1. Key Concept: Non-Trivial Solutions for Homogeneous Systems

A system of linear equations is called a homogeneous system if all the constant terms are zero. For the given system: $\begin{aligned} x + \lambda y - z &= 0 \\ \lambda x - y - z &= 0 \\ x + y - \lambda z &= 0 \end{aligned}$ Notice that all equations are set to zero. Such a system always has at least one solution, which is the trivial solution: $x=0, y=0, z=0$.

For a homogeneous system to have non-trivial solutions (i.e., solutions where at least one of $x, y, z$ is non-zero), a specific condition must be met. This condition is that the determinant of the coefficient matrix must be equal to zero. If we represent the system in matrix form as $AX = 0$, where $A$ is the coefficient matrix and $X$ is the column vector of variables, then non-trivial solutions exist if and only if $\det(A) = 0$.

2. Setting Up the Problem

First, let's write down the coefficient matrix $A$ for the given system of equations: $A = \begin{pmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{pmatrix}$

3. Applying the Condition for Non-Trivial Solutions

As established, for the system to have a non-trivial solution, the determinant of the coefficient matrix $A$ must be zero. $\det(A) = \begin{vmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{vmatrix} = 0$

4. Step-by-Step Calculation of the Determinant

We will expand the $3 \times 3$ determinant along the first row. The formula for expanding a $3 \times 3$ determinant $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$ is $a(ei - fh) - b(di - fg) + c(dh - eg)$.

Applying this to our determinant: $\begin{vmatrix} 1 & \lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -\lambda \end{vmatrix} = 1 \cdot \left( (-1)(-\lambda) - (-1)(1) \right) - \lambda \cdot \left( (\lambda)(-\lambda) - (-1)(1) \right) + (-1) \cdot \left( (\lambda)(1) - (-1)(1) \right)$

Now, let's simplify each term:

• First term: $1 \cdot (\lambda + 1)$
  • Here, $(-1)(-\lambda) = \lambda$ and $(-1)(1) = -1$. So, $\lambda - (-1) = \lambda + 1$.
• Second term: $-\lambda \cdot (-\lambda^2 + 1)$
  • Here, $(\lambda)(-\lambda) = -\lambda^2$ and $(-1)(1) = -1$. So, $-\lambda^2 - (-1) = -\lambda^2 + 1$.
• Third term: $-1 \cdot (\lambda + 1)$
  • Here, $(\lambda)(1) = \lambda$ and $(-1)(1) = -1$. So, $\lambda - (-1) = \lambda + 1$.

Substitute these simplified terms back into the determinant equation: $1(\lambda + 1) - \lambda(-\lambda^2 + 1) - 1(\lambda + 1) = 0$

5. Solving the Resulting Equation for $\lambda$

Now, we simplify the equation further: $(\lambda + 1) - \lambda(-\lambda^2 + 1) - (\lambda + 1) = 0$ Notice that the first term $(\lambda+1)$ and the third term $-(\lambda+1)$ cancel each other out. $-\lambda(-\lambda^2 + 1) = 0$ We can multiply both sides by $-1$ to make it easier to work with: $\lambda(-\lambda^2 + 1) = 0$ Rearrange the term inside the parenthesis: $\lambda(1 - \lambda^2) = 0$ Now, factorize the term $(1 - \lambda^2)$ using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$): $\lambda(1 - \lambda)(1 + \lambda) = 0$ For this product to be zero, at least one of the factors must be zero. This gives us the possible values for $\lambda$: $\lambda = 0 \quad \text{or} \quad 1 - \lambda = 0 \quad \text{or} \quad 1 + \lambda = 0$ Solving each of these simple equations: $\lambda = 0, \quad \lambda = 1, \quad \lambda = -1$

6. Interpreting the Results and Conclusion

We have found three distinct values of $\lambda$ for which the system of linear equations has a non-trivial solution: $\lambda = -1, 0, 1$. Therefore, there are exactly three values of $\lambda$.

Comparing this with the given options: (A) infinitely many values of $\lambda$. (B) exactly one value of $\lambda$. (C) exactly two values of $\lambda$. (D) exactly three values of $\lambda$.

The correct option is (D).

Tips for Success and Common Mistakes:

• Understand the Condition: Always remember that for a homogeneous system, $\det(A) = 0$ is the necessary and sufficient condition for non-trivial solutions. If $\det(A) \ne 0$, only the trivial solution ($x=y=z=0$) exists.
• Determinant Expansion Accuracy: Be very careful with signs when expanding determinants, especially for $3 \times 3$ matrices. A common mistake is forgetting the alternating signs ($+ - +$) for the terms in the expansion.
• Algebraic Simplification: Take your time with the algebraic simplification of the polynomial equation. Look for opportunities to factor or cancel terms, as shown in step 5 where $(\lambda+1)$ terms cancelled out. This can significantly simplify the process of finding roots.
• Don't Miss $\lambda=0$: It's easy to overlook $\lambda=0$ if you jump straight to $\lambda^2 - 1 = 0$ after seeing $\lambda(\lambda^2 - 1) = 0$. Always factor out common terms completely.

Summary / Key Takeaway:

For a homogeneous system of linear equations ($AX=0$), non-trivial solutions exist if and only if the determinant of the coefficient matrix $A$ is zero. The process involves forming the determinant, setting it equal to zero, and then solving the resulting algebraic equation for the unknown parameter(s). In this problem, solving the determinant equation $\lambda(1-\lambda^2)=0$ led to three distinct values for $\lambda$.