Precalculus · Chapter 10 — Systems of Equations and Inequalities
Precalculus · Chapter 10

Systems of Equations and Inequalities

SECTION 10.1

Systems of Linear Equations in Two Variables

System of Equations

A system of equations is a set of equations that involve the same variables. A system of linear equations is one in which every equation is linear. A solution of a system is an assignment of values to the variables that makes every equation true. To solve a system means to find all of its solutions. Graphically, each linear equation is a line, and a solution is a point lying on every line in the system — a point of intersection.

Substitution Method

A strategy for solving a system in which you isolate one variable in one equation and substitute the resulting expression into the other equation, reducing the system to a single equation in one variable. Step 1 — Solve for one variable in terms of the other. Step 2 — Substitute that expression into the other equation and solve. Step 3 — Back-substitute the value found into the expression from Step 1 to get the remaining variable.

Elimination Method

A strategy that combines the equations (by adding or subtracting) so one variable cancels. Step 1 — Adjust the coefficients by multiplying equations so that the coefficient of one variable in one equation is the negative of its coefficient in the other. Step 2 — Add the equations to eliminate that variable and solve. Step 3 — Back-substitute into an original equation to find the remaining variable.

Graphical Method

Using a graphing device, express each equation by solving for \(y\), graph both on the same screen, and read off the coordinates of the point(s) of intersection. Useful for approximate solutions with awkward coefficients.

Number of Solutions of a Linear System in Two Variables

Because two lines either cross once, run parallel, or coincide, exactly one of the following is always true: the system has exactly one solution, no solution, or infinitely many solutions. A system with no solution is inconsistent; a system with infinitely many solutions is dependent.

Modeling with Linear Systems

Guidelines for translating a word problem into a system: (1) identify the variables (what the problem asks you to find), (2) express all unknown quantities in terms of those variables, (3) set up a system that captures the relationships in the problem, and (4) solve the system, check, and state the answer in a sentence.

Linear Equation in Two Variables
\[ ax + by = c \]
A System of Two Linear Equations (general form)
\[ \begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} \]
Dependent System — Parametric Solution
\[ x = t, \qquad y = \tfrac{1}{2}t - 2, \qquad t \in \mathbb{R} \]

(Form used to express the infinitely many solutions of a dependent two-variable system, as in Example 6: the ordered pair \(\left(t, \tfrac{1}{2}t - 2\right)\) for any real \(t\).)

