Unique, inconsistent and dependent systems

Example

For \(x+y+z=4\), \(2x+2y+2z=8\) and \(x-y+z=0\), the second equation repeats the first. Subtracting the third from the first gives \(2y=4\), so \(y=2\) and \(x+z=2\).

The full solution is \((x,y,z)=(2-t,2,t)\), \(t\in\mathbb R\). A single choice such as \((2,2,0)\) does not describe all solutions.

Replacing the second equation by \(2x+2y+2z=9\) gives \(0=1\) after subtracting twice the first equation, so there is no solution.

Technology check

A row of zeros in a reduced matrix indicates a redundant equation; a zero coefficient row with a nonzero final entry indicates inconsistency. A solver error alone does not distinguish these cases. Inspect the reduced equations or perform elimination, and report the mathematical conclusion.

Parameter cases

If an elimination step would divide by a parameter expression, separate its zero case first. A zero denominator in a derived formula identifies a case to investigate; it does not by itself prove that there are no solutions.

For the illustrative system \(x+y=2\), \(2x+ay=b\), subtract twice the first equation to obtain \((a-2)y=b-4\). If \(a\ne2\), then \(y=\dfrac{b-4}{a-2}\) and \(x=2-y\). If \(a=2,b=4\), every pair \((2-t,t)\), \(t\in\mathbb R\), is a solution. If \(a=2,b\ne4\), the equations contradict one another.

For three unknowns, give one parameter for each free variable and substitute the full parameterisation into every original equation. A dependent system may have one or two free variables; never choose arbitrary zero values and report a single point as the general solution.