A system AX=B is homogeneous when every constant bi=0, i.e. AX=O. Since x1=x2=⋯=xn=0 (the trivial solution) always satisfies it, ρ(A)=ρ([A∣O]) automatically -- a homogeneous system is always consistent; the only real question is whether it has a non-trivial (non-zero) solution too.
Let A be the n×n coefficient matrix of a homogeneous system in n unknowns.
If ρ(A)=n (equivalently ∣A∣=0, A non-singular), the system has only the trivial solution.
If ρ(A)<n (equivalently ∣A∣=0, A singular), the system has infinitely many non-trivial solutions, forming an (n−ρ(A))-parameter family.
So for a square coefficient matrix, the entire question collapses to one determinant check: a non-trivial solution exists exactly when ∣A∣=0. (If there are more unknowns than equations, ρ(A)<n automatically, so a non-trivial solution is guaranteed without even computing a determinant.)
Worked illustration. For x+y+z=0,2x−y+z=0,x−2y=0: ∣A∣=1211−1−2110=1(0+2)−1(0−1)+1(−4+1)=2+1−3=0, so a non-trivial solution exists; row-reducing [A∣O] recovers it as a one-parameter family.
A problem with an unknown parameter λ in the coefficients ("find λ so the system has a non-trivial solution") reduces to solving ∣A(λ)∣=0 for λ -- an ordinary polynomial equation in λ, often factored using the row/column operations that create zeros before expanding.
Application -- balancing a chemical equation. Demanding the same count of each atom on both sides of a reaction, with the stoichiometric coefficients as unknowns, is itself a homogeneous linear system: it is automatically consistent (there's always the physically useless all-zero "solution"), and Gaussian elimination on it typically leaves one coefficient free -- fixed at the smallest value that makes every coefficient a positive integer, giving the balanced equation.
Note
The trivial solution is never itself "the answer" to a homogeneous word problem (e.g. a balanced chemical equation with every coefficient 0 is meaningless) -- the entire point of testing ∣A∣=0 is to locate the non-trivial solutions that actually matter.
For a homogeneous system, x=y=z=0 (the trivial solution) always exists; a non-trivial solution exists only when detA=0. Check detA in each part.
(i) detA=0⇒ non-trivial solutions exist.
(ii) detA=−33=0⇒ only the trivial solution.
✓Final answer
x=k,y=2k,z=−k,k∈R (infinitely many non-trivial solutions).
Since detA=0, the system has infinitely many non-trivial solutions besides x=y=z=0.
Step 2. Part (i): find the rank and the free variable. The 2×2 minor from rows 1–2, columns x,y is 342−3=−9−8=−17=0, so ρ(A)=2 (row 3 is a combination of rows 1–2, so it drops out). Use the two independent equations
3x+2y=−7z,4x−3y=2z.
Step 3. Part (i): solve for x,y in terms of z. Multiply the first by 3 and the second by 2, then add: 9x+6y=−21z and 8x−6y=4z give 17x=−17z⇒x=−z. Back-substitute: 3(−z)+2y=−7z⇒2y=−4z⇒y=−2z.
Step 4. Part (i): write the general solution. So x=−z,y=−2z for any z. Setting the free parameter as z=−k (i.e. k=−z) turns this into the tidy form x=k,y=2k,z=−k,k∈R — infinitely many non-trivial solutions along this one direction, with the trivial solution recovered at k=0.
Step 6. Part (ii): conclude. Since detA=−33=0, ρ(A)=3=n, so the only solution is the trivial solutionx=y=z=0.
✓Final answer
x=k,y=2k,z=−k,k∈R (infinitely many non-trivial solutions).
x=y=z=0 only.
Homogeneous system — detA=0⇒ non-trivial solutions via rank-deficient row reduction; detA=0⇒ trivial only
Saying a homogeneous system 'has no solution' when detA=0 — it always has the trivial solution x=y=z=0, and that IS the (unique) solution.
Leaving the answer as x=−z,y=−2z without checking it matches a rescaled parametrisation like x=k,y=2k,z=−k.
Using all three equations to solve for x,y,z when detA=0 — only ρ(A) of them are independent, and the extra equation is redundant, not a fresh constraint.