EXAMPLE 1 Substitution Method
Find all solutions of the system \(\;\begin{cases} 2x + y = 1 \\ 3x + 4y = 14 \end{cases}\)
Solution
Solve for one variable. From Equation 1, \( y = 1 - 2x \).
Substitute into Equation 2: \( 3x + 4(1 - 2x) = 14 \Rightarrow 3x + 4 - 8x = 14 \Rightarrow -5x + 4 = 14 \Rightarrow -5x = 10 \Rightarrow x = -2 \).
Back-substitute. \( y = 1 - 2(-2) = 5 \).
Solution: the ordered pair \((-2,\,5)\).
EXAMPLE 2 Elimination Method
Find all solutions of the system \(\;\begin{cases} 3x + 2y = 14 \\ \phantom{0}x - 2y = 2 \end{cases}\)
Solution
The \(y\)-coefficients are negatives of each other, so add the equations: \( 4x = 16 \Rightarrow x = 4 \).
Back-substitute \(x = 4\) into the simpler equation \(x - 2y = 2\): \( 4 - 2y = 2 \Rightarrow -2y = -2 \Rightarrow y = 1 \).
Solution: \((4,\,1)\).
EXAMPLE 3 Graphical Method
Find all solutions of \(\;\begin{cases} 1.35x - 2.13y = 2.36 \\ 2.16x + 0.32y = 1.06 \end{cases}\)
Solution
Solve each equation for \(y\): \( y = 0.63x - 1.11 \) and \( y = -6.75x + 3.31 \).
Graph both lines and zoom in on the intersection. The solution is approximately \((0.30,\,-1.30)\).
EXAMPLE 4 A Linear System with One Solution
Solve and graph the lines: \(\;\begin{cases} 3x - y = 0 \\ 5x + 2y = 22 \end{cases}\)
Solution
Multiply Equation 1 by \(2\): \(6x - 2y = 0\). Add to Equation 2: \( 11x = 22 \Rightarrow x = 2 \).
Back-substitute into \(3x - y = 0\): \( 6 - y = 0 \Rightarrow y = 6 \).
Solution: \((2,\,6)\). The two lines intersect at a single point.
EXAMPLE 5 A Linear System with No Solution
Solve \(\;\begin{cases} 8x - 2y = 5 \\ -12x + 3y = 7 \end{cases}\)
Solution
Multiply Equation 1 by \(3\) and Equation 2 by \(2\): \( 24x - 6y = 15 \) and \( -24x + 6y = 14 \).
Add: \( 0 = 29 \), which is false. The system has no solution — the lines are parallel, so the system is inconsistent.
EXAMPLE 6 A Linear System with Infinitely Many Solutions
Solve \(\;\begin{cases} 3x - 6y = 12 \\ 4x - 8y = 16 \end{cases}\)
Solution
Both equations describe the same line. In slope-intercept form, \( y = \tfrac{1}{2}x - 2 \).
Let \(t\) be any real number. The complete solution is \( x = t,\; y = \tfrac{1}{2}t - 2 \), i.e. the ordered pair \(\left(t,\,\tfrac{1}{2}t - 2\right)\). The system is dependent.
EXAMPLE 7 A Distance–Speed–Time Problem
A woman rows upstream 4 mi in \(1\tfrac{1}{2}\) h; the return trip downstream takes 45 min. Find her rowing speed relative to the water and the speed of the current.
Solution
Let \(x =\) rowing speed (mi/h) and \(y =\) current speed (mi/h). Speed upstream is \(x - y\); downstream is \(x + y\).
Using distance = speed × time with distance 4 mi each way: \( (x - y)\tfrac{3}{2} = 4 \) and \( (x + y)\tfrac{3}{4} = 4 \).
Clear denominators: \( \begin{cases} 3x - 3y = 8 \\ 3x + 3y = 16 \end{cases} \). Add: \( 6x = 24 \Rightarrow x = 4 \).
Back-substitute: \( 3(4) - 3y = 8 \Rightarrow 3y = 4 \Rightarrow y = \tfrac{4}{3} \).
Answer: She rows at 4 mi/h, and the current flows at \(1\tfrac{1}{3}\) mi/h.
EXAMPLE 8 A Mixture Problem
A vintner fortifies wine containing 10% alcohol by adding a 70% alcohol solution. The mixture is 16% alcohol and fills 1000 one-liter bottles. How many liters of wine and of the alcohol solution are used?
Solution
Let \(x =\) liters of wine, \(y =\) liters of alcohol solution. Total volume: \( x + y = 1000 \).
Alcohol balance: \( 0.10x + 0.70y = 0.16(1000) = 160 \). Multiply by 10: \( x + 7y = 1600 \).
Subtract \(x + y = 1000\) from \(x + 7y = 1600\): \( 6y = 600 \Rightarrow y = 100 \), then \( x = 900 \).
Answer: 900 L of wine and 100 L of the alcohol solution.
A linear system in two variables; the solution is the intersection point of the two lines.
Figure 10.1.1
The three possible outcomes for a linear system in two variables: one solution, no solution (parallel), infinitely many solutions (coincident).
Figure 10.1.2
ADMIN NOTE
Pick your method by what the equations hand you. If one variable already has coefficient \(1\) or \(-1\), substitution is fastest. If the coefficients of one variable are already negatives or easy multiples of each other, elimination wins. Reach for the graphical method only when coefficients are decimals/awkward and an approximate answer is acceptable. When you finish, always plug your pair back into both original equations — a 10-second check that catches most sign errors.
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SECTION 10.2

Systems of Linear Equations in Several Variables

Linear Equation in n Variables

An equation that can be written as \( a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = c \), where the \(a_i\) and \(c\) are real numbers. Equations are linear when each term contains at most one variable to the first power — no squares, square roots, or products of variables. With three or four variables we usually write \(x, y, z, w\) instead of \(x_1, x_2, x_3, x_4\).

Triangular Form & Back-Substitution

A system is in triangular form when each successive equation drops a variable (the second equation has no \(x\), the third has neither \(x\) nor \(y\), etc.). Such systems are solved easily by back-substitution: solve the last equation, then work upward, substituting known values as you go.

Gaussian Elimination

The process of using equation operations to convert a linear system into an equivalent triangular system, which is then solved by back-substitution. Two systems are equivalent when they have the same solutions.

Number of Solutions / Parameters

A linear system has exactly one solution, no solution, or infinitely many solutions. No solution means it is inconsistent (Gaussian elimination yields a false statement). Infinitely many means it is dependent; the solution is described using one or more parameters — free variables that may take any real value.

