Lecture 2: Gaussian Elimination and Solution Sets

Solve $\;x_1 + 2x_2 + x_3 = 2$, $\;2x_1 + 5x_2 + 3x_3 = 5$, $\;x_1 + 3x_2 + 3x_3 = 4$. Working on the augmented matrix: $$\begin{pmatrix} 1 & 2 & 1 & 2 \\ 2 & 5 & 3 & 5 \\ 1 & 3 & 3 & 4 \end{pmatrix} \xrightarrow{\substack{R_2 - 2R_1 \\ R_3 - R_1}} \begin{pmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 2 & 2 \end{pmatrix} \xrightarrow{R_3 - R_2} \begin{pmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{pmatrix}$$ The matrix is now in row echelon form with three pivots. Continue to the reduced row echelon form by clearing the entries above each pivot, from the rightmost pivot leftward: $$\xrightarrow{\substack{R_2 - R_3 \\ R_1 - R_3}} \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix} \xrightarrow{R_1 - 2R_2} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix}$$ Each row of the reduced form names a single variable: $x_1 = 1$, $x_2 = 0$, $x_3 = 1$. The unique solution $\mathbf{x} = \begin{pmatrix} 1 & 0 & 1 \end{pmatrix}^{T}$ (in the transpose notation) is read straight off the last column; substituting it into all three original equations confirms it.

Let $b_1, b_2, b_3$ be any constants and consider the system $$\left\{ \begin{aligned} x_1 + x_2 + x_3 + x_4 + x_5 + x_6 &= b_1 \\ x_3 - x_4 + 2x_5 + x_6 &= b_2 \\ x_5 + 3x_6 &= b_3 \end{aligned} \right.$$ The augmented matrix is already in echelon form, with pivots in columns $1$, $3$, $5$: the basic variables are $x_1, x_3, x_5$, and the remaining variables $x_2, x_4, x_6$ are free — six columns, three pivots, $6 - 3 = 3$ free variables. The echelon form tells us which variables are free; to read off all the solutions, continue to the reduced row echelon form by clearing the entries above each pivot, working from the rightmost pivot leftward: $$\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & b_1 \\ 0 & 0 & 1 & -1 & 2 & 1 & b_2 \\ 0 & 0 & 0 & 0 & 1 & 3 & b_3 \end{pmatrix} \xrightarrow{\substack{R_1 - R_3 \\ R_2 - 2R_3 \\ R_1 - R_2}} \begin{pmatrix} 1 & 1 & 0 & 2 & 0 & 3 & b_1 - b_2 + b_3 \\ 0 & 0 & 1 & -1 & 0 & -5 & b_2 - 2b_3 \\ 0 & 0 & 0 & 0 & 1 & 3 & b_3 \end{pmatrix}$$ In the reduced form each pivot row contains exactly one basic variable, so every basic variable is expressed directly in terms of the free ones — nothing left to solve. Solving each pivot row for its basic variable gives the parametric form $$\begin{aligned} x_1 &= b_1 - b_2 + b_3 - x_2 - 2x_4 - 3x_6, \\ x_3 &= b_2 - 2b_3 + x_4 + 5x_6, \\ x_5 &= b_3 - 3x_6, \end{aligned}$$ with $x_2, x_4, x_6 \in \mathbb{R}$ arbitrary. There is no need to rename the free variables: they are already names, and every choice of the three numbers $x_2, x_4, x_6$ produces exactly one solution.

To get the parametric vector form, first assemble the solution into a single column, using the parametric form for the basic entries and the free variables themselves for the rest: $$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{pmatrix} = \begin{pmatrix} b_1 - b_2 + b_3 - x_2 - 2x_4 - 3x_6 \\ x_2 \\ b_2 - 2b_3 + x_4 + 5x_6 \\ x_4 \\ b_3 - 3x_6 \\ x_6 \end{pmatrix}.$$ Now split that one column into the constant part plus the part carried by each free variable: $$\mathbf{x} = \begin{pmatrix} b_1 - b_2 + b_3 \\ 0 \\ b_2 - 2b_3 \\ 0 \\ b_3 \\ 0 \end{pmatrix} + x_2 \begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} -2 \\ 0 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_6 \begin{pmatrix} -3 \\ 0 \\ 5 \\ 0 \\ -3 \\ 1 \end{pmatrix},$$ again with $x_2, x_4, x_6 \in \mathbb{R}$ arbitrary. Every solution appears exactly once as $x_2, x_4, x_6$ range over $\mathbb{R}$: one particular solution built from the constants, plus one direction vector per free variable. Geometrically the solution set is a three-dimensional flat sheet inside $\mathbb{R}^6$. Note also that this works for every choice of $b_1, b_2, b_3$ — each row carries a pivot, a fact whose meaning is part of the theorem below.

