<p>The greatest value of \(c \in R\) for which the system of linear equations<br>\(x - cy - cz = 0\)<br>\(cx - y + cz = 0\)<br>\(cx + cy - z = 0\)<br>has a non-trivial solution, is:</p>
Step-by-Step Solution
Key Concept: A homogeneous system has non-trivial solutions if and only if the determinant of the coefficient matrix equals zero. Find the determinant, set it to zero, and solve for the parameter c to identify which value gives the maximum.
<p><strong>Step 1:</strong> Write the coefficient matrix for the homogeneous system:</p><p>$$A = \begin{pmatrix} 1 & -c & -c \\ c & -1 & c \\ c & c & -1 \end{pmatrix}$$</p><p><strong>Step 2:</strong> For non-trivial solutions, we need det(A) = 0. Expand along the first row:</p><p>$$\det(A) = 1\begin{vmatrix} -1 & c \\ c & -1 \end{vmatrix} + c\begin{vmatrix} c & c \\ c & -1 \end{vmatrix} - c\begin{vmatrix} c & -1 \\ c & c \end{vmatrix}$$</p><p><strong>Step 3:</strong> Calculate each 2×2 determinant:</p><p>$$= 1(1 - c^2) + c(-c - c^2) - c(c^2 + c)$$</p><p>$$= 1 - c^2 - c^2 - c^3 - c^3 - c^2$$</p><p>$$= 1 - 3c^2 - 2c^3$$</p><p><strong>Step 4:</strong> Set det(A) = 0:</p><p>$$2c^3 + 3c^2 - 1 = 0$$</p><p><strong>Step 5:</strong> Test c = 1: $$2(1) + 3(1) - 1 = 4 ≠ 0$$. Test c = -1: $$2(-1) + 3(1) - 1 = 0$$ ✓</p><p>Factor: $$(c + 1)(2c^2 + c - 1) = 0$$</p><p>$$(c + 1)(2c - 1)(c + 1) = 0$$</p><p>$$(c + 1)^2(2c - 1) = 0$$</p><p><strong>Step 6:</strong> Solutions are $$c = -1$$ (double root) and $$c = \frac{1}{2}$$</p><p>∴ Greatest value of c is: <strong>1/2</strong></p>
Correct Answer: D