Operations That Yield an Equivalent System

1. Add a nonzero multiple of one equation to another.
2. Multiply an equation by a nonzero constant.
3. Interchange the positions of two equations.

Linear Equation in n Variables
\[ a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = c \]
EXAMPLE 1 Solving a Triangular System by Back-Substitution
Solve \(\;\begin{cases} x - 2y - z = 1 \\ \phantom{x - 2}y + 2z = 5 \\ \phantom{x - 2y + 2}z = 3 \end{cases}\)
Solution
From Equation 3, \( z = 3 \).
Back-substitute into Equation 2: \( y + 2(3) = 5 \Rightarrow y = -1 \).
Back-substitute into Equation 1: \( x - 2(-1) - 3 = 1 \Rightarrow x = 2 \).
Solution: \((2,\,-1,\,3)\).
EXAMPLE 2 Three Equations in Three Variables (Gaussian Elimination)
Solve \(\;\begin{cases} x - 2y + 3z = 1 \\ x + 2y - z = 13 \\ 3x + 2y - 5z = 3 \end{cases}\)
Solution
Eliminate \(x\) from Equation 2 (\(E_2 - E_1\)) and Equation 3 (\(E_3 - 3E_1\)), then simplify, reaching the triangular system \( \begin{cases} x - 2y + 3z = 1 \\ y - 3z = -3 \\ z = 4 \end{cases} \).
Back-substitute: \( z = 4 \Rightarrow y = -1(... ) \). Working upward gives \( y = -7,\; x = 3 \).
Solution: \((3,\,-7,\,4)\).
EXAMPLE 3 A System with No Solution
Solve \(\;\begin{cases} x + 2y - 2z = 1 \\ 2x + 2y - z = 6 \\ 3x + 4y - 3z = 5 \end{cases}\)
Solution
Eliminate \(x\) from Equations 2 and 3, then eliminate \(y\) from the third equation.
The third row reduces to \( 0 = -2 \), which is false. The system has no solution (inconsistent).
EXAMPLE 4 A System with Infinitely Many Solutions
Solve \(\;\begin{cases} x - y - 5z = -2 \\ 2x + y - 4z = 2 \\ 2x + 4y + 2z = 8 \end{cases}\)
Solution
Eliminating \(x\) and then \(y\) reduces the third equation to \( 0 = 0 \) — true but redundant, so drop it.
Solve the remaining equations for \(x\) and \(y\) in terms of \(z\). Let \( z = t \): \( x = 3t,\; y = 2t + 2,\; z = t \).
Solution: the ordered triple \((3t,\, 2t + 2,\, t)\) for any real \(t\) (a dependent system).
EXAMPLE 5 Modeling a Financial Problem
Jason invests $50,000 across a money-market fund (5% return), a blue-chip fund (9%), and a high-tech fund (16%). He wants a total first-year return of $4000 and decides to put three times as much in the money-market fund as in the high-tech fund. How much goes in each?
Solution
Let \(x, y, z\) be the amounts in the money-market, blue-chip, and high-tech funds. Then \( \begin{cases} x + y + z = 50000 \\ 0.05x + 0.09y + 0.16z = 4000 \\ x = 3z \end{cases} \).
Solve by Gaussian elimination: \( x = 30000,\; y = 10000,\; z = 10000 \).
Answer: $30,000 in the money-market fund, $10,000 in the blue-chip fund, and $10,000 in the high-tech fund.
A linear equation in three variables is a plane. Three planes may meet at one point (one solution), along a line (infinitely many), or share no common point (no solution).
Figure 10.2.1
ADMIN NOTE
Always march toward triangular form by clearing the lower-left corner first: kill \(x\) from rows 2 and 3, then kill \(y\) from row 3. When you hit a row that reads \(0 = (\text{nonzero})\) the system is inconsistent — stop. When you hit \(0 = 0\), the system is dependent — drop that row and introduce a parameter \(t\). In engineering work, dependent systems usually mean one of your "facts" was actually a consequence of the others.
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SECTION 10.3

Matrices and Systems of Linear Equations

Matrix and Dimension

A matrix is a rectangular array of numbers. An \(m \times n\) matrix has \(m\) rows and \(n\) columns; we say it has dimension \(m \times n\). The number \(a_{ij}\) (the entry in row \(i\), column \(j\)) carries a double subscript indicating its position.

