1. Key Concept: Homogeneous System of Linear Equations and Conditions for Solutions

A system of linear equations is classified as homogeneous if all the constant terms on the right-hand side of the equations are zero. Such a system can be concisely written in matrix form as $AX = 0$, where:

• $A$ is the $n \times n$ coefficient matrix (containing the coefficients of the variables).
• $X$ is the column matrix of variables (e.g., $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$).
• $0$ is the zero column matrix (representing the constant terms).

For a homogeneous system $AX = 0$, there are two fundamental possibilities for its solutions:

• Unique (Trivial) Solution: If the determinant of the coefficient matrix, $\det(A)$, is non-zero ($\det(A) \neq 0$), the system has only one solution: the trivial solution, where all variables are zero ($x=0, y=0, z=0$).
• Infinitely Many (Non-Trivial) Solutions: If the determinant of the coefficient matrix, $\det(A)$, is zero ($\det(A) = 0$), the system has infinitely many solutions. These solutions include the trivial solution, but also non-trivial solutions where at least one variable is non-zero.

Why this is important: The question asks for the number of real values of $\lambda$ for which the given system has infinitely many solutions. Based on the concept above, this condition is met precisely when the determinant of the system's coefficient matrix is equal to zero. Our strategy will be to form the coefficient matrix, calculate its determinant, and then solve the resulting equation $\det(A)=0$ for $\lambda$.

2. Forming the Coefficient Matrix

The given system of linear equations is:

$$\begin{align*} 2x + 4y - \lambda z &= 0 \\ 4x + \lambda y + 2z &= 0 \\ \lambda x + 2y + 2z &= 0 \end{align*}$$

To apply the condition $\det(A) = 0$, we first need to construct the coefficient matrix $A$. This matrix is formed by arranging the coefficients of $x, y,$ and $z$ from each equation into rows.

Why this step is taken: Representing the system in matrix form allows us to utilize the powerful tools of matrix algebra and determinants to analyze the nature of its solutions.

The coefficient matrix $A$ for this system is:

$$A = \begin{pmatrix} 2 & 4 & -\lambda \\ 4 & \lambda & 2 \\ \lambda & 2 & 2 \end{pmatrix}$$

3. Calculating the Determinant of the Coefficient Matrix

Next, we calculate the determinant of the coefficient matrix $A$. We will use the cofactor expansion method along the first row (or Sarrus' rule for a $3 \times 3$ matrix).

Why this step is taken: The condition for infinitely many solutions is $\det(A) = 0$. Therefore, calculating the determinant is a crucial step towards finding the values of $\lambda$.

The determinant of $A$ is:

$$\det(A) = \begin{vmatrix} 2 & 4 & -\lambda \\ 4 & \lambda & 2 \\ \lambda & 2 & 2 \end{vmatrix}$$

Expanding along the first row:

$$\det(A) = 2 \begin{vmatrix} \lambda & 2 \\ 2 & 2 \end{vmatrix} - 4 \begin{vmatrix} 4 & 2 \\ \lambda & 2 \end{vmatrix} + (-\lambda) \begin{vmatrix} 4 & \lambda \\ \lambda & 2 \end{vmatrix}$$

Explanation: For a $3 \times 3$ matrix $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, its determinant is $a(ei-fh) - b(di-fg) + c(dh-eg)$.

