The number of real values of for which the system of linear equations 2x+4y-`lambdaz` =0 , `4x+lambday +2z=0` , `lambdax + 2y+2z=0` has infinitely many solutions , is :

Options

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

Answer

To determine the number of real values of \( \lambda \) for which the given system of linear equations has infinitely many solutions, we need to analyze the conditions under which a 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 system of equations is:
1. \( 2x + 4y - \lambda z = 0 \)
2. \( 4x + \lambda y + 2z = 0 \)
3. \( \lambda x + 2y + 2z = 0 \)

We can represent this system in matrix form as follows:

\[ \begin{bmatrix} 2 & 4 & -\lambda \\ 4 & \lambda & 2 \\ \lambda & 2 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix

To find the values of \( \lambda \) for which the system has infinitely many solutions, we need to set the determinant of the coefficient matrix to zero.

\[ \text{Det} = \begin{vmatrix} 2 & 4 & -\lambda \\ 4 & \lambda & 2 \\ \lambda & 2 & 2 \end{vmatrix} \]

Calculating this determinant using the rule of Sarrus or cofactor expansion:

\[ \text{Det} = 2 \begin{vmatrix} \lambda & 2 \\ 2 & 2 \end{vmatrix} - 4 \begin{vmatrix} 4 & 2 \\ \lambda & 2 \end{vmatrix} - \lambda \begin{vmatrix} 4 & \lambda \\ \lambda & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:

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

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

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

Substituting these back into the determinant:

\[ \text{Det} = 2(2\lambda - 4) - 4(8 - 2\lambda) - \lambda(8 - \lambda^2) \]

Expanding this gives:

\[ \text{Det} = 4\lambda - 8 - 32 + 8\lambda - 8 + \lambda^3 \]

Combining like terms:

\[ \text{Det} = \lambda^3 + 12\lambda - 48 \]

### Step 3: Set the determinant to zero

To find the values of \( \lambda \) for which the system has infinitely many solutions, we set the determinant equal to zero:

\[ \lambda^3 + 12\lambda - 48 = 0 \]

### Step 4: Find the roots of the cubic equation

To find the roots of the cubic equation, we can use the Rational Root Theorem or synthetic division. Testing for potential rational roots, we can find that \( \lambda = 4 \) is a root.

Using synthetic division to factor the cubic polynomial:

\[ \lambda^3 + 12\lambda - 48 = (\lambda - 4)(\lambda^2 + 4\lambda + 12) \]

Now, we need to determine the roots of \( \lambda^2 + 4\lambda + 12 \). The discriminant of this quadratic is:

\[ D = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 12 = 16 - 48 = -32 \]

Since the discriminant is negative, this quadratic has no real roots.

### Conclusion

The only real solution to the original cubic equation is \( \lambda = 4 \). Therefore, the number of real values of \( \lambda \) for which the system has infinitely many solutions is:

\[ \boxed{1} \]