Linear systems and consistency

Setting up the problem. A system of $m$ linear equations in $n$ unknowns is written in matrix form as $A\mathbf x=\mathbf b$, where $A$ is an $m\times n$ coefficient matrix, $\mathbf x=(x_1,\dots,x_n)^T$ is the unknown vector, and $\mathbf b\in\mathbb R^m$ collects the right-hand sides. The augmented matrix $[A\,|\,\mathbf b]$ is formed by tacking $\mathbf b$ on as one extra column; row-reducing this single matrix reveals everything about both $A$ and the system at once.

The consistency criterion. The Rouché–Capelli theorem compares $\operatorname{rank}(A)$ with $\operatorname{rank}([A\,|\,\mathbf b])$:

where: $\operatorname{rank}(A)$ is the number of linearly independent rows (equivalently columns) of $A$, obtained by counting nonzero rows after row-reducing $A$ to echelon form, and $\operatorname{rank}([A\,|\,\mathbf b])$ is the same count for the augmented matrix. Appending a column can only keep the rank the same or increase it by exactly one; it increases by one precisely when $\mathbf b$ is not expressible using the row structure already present in $A$, which shows up in row-reduced form as a row reading $0=c$ for some nonzero constant $c$ — an equation no values of $x_1,\dots,x_n$ can satisfy.

Reading off the solution count. Once consistency holds, let $r=\operatorname{rank}(A)=\operatorname{rank}([A\,|\,\mathbf b])$. If $r=n$ (the number of unknowns), every unknown is pinned down by a leading entry in the reduced system, so the solution is unique. If $r<n$, there are $n-r$ unknowns left free to take any value, and the solution set is an $(n-r)$-parameter family — a line, plane, or higher-dimensional flat, depending on $n-r$.

Square systems as a shortcut. When $m=n$, the determinant gives a fast route to the same conclusion. If $\det A\neq0$, then $\operatorname{rank}(A)=n$ automatically, the system is consistent for every $\mathbf b$, and the unique solution is $\mathbf x=A^{-1}\mathbf b$. If $\det A=0$, the determinant alone cannot tell you whether the system is inconsistent or has infinitely many solutions — you must fall back on comparing the two ranks directly.

Non-square systems. The picture below makes concrete why shape matters: rank tells the whole story, but the number of equations versus unknowns sets a ceiling on what is even possible. When $m<n$ (fewer equations than unknowns), $\operatorname{rank}(A)\le m<n$ automatically, so a consistent system can never have a unique solution — it is always an infinite family. When $m>n$ (an overdetermined system), consistency becomes the tight constraint: the extra equations generically conflict, and the system is inconsistent unless the extra rows are combinations of the others.

Determine consistency and solve, if possible:

Step 1. Write $A=\begin{pmatrix}1&1&1\\2&1&-1\\1&2&3\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}6\\1\\14\end{pmatrix}$, and form the augmented matrix $[A\,|\,\mathbf b]$.

Step 2. Eliminate below the first pivot: $R_2\to R_2-2R_1$ gives $(0,-1,-3\,|\,-11)$, and $R_3\to R_3-R_1$ gives $(0,1,2\,|\,8)$, leaving

Step 3. Eliminate below the second pivot: $R_3\to R_3+R_2$ gives $(0,0,-1\,|\,-3)$, so

Both $A$ and $[A\,|\,\mathbf b]$ now have three nonzero rows, so $\operatorname{rank}(A)=\operatorname{rank}([A\,|\,\mathbf b])=3=n$: the system is consistent with a unique solution.

Step 4. Back-substitute. Row 3 gives $-z=-3\Rightarrow z=3$. Row 2 gives $-y-3(3)=-11\Rightarrow -y=-2\Rightarrow y=2$. Row 1 gives $x+2+3=6\Rightarrow x=1$. The unique solution is $(x,y,z)=(1,2,3)$.

Competitive stretch

Now test a superficially similar system for consistency:

Row reduction gives $R_2\to R_2-2R_1$: $(0,0,0\,|\,-1)$, and $R_3\to R_3-R_1$: $(0,-3,2\,|\,-2)$. The row $(0,0,0\,|\,-1)$ already signals trouble: it reads $0=-1$, impossible for any $x,y,z$. Here $\operatorname{rank}(A)=2$ (the coefficient rows reduce to two independent nonzero rows) but $\operatorname{rank}([A\,|\,\mathbf b])=3$, so $\operatorname{rank}(A)\neq\operatorname{rank}([A\,|\,\mathbf b])$: the system is inconsistent — no solution exists, even though it looks just as solvable as the first at a glance.

• Checking only $\det A=0$ for a non-square or rank-deficient square system and stopping there — you must compare $\operatorname{rank}(A)$ with $\operatorname{rank}([A\,|\,\mathbf b])$, not just inspect $A$ alone.
• Forgetting that a homogeneous system $A\mathbf x=\mathbf 0$ is always consistent (since $\mathbf x=\mathbf 0$ always works); the real question is whether a non-trivial solution exists, decided by $\det A=0$ in the square case or by rank versus $n$ in general.
• Miscounting the free parameters in an infinite family: it is $n-r$ where $r=\operatorname{rank}(A)$, not $n-m$ — the number of equations does not directly enter the count.
• Arithmetic slips while row-reducing that create a spurious $0=c$ row (or erase a genuine one) — always double-check a claimed inconsistency by confirming the row truly reads $0=\text{nonzero}$, not a miscopied coefficient.

Ask the tutor: Not sure whether your system has one, none, or infinitely many solutions before you even start solving — ask the AI tutor to walk through the rank comparison with you.