Homogeneous System of Linear Equations

A system $AX=B$ is homogeneous when $B=O$ (every constant is $0$): $AX=O$. Since $x_1=x_2=\cdots=x_n=0$ (the trivial solution) satisfies $AX=O$ automatically, $\rho(A)=\rho([A|O])$ always holds -- a homogeneous system is always consistent; the only real question is whether a non-trivial (some $x_i\ne0$) solution also exists.

Let $A$ be $n\times n$ (square coefficient matrix, $n$ unknowns).

Case $\rho(A)=n$. The system has the unique solution, which must be the trivial one: since $\rho(A)=n\Leftrightarrow|A|\ne0$, only $X=O$ satisfies $AX=O$.

Case $\rho(A)<n$. The system has a non-trivial solution too, forming an $(n-\rho(A))$-parameter family; since $\rho(A)<n\Leftrightarrow|A|=0$.

Governing rule (square case). A homogeneous system with a square coefficient matrix has a non-trivial solution exactly when $|A|=0$.

If there are more unknowns than equations ($m<n$), then automatically $\rho(A)\le m<n$, so a non-trivial solution is guaranteed without even computing a determinant.

Worked illustration. For $x+2y+3z=0,\ 2x+3y+z=0,\ 3x+y+2z=0$: $|A|=\begin{vmatrix}1&2&3\\2&3&1\\3&1&2\end{vmatrix}=1(6-1)-2(4-3)+3(2-9)=5-2-21=-18\ne0$, so only the trivial solution exists.

A parametrised homogeneous system. "Find $\lambda$ so that the system has a non-trivial solution" reduces to solving the polynomial equation $|A(\lambda)|=0$ for $\lambda$ -- often factored cleanly using row/column operations (adding all rows/columns together, or exploiting a common factor) before expanding. …