Q.Test the consistency of the system $x+y+z=6$, $2x+2y+2z=10$, $x-y+2z=5$ using the rank method.

A genuine numeric contradiction between two otherwise-proportional equations shows up as a mismatch between $\rho(A)$ and $\rho([A|B])$.

Row-reducing the augmented matrix $[A|B]=\begin{pmatrix}1&1&1&|&6\\2&2&2&|&10\\1&-1&2&|&5\end{pmatrix}$

$R_2\to R_2-2R_1$: $[0,0,0\,|\,10-12]=[0,0,0\,|\,-2]$ — the coefficients vanish but the constant does NOT, i.e. the row reads $0=-2$, a direct contradiction.

$R_3\to R_3-R_1$ gives $[0,-2,1\,|\,-1]$, an independent non-zero row.

In the coefficient matrix $A$ alone (ignoring the constants column), only rows $[1,1,1]$ and $[0,-2,1]$ are non-zero and independent, so $\rho(A)=2$. But in the augmented matrix, the row $[0,0,0\,|\,-2]$ is itself a non-zero row (because of its non-zero constant entry), making $\rho([A|B])=3$. …

Treating a row like $[0,0,0\,|\,-2]$ as a zero row and discarding it — a row that is all-zero in the COEFFICIENTS but non-zero in the CONSTANT is exactly the signatu …