Systems of Equations and Inequalities: Chapter Review

Vocabulary and notation

Solution set of a system

Every tuple satisfying all equations at once, as one object. One equation gives a line in two variables, a plane in three; a system's set is their intersection.

Consistent / inconsistent

Consistent means at least one solution exists. Inconsistent means none.

Independent / dependent

Two ways a consistent system ends. Independent: exactly one solution. Dependent: some equation restates what the others say, so infinitely many.

Pivot / free variable

A pivot is the first variable still present in a row after elimination, its value forced by back-substitution. A variable with no pivot is free, taking any value.

Parameterize

To hand each free variable a letter and write the pivots in terms of them, naming the whole infinite set.

Augmented matrix

Each equation as a row of coefficients, constants in a final column behind a bar; without that column it is the coefficient matrix. An $m \times n$ matrix has $m$ rows, $n$ columns, and $a_{ij}$ is row $i$, column $j$.

Row-echelon (triangular) form

Each row's pivot sits strictly right of the one above it, and all-zero rows have sunk to the bottom.

Half-plane

One of the two sides the line $ax + by = c$ cuts the plane into; the solution set of one linear inequality.

Feasible region

The overlap of all the half-planes of a system of inequalities: a bounded polygon, unbounded, or empty.

Objective function

The linear expression maximized or minimized over a feasible region, the target of a linear programming problem.

Formulas and theorems

• Classifying a two-variable system by coefficient ratios

  $$\begin{gathered} \tfrac{a_1}{a_2} \ne \tfrac{b_1}{b_2}: \text{one} \\ \tfrac{a_1}{a_2} = \tfrac{b_1}{b_2} \ne \tfrac{c_1}{c_2}: \text{none} \\ \tfrac{a_1}{a_2} = \tfrac{b_1}{b_2} = \tfrac{c_1}{c_2}: \text{infinitely many} \end{gathered}$$

  Use when Standard form $a_ix + b_iy = c_i$ with $a_2, b_2, c_2 \neq 0$ so the ratios exist. If a denominator is $0$, cross-multiply instead: $a_1b_2 \neq a_2b_1$ gives one solution; when they are equal, $a_1c_2 = a_2c_1$ and $b_1c_2 = b_2c_1$ together give infinitely many and either one failing gives none. Two variables only.

  e.g. $3x - 6y = 9$ and $x - 2y = 4$: $\tfrac{3}{1} = \tfrac{-6}{-2} \ne \tfrac{9}{4}$, so no solution.

• What each elimination ending means

  $$\begin{gathered} px = q \ (p \ne 0) \Rightarrow \text{one} \\ 0 = k \ (k \ne 0) \Rightarrow \text{none} \\ 0 = 0 \Rightarrow \text{infinitely many} \end{gathered}$$

  Use when Read after a variable cancels via adding a multiple of one equation to another, which never changes the solution set. Past two unknowns, $0 = 0$ frees a variable only if no row is false: a false row overrides every vanishing row.

• Pivot count fixes the shape (three variables)

  $$\begin{gathered} \text{free variables} = 3 - r \\ r = \text{number of pivots} \end{gathered}$$

  Use when No contradiction row, or the set is empty whatever $r$ is. $r = 3$ gives a point, $r = 2$ a line, $r = 1$ a plane. Two solutions exactly is impossible: one free variable already delivers infinitely many.

• The three row operations

  $$\begin{gathered} R_i \leftrightarrow R_j \\ R_i \to kR_i \\ R_i \to R_i + kR_j \end{gathered}$$

  Use when $k \ne 0$ for the scaling move: multiplying by $0$ discards a constraint irreversibly. In the third move every entry of $R_i$ changes, the constant included, while $R_j$ is untouched. These three alone preserve the solution set.

• Reading a reduced augmented matrix

  $$\begin{gathered} [\,0 \cdots 0 \mid c\,],\ c \ne 0 \Rightarrow \text{no solution} \\ [\,0 \cdots 0 \mid 0\,] \Rightarrow \text{no constraint} \end{gathered}$$

  Use when With no contradiction row, the solution is unique when every variable column carries a pivot and infinitely many when some column does not, one parameter per pivot-free column.

• Graphing one linear inequality

  $$\begin{gathered} \le \ \text{or} \ \ge: \text{solid boundary} \\ < \ \text{or} \ >: \text{dashed boundary} \end{gathered}$$

  Text description

  The same shaded half-plane twice: solid boundary with a filled dot for a non-strict inequality, dashed boundary with an open dot for a strict one.

