 4: Solve for \( \lambda \)
Solving the equation:

\[ -8\lambda = -24 \]
\[ \lambda = \frac{24}{8} = 3 \]

### Conclusion
Thus, the value of \( \lambda \) for which the system has infinitely many solutions is \( \lambda = 3 \).