Gaussian Elimination and Row Reduction: Examples and Counterexamples

Its augmented matrix reduces as follows: subtract twice row $0$ from row $1$, subtract row $0$ from row $2$, swap rows $1,2$, add three times row $1$ to row $2$, scale row $2$ by $-1\mathrm{/}7$, clear above its pivot, and then clear above the pivot in column $1$; the result is $[I_3\mid (1,2,3{)}^{\mathsf{T}}]$.

The RREF gives the unique solution. Direct substitution yields $1+2+3=6$, $2-2+3=3$, and $1+4-3=2$.

Subtracting twice row $0$ from row $1$ gives the row $(0,0,0\mid 1)$, which normalizes to a pivot in the augmented column and represents $0=1$.

The coefficient matrix has rank $2$ because rows $(1,1,1)$ and $(1,-1,0)$ are independent, while the augmented matrix has the additional pivot and rank $3$. Thus [L2] independently gives inconsistency.

Subtracting twice the first row from the second leaves one nonzero row $x+2y-z=3$. Put $y=s$ and $z=t$; then the pivot variable is $x=3-2s+t$.

This is the displayed affine form. The base point maps to $3$, while each direction maps to $0$, so direct multiplication verifies that $(3,0,0)$ is particular and the two direction vectors lie in the nullspace, as [L2] requires.

Starting from $[A\mid I_3]$, replace row $1$ by row $1$ minus row $2$, then replace row $0$ by row $0$ minus the new row $1$. The left block becomes $I_3$ and the right block becomes the displayed matrix $B$.

Direct multiplication gives $AB=I_3$ and $BA=I_3$: the only off-diagonal sums are $-1+1$, $1-1$, and $-1+1$. Hence $B=A^{-1}$.

The replacement $R_1\leftarrow R_1+(-1)R_0$ is an elementary row operation of the third kind and takes $A$ to $R$; the scaling $R_1\leftarrow (-1)R_1$ is one of the second kind, $-1$ being nonzero in $\mathbb{Q}$, and takes $R$ to $S$. So $A$ is row equivalent to each of $R$ and $S$ by [L3]. Each has no zero row, leading entries in columns $0$ and $1$ respectively, and $0$ below the leading entry of its first row, so each is in row echelon form by [L2]. Finally $R\mathrm{\ne }S$ because $-1\mathrm{\ne }1$ in $\mathbb{Q}$. The ground field matters here and not elsewhere in the example: over ${\mathbb{F}}_2$ one has $-1=1$, so $R$ and $S$ coincide and this matrix exhibits no failure of uniqueness.

From $S$, the replacement $R_0\leftarrow R_0+(-1)R_1$ is again elementary of the third kind and gives $I_2$, so $I_2$ is row equivalent to $A$ by [L3] and step 1.1; its two leading entries equal $1$ and each is alone in its column, so it is in reduced row echelon form by [L2]. Applying the same replacement after the scaling of step 1.1 carries $R$ to $I_2$ as well, in agreement with the uniqueness of [L1].

Subtract row $0$ from row $1$ without dividing by the parameter; the augmented matrix becomes $\left(\begin{array}{ccc}1& 1& 2\\ 0& t-1& 0\end{array}\right)$.

If $t\mathrm{\ne }1$, the second row has a pivot, so the rank is $2$, $y=0$, and $x=2$.

If $t=1$, the second row is zero, so the rank is $1$ and $y=s$ is free with $x=2-s$.

The alternatives $t=1$ and $t\mathrm{\ne }1$ are exhaustive and give the stated solution behaviour.

Subtract row $2$ from row $1$, then subtract the new row $1$ from row $0$; these elementary operations reduce $A$ to $I_3$, so their matrices satisfy $E_2{E}_{1}A=I_3$.

Reversing the operations gives the two displayed row-addition matrices. Their direct product is $A$, verifying the factorisation entry by entry.

Swapping the two rows of $A$ gives $B$, so the matrices are row equivalent.

Yet $\mathrm{Col}(A)=\mathrm{span}\{(1,0{)}^{\mathsf{T}}\}$ and $\mathrm{Col}(B)=\mathrm{span}\{(0,1{)}^{\mathsf{T}}\}$, which are distinct one-dimensional subspaces. Their column ranks are both $1$, exactly as [L1] predicts.

Exhausting ${\mathbb{F}}_2^2$, the sums are $0+0=0$, $0+1=1$, $1+0=1$, and $1+1=0$; hence precisely $(0,0)$ and $(1,1)$ solve the equation.

The solution set therefore has exactly two elements, so it is neither a singleton nor infinite. This refutes the extension and shows why [L1] requires an infinite field.