  Use when Boundary: the line $ax + by = c$. A test point OFF the line settles the shaded side, never the symbol, since $b < 0$ swaps the sides of $<$ and $>$. Use $(0,0)$ unless it lies on the boundary.

• Corner point principle

  $$\begin{gathered} \text{optimum of } P = \alpha x + \beta y \\ \text{over a feasible region} \\ \Rightarrow \text{attained at a vertex} \end{gathered}$$

  Text description

  A shaded feasible polygon with every corner dotted, and parallel objective lines sliding across it until the last one meets it at a single corner, the optimum.

  Use when Needs a closed region (boundaries solid) with a corner. Bounded and closed guarantees a maximum and a minimum; unbounded may give one, both, or neither, and whichever exists still sits at a vertex. Empty gives nothing. An edge parallel to the objective line ties, endpoints included.

• $2 \times 2$ determinant

  $$D = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$$

  Use when Main diagonal first, then subtract the anti-diagonal; $bc - ad$ flips the sign of everything downstream. The bars are not absolute value: a determinant is signed and often negative.

  e.g. $\begin{vmatrix} 2 & 5 \\ 3 & 4 \end{vmatrix} = 8 - 15 = -7$.

• $3 \times 3$ determinant by cofactor expansion

  $$\begin{gathered} \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \\ = a(ei - fh) - b(di - fg) \\ + \ c(dh - eg) \end{gathered}$$

  Use when First-row expansion, signs $+$, $-$, $+$. Each bracket is that entry's minor: the $2 \times 2$ determinant left after deleting its own row and column. A first-row zero kills its term.

  e.g. $\begin{vmatrix} 2 & 1 & 0 \\ 3 & 2 & 1 \\ 1 & 4 & 2 \end{vmatrix} = 2(0) - 1(5) + 0 = -5$.

• Cramer's rule

  $$x = \frac{D_x}{D}, \qquad y = \frac{D_y}{D}, \qquad z = \frac{D_z}{D}$$

  Use when REQUIRES $D \ne 0$; at $D = 0$ it yields nothing. $D$ is the coefficient determinant; $D_x$ replaces its $x$-column with the constants, $D_y$ its $y$-column, $D_z$ its $z$-column, after lining the variables up in one order with the constants alone on the right. The swapped determinant goes on top, never $D/D_x$. Best for one unknown or for theory; row reduction is far cheaper on large systems.

  e.g. $2x + 3y = 8$, $x + 4y = 9$: $D = 5$, $D_x = 5$, $D_y = 10$, so $(x, y) = (1, 2)$.

• Determinant test for a unique solution

  $$D \ne 0 \iff \text{exactly one solution}$$

  Use when Square systems, equations matching unknowns. $D = 0$ says only that the solution is not unique: none or infinitely many, and $D$ never distinguishes those, so compare the constants or reduce.

  e.g. $\begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} = k^2 - 1$, so $kx + y = 1$, $x + ky = 3$ is unique for every $k$ except $\pm 1$.

Problem types, step by step

Exam traps

• Trap Elimination ends at $0 = 0$ and the answer is written as $x = 0$, or as "no solution".

  Fix $0 = 0$ is always true, so that equation was redundant. Past two EQUATIONS it does NOT settle the count on its own: the equations still standing decide, and every count stays open. A single $0 = k$ with $k \ne 0$ anywhere overrides every vanished row and means none, while $x = 1$ with $y = 2$ and $x + y = 3$ vanishes a row and still pins one pair.

• Trap Answering "infinitely many" and stopping there.

  Fix That names the case without describing the set. Solve for the pivots and give the family, such as $(x,\ 2x - 3)$ for all real $x$.

• Trap Taking "no equation is a multiple of another" as proof that a three-variable system is consistent or unique.

  Fix It proves neither. $x + y + 2z = 4$ plus $2x + 3y + z = 5$ gives $3x + 4y + 3z = 9$, so a third equation $3x + 4y + 3z = 12$ is inconsistent with them, yet no two are proportional.

• Trap Concluding that $D = 0$ means the system has no solution.

  Fix It means not unique, so either none or infinitely many. $D$ cannot tell those apart; compare the constants or reduce.

• Trap Adding all three cofactor terms, so $1(-24) - 2(-20) + 3(-5)$ comes out as $-24 - 40 - 15 = -79$ instead of $1$.

  Fix The first-row signs are $+$, $-$, $+$. The middle term is subtracted, so a negative minor there ADDS.

• Trap Shading a half-plane by the direction of the symbol.

  Fix Substitute a test point: $-x - y > -4$ shades the side holding the origin, since $0 > -4$ is true. A point on the boundary gives equality, deciding nothing.