Linear Systems: Matrix Form & Solvability

A system of $m$ linear equations in $n$ unknowns $a_{11}x_1+\cdots+a_{1n}x_n=b_1,\ \ \ldots,\ \ a_{m1}x_1+\cdots+a_{mn}x_n=b_m$ is packaged as a single matrix equation $Ax=b$. The $m\times n$ matrix $A$ of coefficients is the coefficient matrix (Def. 5.5): its rows count the equations ($m$) and its columns count the unknowns ($n$). Appending the right-hand side as one extra column gives the $m\times(n+1)$ augmented matrix (Def. 5.6) $[A\,|\,b]=\left[\begin{array}{cccc|c} a_{11} & a_{12} & \cdots & a_{1n} & b_1\\ \vdots & \vdots & & \vdots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} & b_m \end{array}\right].$ If $b\neq 0$ the system is non-homogeneous; the companion system $Ax=0$ is its associated homogeneous system.

The solution set is $\operatorname{sol}(A,b)=\{x\in\mathbb{R}^n : Ax=b\}$ (Def. 5.7). Reading $A$ as a linear map, this is exactly the set of inputs that $A$ sends to $b$. A system is consistent when $\operatorname{sol}(A,b)\neq\varnothing$ and inconsistent when $\operatorname{sol}(A,b)=\varnothing$ (Def. 5.8). Theorem 5.4(2) gives three equivalent ways to test consistency: $Ax=b$ has a solution $\iff b\in\operatorname{im}(A)\iff\operatorname{rank}(A)=\operatorname{rank}([A\,|\,b])$. Since the columns of $A$ span $\operatorname{im}(A)$, adjoining $b$ cannot lower the rank; it raises it precisely when $b$ is not a combination of those columns. So a system is inconsistent exactly when $\operatorname{rank}([A\,|\,b])>\operatorname{rank}(A)$ (Rem. 5.3).

Theorem 5.4(1) says $\operatorname{sol}(A,0)=\ker(A)$, which is a subspace of $\mathbb{R}^n$. For a consistent non-homogeneous system, Theorem 5.4(3) says: pick any one solution $w$ (a particular solution); then every solution is $w$ plus something in the kernel, $\operatorname{sol}(A,b)=w+\ker(A)=\{w+x : x\in\ker(A)\}.$ So you solve a non-homogeneous system by finding the general homogeneous solution plus a single particular solution. The dimension of the solution set is the nullity, $\dim\operatorname{sol}(A,0)=n-\operatorname{rank}(A)$ (rank–nullity, Rem. 5.3). A homogeneous solution set always contains $0$ and is a subspace; a non-homogeneous one (with $b\neq0$) does not contain $0$ (because $A0=0\neq b$) and is an affine subspace — a copy of $\ker(A)$ shifted off the origin.