Discussion
\[ \begin{aligned} 2x + 3y + z &= 0 \\ x + y &= 0 \\ y + z &= 0 \end{aligned} \]
The given system of equations has
Answer: B
The system has infinitely many solutions.
It is homogeneous (all right-hand sides zero), so \(x = y = z = 0\) is always a solution, and there are other solutions exactly when the coefficient determinant vanishes:
\[ \begin{aligned} \det &= 2(1 \cdot 1 - 0) - 3(1 \cdot 1 - 0) + 1(1 - 0) \\ &= 2 - 3 + 1 = 0 \end{aligned} \]
Indeed the first equation is the sum of the others plus one copy of the second: \((x + y) + (x + y) + (y + z) = 2x + 3y + z\). With y = t, x = −t and z = −t for any t. A homogeneous system always has the zero solution, so 'no solution' is impossible, and a linear system cannot have a finite number greater than one.
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