Question:
The greatest value of \(c \in R\) for which the system of linear equations
\(x - cy - cz = 0\)
\(cx - y + cz = 0\)
\(cx + cy - z = 0\)
has a non-trivial solution, is :
The greatest value of \(c \in R\) for which the system of linear equations
\(x - cy - cz = 0\)
\(cx - y + cz = 0\)
\(cx + cy - z = 0\)
has a non-trivial solution, is :
\(x - cy - cz = 0\)
\(cx - y + cz = 0\)
\(cx + cy - z = 0\)
has a non-trivial solution, is :
Show Hint
A homogeneous system has a non-trivial solution exactly when the coefficient determinant is zero.
Updated On: Nov 5, 2026
- \(-1\)
- \(0\)
- \(\frac{1}{2}\)
- \(2\)
Show Solution
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The Correct Option is C
Solution and Explanation
A homogeneous system has a non-trivial solution iff the determinant of coefficients is 0.
\(\begin{vmatrix}1 & -c & -c\\ c & -1 & c\\ c & c & -1\end{vmatrix} = 1(1 - c^2) + c(-c - c^2) - c(c^2 + c)\)
\(= 1 - c^2 - c^2 - c^3 - c^3 - c^2 = 1 - 3c^2 - 2c^3\)
Set \(2c^3 + 3c^2 - 1 = 0\). Since \(c = -1\) is a root: \(2c^3 + 3c^2 - 1 = (c + 1)(2c^2 + c - 1) = (c + 1)^2(2c - 1)\).
So \(c = -1\) or \(c = \frac{1}{2}\). The greatest value is \(c = \frac{1}{2}\).
Answer: (3) \(\frac{1}{2}\)
\(\begin{vmatrix}1 & -c & -c\\ c & -1 & c\\ c & c & -1\end{vmatrix} = 1(1 - c^2) + c(-c - c^2) - c(c^2 + c)\)
\(= 1 - c^2 - c^2 - c^3 - c^3 - c^2 = 1 - 3c^2 - 2c^3\)
Set \(2c^3 + 3c^2 - 1 = 0\). Since \(c = -1\) is a root: \(2c^3 + 3c^2 - 1 = (c + 1)(2c^2 + c - 1) = (c + 1)^2(2c - 1)\).
So \(c = -1\) or \(c = \frac{1}{2}\). The greatest value is \(c = \frac{1}{2}\).
Answer: (3) \(\frac{1}{2}\)
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