Now, we evaluate each $2 \times 2$ determinant:

• $2(\lambda \cdot 2 - 2 \cdot 2) = 2(2\lambda - 4)$
• $-4(4 \cdot 2 - 2 \cdot \lambda) = -4(8 - 2\lambda)$
• $-\lambda(4 \cdot 2 - \lambda \cdot \lambda) = -\lambda(8 - \lambda^2)$

Substitute these simplified expressions back into the determinant equation:

$$\det(A) = 2(2\lambda - 4) - 4(8 - 2\lambda) - \lambda(8 - \lambda^2)$$

Distribute and simplify:

$$\det(A) = 4\lambda - 8 - 32 + 8\lambda - 8\lambda + \lambda^3$$

Combine like terms:

$$\det(A) = \lambda^3 + (4\lambda + 8\lambda - 8\lambda) - (8 + 32)$$ $$\det(A) = \lambda^3 + 4\lambda - 40$$

4. Solving for $\lambda$

For the system to have infinitely many solutions, we must set $\det(A) = 0$. So, we have the cubic equation:

$$\lambda^3 + 4\lambda - 40 = 0$$

Why this step is taken: This equation directly provides the values of $\lambda$ that satisfy the condition for infinitely many solutions, as established in Section 1.

Let's analyze the function $f(\lambda) = \lambda^3 + 4\lambda - 40$ to determine the number of its real roots. First, we find the derivative of $f(\lambda)$:

$$f'(\lambda) = \frac{d}{d\lambda}(\lambda^3 + 4\lambda - 40) = 3\lambda^2 + 4$$

Explanation: The derivative tells us about the slope and monotonicity of the function. Since $\lambda^2 \ge 0$ for all real values of $\lambda$, it follows that $3\lambda^2 \ge 0$. Therefore, $f'(\lambda) = 3\lambda^2 + 4 \ge 4$. This means $f'(\lambda)$ is always positive for all real $\lambda$ (i.e., $f'(\lambda) > 0$).

Why this analysis is important: A function whose derivative is always positive is strictly increasing over its entire domain. A strictly increasing continuous function can intersect the horizontal axis (i.e., have a root) at most once. Since $f(\lambda)$ is a cubic polynomial, its value approaches $-\infty$ as $\lambda \to -\infty$ and $+\infty$ as $\lambda \to +\infty$. By the Intermediate Value Theorem, it must cross the x-axis exactly once.

Thus, the equation $\lambda^3 + 4\lambda - 40 = 0$ has exactly one real root.

We can approximate this root:

• $f(3) = 3^3 + 4(3) - 40 = 27 + 12 - 40 = 39 - 40 = -1$
• $f(4) = 4^3 + 4(4) - 40 = 64 + 16 - 40 = 80 - 40 = 40$ Since $f(3)$ is negative and $f(4)$ is positive, the unique real root lies between $3$ and $4$.

5. Analyzing the Solutions for $\lambda$

Our analysis of the cubic equation $\lambda^3 + 4\lambda - 40 = 0$ clearly shows that there is exactly one real value of $\lambda$ for which the determinant of the coefficient matrix is zero.

6. Conclusion and Final Answer

The number of real values of $\lambda$ for which the given system of linear equations has infinitely many solutions is 1.

The final answer is $\boxed{\text{1}}$.

7. Tips for Success & Common Mistakes

• Double-Check Determinant Calculation: Errors in calculating the determinant are very common. Always double-check your signs and arithmetic, especially for $3 \times 3$ matrices. Using both cofactor expansion and Sarrus' rule (if applicable) can help verify your result.
• Understanding Homogeneous Systems: Remember the specific conditions for homogeneous systems ($AX=0$): $\det(A) \neq 0$ implies a unique trivial solution, while $\det(A) = 0$ implies infinitely many solutions (which include non-trivial ones). Do not confuse this with non-homogeneous systems ($AX=B$ where $B \neq 0$), which have different conditions for unique, no, or infinitely many solutions.
• Analyzing Polynomial Roots: For cubic equations, always remember that they have at least one real root. To determine the exact number of real roots, analyzing the first derivative of the polynomial can be very effective. If the derivative is always positive or always negative, the cubic has only one real root. If the derivative has two real roots, the cubic can have one, two, or three real roots.
• Rational Root Theorem: For integer coefficient polynomials, testing integer factors of the constant term can help find rational roots, which can then be used to factorize the polynomial. In our case, the root was not an integer.