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

For a system of linear equations to have a non-trivial solution (meaning, a solution where not all variables are zero), a fundamental condition must be met. Specifically, for a homogeneous system of linear equations (where all constant terms are zero), such as:

$$\begin{cases} a_1x + b_1y + c_1z = 0 \\ a_2x + b_2y + c_2z = 0 \\ a_3x + b_3y + c_3z = 0 \end{cases}$$

the system possesses non-trivial solutions if and only if the determinant of its coefficient matrix is equal to zero. If the determinant is non-zero, the only solution is the trivial solution ($x=0, y=0, z=0$).

2. Forming the Coefficient Matrix

First, let's write down the given system of linear equations:

1. $x - cy - cz = 0$
2. $cx - y + cz = 0$
3. $cx + cy - z = 0$

We can represent this system in matrix form as $AX = O$, where $A$ is the coefficient matrix, $X$ is the column vector of variables $(x, y, z)^T$, and $O$ is the zero vector. The coefficient matrix $A$ is formed by taking the coefficients of $x, y,$ and $z$ from each equation:

$$A = \begin{pmatrix} 1 & -c & -c \\ c & -1 & c \\ c & c & -1 \end{pmatrix}$$

3. Setting up the Determinant Equation

As established by the key concept, for the system to have a non-trivial solution, the determinant of the coefficient matrix $A$ must be zero. So, we set $\det(A) = 0$:

$$\left| \begin{matrix} 1 & -c & -c \\ c & -1 & c \\ c & c & -1 \end{matrix} \right| = 0$$

4. Step-by-Step Determinant Expansion

Now, we expand the $3 \times 3$ determinant. We'll use the cofactor expansion method along the first row for clarity. The formula for a $3 \times 3$ determinant $\left| \begin{smallmatrix} a & b & c \\ d & e & f \\ g & h & i \end{smallmatrix} \right|$ is $a(ei - fh) - b(di - fg) + c(dh - eg)$.

Applying this to our determinant:

$$1 \cdot ((-1)(-1) - (c)(c)) - (-c) \cdot ((c)(-1) - (c)(c)) + (-c) \cdot ((c)(c) - (-1)(c)) = 0$$

Let's simplify each term:

• First term: $1 \cdot (1 - c^2) = 1 - c^2$
  • Explanation: We multiply the element $1$ by the determinant of the $2 \times 2$ submatrix obtained by removing its row and column.
• Second term: $- (-c) \cdot (-c - c^2) = c(-c - c^2) = -c^2 - c^3$
  • Explanation: We multiply the element $-c$ by the determinant of its corresponding $2 \times 2$ submatrix, and then subtract this product (due to the sign pattern of the cofactor expansion, which is $\begin{smallmatrix} + & - & + \\ - & + & - \\ + & - & + \end{smallmatrix}$).
• Third term: $-c \cdot (c^2 + c) = -c^3 - c^2$
  • Explanation: We multiply the element $-c$ by the determinant of its corresponding $2 \times 2$ submatrix, and then add this product.

Combining these simplified terms, we get:

$$(1 - c^2) + (-c^2 - c^3) + (-c^3 - c^2) = 0$$

5. Algebraic Simplification and Factorization

Now, we combine like terms to simplify the polynomial equation in $c$:

$$1 - c^2 - c^2 - c^3 - c^3 - c^2 = 0$$ $$1 - 3c^2 - 2c^3 = 0$$

Rearranging the terms in descending powers of $c$:

$$-2c^3 - 3c^2 + 1 = 0$$

Multiply by $-1$ to make the leading coefficient positive (optional, but often makes factorization easier):

$$2c^3 + 3c^2 - 1 = 0$$

This is a cubic polynomial equation. To find its roots, we can try to factor it. A common strategy for cubic polynomials in competitive exams is to test for simple integer or rational roots using the Rational Root Theorem. We can test $c = \pm 1, \pm 1/2$.

Let's test $c = -1$: $2(-1)^3 + 3(-1)^2 - 1 = 2(-1) + 3(1) - 1 = -2 + 3 - 1 = 0$. Since $c = -1$ is a root, $(c - (-1))$, which is $(c+1)$, is a factor of the polynomial.

Now, we can perform polynomial division or synthetic division to find the other factor(s). Dividing $2c^3 + 3c^2 - 1$ by $(c+1)$:

$$(2c^3 + 3c^2 - 1) \div (c+1) = 2c^2 + c - 1$$

So, the equation becomes:

$$(c+1)(2c^2 + c - 1) = 0$$

Next, we factor the quadratic term $2c^2 + c - 1$. We can use the splitting the middle term method: $2c^2 + 2c - c - 1 = 2c(c+1) - 1(c+1) = (2c-1)(c+1)$.

Substituting this back into the equation:

$$(c+1)(2c-1)(c+1) = 0$$

This can be written more compactly as:

$$(c+1)^2 (2c-1) = 0$$

6. Finding the Values of c

From the factored form $(c+1)^2 (2c-1) = 0$, we can find the possible values of $c$ that satisfy the equation:

• $c+1 = 0 \Rightarrow c = -1$ (this root has a multiplicity of 2)
• $2c-1 = 0 \Rightarrow 2c = 1 \Rightarrow c = \frac{1}{2}$

So, the values of $c$ for which the system has a non-trivial solution are $c = -1$ and $c = \frac{1}{2}$.

7. Identifying the Greatest Value

We are asked for the greatest value of $c$. Comparing the two values:

• $c_1 = -1$
• $c_2 = \frac{1}{2}$

Clearly, $\frac{1}{2}$ is greater than $-1$.

Therefore, the greatest value of $c$ is $\frac{1}{2}$.

8. Tips and Common Mistakes

• Sign Errors: Be extremely careful with signs when expanding determinants, especially with the negative terms. A common mistake is forgetting the negative sign in front of the second term in the $3 \times 3$ determinant expansion.
• Algebraic Simplification: Ensure careful algebraic manipulation when combining terms and factoring polynomials. Double-check your work, especially when dealing with cubic equations.
• Factoring Cubics: For cubic polynomials in JEE problems, try testing small integer roots like $\pm 1, \pm 2$ or simple rational roots like $\pm 1/2, \pm 1/3$. This often leads to a quick factorization.
• Understanding "Non-trivial": Remember that "non-trivial solution" for a homogeneous system always means the determinant of the coefficient matrix must be zero. If the system were non-homogeneous, the conditions for unique, no, or infinite solutions would be different (using Cramer's rule or rank method).

9. Summary and Key Takeaway

The problem demonstrates a core concept in linear algebra: a homogeneous system of linear equations has non-trivial solutions if and only if the determinant of its coefficient matrix is zero. The process involves:

1. Constructing the coefficient matrix.
2. Setting its determinant to zero.
3. Carefully expanding and simplifying the determinant to form a polynomial equation.
4. Solving the polynomial equation to find the possible values of the unknown parameter ($c$ in this case).
5. Finally, selecting the required value (the greatest in this question). Mastery of determinant calculation and polynomial factorization is crucial for such problems.