Question:
The following system of linear equations
\(7x + 6y - 2z = 0\)
\(3x + 4y + 2z = 0\)
\(x - 2y - 6z = 0\), has
The following system of linear equations
\(7x + 6y - 2z = 0\)
\(3x + 4y + 2z = 0\)
\(x - 2y - 6z = 0\), has
\(7x + 6y - 2z = 0\)
\(3x + 4y + 2z = 0\)
\(x - 2y - 6z = 0\), has
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Check whether the coefficient determinant is zero; if so, solve two equations to get the relation between x, y, z and test it in the third.
Updated On: Nov 5, 2026
- no solution
- infinitely many solutions, (x, y, z) satisfying y = 2z
- infinitely many solutions, (x, y, z) satisfying x = 2z
- only the trivial solution
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The Correct Option is C
Solution and Explanation
The system is homogeneous, so it always has the trivial solution (no solution is impossible).
\(\Delta = \begin{vmatrix} 7 & 6 & -2 \\ 3 & 4 & 2 \\ 1 & -2 & -6 \end{vmatrix} = 7(-24 + 4) - 6(-18 - 2) + (-2)(-6 - 4) = -140 + 120 + 20 = 0\).
So there are infinitely many solutions.
Adding equations 1 and 2: \(10x + 10y = 0\), so \(y = -x\). Put in equation 1: \(7x - 6x - 2z = 0\), so \(x = 2z\) and \(y = -2z\).
Check equation 3: \(2z + 4z - 6z = 0\), true.
The solutions satisfy \(x = 2z\) (and \(y = -2z\), not \(y = 2z\)).
Answer: (3) infinitely many solutions, (x, y, z) satisfying x = 2z
\(\Delta = \begin{vmatrix} 7 & 6 & -2 \\ 3 & 4 & 2 \\ 1 & -2 & -6 \end{vmatrix} = 7(-24 + 4) - 6(-18 - 2) + (-2)(-6 - 4) = -140 + 120 + 20 = 0\).
So there are infinitely many solutions.
Adding equations 1 and 2: \(10x + 10y = 0\), so \(y = -x\). Put in equation 1: \(7x - 6x - 2z = 0\), so \(x = 2z\) and \(y = -2z\).
Check equation 3: \(2z + 4z - 6z = 0\), true.
The solutions satisfy \(x = 2z\) (and \(y = -2z\), not \(y = 2z\)).
Answer: (3) infinitely many solutions, (x, y, z) satisfying x = 2z
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