For $x_1 + x_2 = 1$, $\;2x_1 + 2x_2 = 3$, the row operation $R_2 - 2R_1$ gives the row $(\,0 \; 0 \mid 1\,)$, i.e., the false equation $0 = 1$. By the existence theorem the system is inconsistent — this is case (ii) of the opening example of Lecture 1, parallel lines, seen through algebra: the left-hand sides are proportional but the right-hand sides are not.

The same principle decides bigger systems. Consider four equations in five unknowns with the augmented matrix below, and run Gaussian elimination column by column. First column: the top-left $1$ is the first pivot; clear everything below it. $$(\,A \mid \mathbf{b}\,) = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 1 & 1 & 1 \\ 2 & 3 & 4 & 2 & 2 & 3 \\ 3 & 4 & 5 & 6 & 1 & 2 \end{pmatrix} \xrightarrow{\substack{R_2 - R_1 \\ R_3 - 2R_1 \\ R_4 - 3R_1}} \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 & 1 \\ 0 & 1 & 2 & 3 & -2 & -1 \end{pmatrix}$$ Second column: the $1$ in row $2$ is the next pivot; clear below it. $$\xrightarrow{\substack{R_3 - R_2 \\ R_4 - R_2}} \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 3 & -2 & -1 \end{pmatrix}$$ Stop right here. We are only halfway to echelon form — rows $3$ and $4$ would still need a swap — but row $3$ already reads $(\,0 \;\, 0 \;\, 0 \;\, 0 \;\, 0 \mid 1\,)$, the impossible equation $0 = 1$. By part (1) of the theorem the system is inconsistent, and no amount of further elimination can change that. This early exit is worth remembering: the moment a row $(\,0 \, \cdots \, 0 \mid c\,)$ with $c \neq 0$ appears, the system has no solution, and the computation is over.

For which values of $h$ and $k$ is the following system consistent? $$x_1 + h\,x_2 + x_3 = 1, \qquad x_2 - x_3 = 2, \qquad x_1 - x_2 + x_3 = k.$$ Row-reduce the augmented matrix, treating $h$ and $k$ as unknown constants: $$\begin{pmatrix} 1 & h & 1 & 1 \\ 0 & 1 & -1 & 2 \\ 1 & -1 & 1 & k \end{pmatrix} \xrightarrow{R_3 - R_1} \begin{pmatrix} 1 & h & 1 & 1 \\ 0 & 1 & -1 & 2 \\ 0 & -1-h & 0 & k-1 \end{pmatrix}$$ $$\xrightarrow{R_3 + (1+h)R_2} \begin{pmatrix} 1 & h & 1 & 1 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & -(1+h) & k+1+2h \end{pmatrix}$$ Now read the last row through the existence–uniqueness theorem.

Case $h \neq -1$. The entry $-(1+h)$ is nonzero, so it is a third pivot, and there are no free variables: the system is consistent with a unique solution — for every value of $k$.

Case $h = -1$. The last row becomes $(\,0 \;\, 0 \;\, 0 \mid k - 1\,)$ (note $k+1+2h = k-1$ here). If $k \neq 1$ this row says $0 = k-1 \neq 0$: inconsistent. If $k = 1$ the row vanishes, leaving two pivots and one free variable: consistent with infinitely many solutions.

Conclusion. The system is consistent exactly when $h \neq -1$ (any $k$; unique solution), or $h = -1$ and $k = 1$ (infinitely many solutions). Note how all three cases of the theorem appear in a single family of systems as the parameters vary.