Q.If $\rho(A) = \rho([A \mid B])$, then the system $AX = B$ of linear equations is :

Theorem (Rouché-Capelli). The system $AX=B$ is consistent (has at least one solution) if and only if the rank of the coefficient matrix equals the rank of the augmented matrix: $\rho(A)=\rho([A\,|\,B])$.

This single test subsumes and explains matrix inversion, Cramer's rule and Gaussian elimination all at once. Row-reduce $[A\,|\,B]$ to echelon form, read off $\rho(A)$ and $\rho([A\,|\,B])$ from the number of non-zero rows in the left block and in the whole augmented matrix, and apply the standing rule (with $n$ = number of unknowns):

1. $\rho(A)=\rho([A|B])=n$ $\Rightarrow$ consistent, unique solution (matches matrix inversion / Cramer's rule being applicable).
2. $\rho(A)=\rho([A|B])=n-k<n$ $\Rightarrow$ consistent, infinitely many solutions, forming a $k$-parameter family (choose $k$ unknowns arbitrarily, solve the rest by back substitution).
3. $\rho(A)\ne\rho([A|B])$ $\Rightarrow$ inconsistent, no solution (the echelon form of $[A|B]$ shows a row like $0=c$ for some $c\ne0$ -- a contradiction).

Worked illustration. For $x+y=2,\ 2x+2y=5$: $[A|B]=\begin{pmatrix}1&1&|&2\\2&2&|&5\end{pmatrix}\xrightarrow{R_2\to R_2-2R_1}\begin{pmatrix}1&1&|&2\\0&0&|&1\end{pmatrix}$. Here $\rho(A)=1$ but $\rho([A|B])=2$ (the bottom row reads $0=1$), so the system is inconsistent -- geometrically, two parallel, non-coincident lines. …