Augmented Matrix

The augmented matrix of a linear system records only the coefficients and the constants of the system, with a vertical bar separating the coefficient columns from the constant column. A missing variable corresponds to a \(0\) entry.

Elementary Row Operations

Operations on the rows of an augmented matrix that mirror the equation operations and do not change the solution set: (1) add a multiple of one row to another, (2) multiply a row by a nonzero constant, (3) interchange two rows. Notation: \(R_i + kR_j \to R_i\), \(kR_i\), and \(R_i \leftrightarrow R_j\).

Row-Echelon & Reduced Row-Echelon Form

A matrix is in row-echelon form when: (1) the first nonzero number in each row (the leading entry) is \(1\); (2) each leading entry sits to the right of the one above it; (3) all-zero rows are at the bottom. It is in reduced row-echelon form if, additionally, (4) every entry above and below each leading \(1\) is \(0\).

Gaussian vs. Gauss-Jordan Elimination

Gaussian elimination reduces the augmented matrix to row-echelon form and finishes with back-substitution. Gauss-Jordan elimination continues to reduced row-echelon form, so the solution can be read directly with no back-substitution. (TI-83/84 commands: ref and rref.)

Leading Variables; Inconsistent & Dependent Systems

A leading variable corresponds to a leading entry in the row-echelon form. From the row-echelon form: a row of the form \(0 = c\) (\(c \neq 0\)) means no solution (inconsistent); if every variable is leading, there is exactly one solution; if some variables are non-leading and the system is not inconsistent, there are infinitely many solutions (dependent), expressed with non-leading variables as parameters. If a system in row-echelon form has \(n\) nonzero equations in \(m\) variables (\(m > n\)), the solution has \(m - n\) non-leading variables.

An m × n Matrix
\[ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \]
Augmented Matrix of a System
\[ \begin{cases} 2x - y = 5 \\ x + 4y = 7 \end{cases} \;\longleftrightarrow\; \left[\begin{array}{cc|c} 2 & -1 & 5 \\ 1 & 4 & 7 \end{array}\right] \]
Elementary Row Operation Notation
\[ R_i + kR_j \to R_i, \qquad kR_i, \qquad R_i \leftrightarrow R_j \]
EXAMPLE 1 Finding the Augmented Matrix
Write the augmented matrix of \(\;\begin{cases} 6x - 2y - z = 4 \\ x + 3z = 1 \\ 7y + z = 5 \end{cases}\)
Solution
Line up variables in columns (insert \(0\) for missing variables):
\[ \left[\begin{array}{ccc|c} 6 & -2 & -1 & 4 \\ 1 & 0 & 3 & 1 \\ 0 & 7 & 1 & 5 \end{array}\right] \]
EXAMPLE 2 Using Elementary Row Operations
Solve \(\;\begin{cases} x - y + 3z = 4 \\ x + 2y - 2z = 10 \\ 3x - y + 5z = 14 \end{cases}\)
Solution
Apply \(R_2 - R_1 \to R_2\) and \(R_3 - 3R_1 \to R_3\), then scale and interchange to reach row-echelon form.
Back-substitution gives \( x = 2,\; y = 7,\; z = 3 \).
Solution: \((2,\,7,\,3)\).
EXAMPLE 3 Solving a System Using Row-Echelon Form
Solve \(\;\begin{cases} 4x + 8y - 4z = 4 \\ 3x + 8y + 5z = -11 \\ -2x + y + 12z = -17 \end{cases}\) by Gaussian elimination.
Solution
Form the augmented matrix, scale row 1 by \(\tfrac14\), then clear below to reach row-echelon form.
The corresponding system is \( \begin{cases} x + 2y - z = 1 \\ y + 4z = -7 \\ z = -2 \end{cases} \).
Back-substitute: \( z = -2 \Rightarrow y = 1 \Rightarrow x = 3 \).
Solution: \((3,\,1,\,-2)\).
EXAMPLE 4 Solving a System Using Reduced Row-Echelon Form
Solve the same system as Example 3 using Gauss-Jordan elimination.
Solution
Continue row-reducing from Example 3 until the left block is the identity matrix.
The reduced form reads \( x = 3,\; y = 1,\; z = -2 \) directly — no back-substitution needed.
EXAMPLE 5 A System with No Solution
Solve \(\;\begin{cases} x - 3y + 2z = 12 \\ 2x - 5y + 5z = 14 \\ x - 2y + 3z = 20 \end{cases}\)
Solution
Row-reduce; the last row becomes \( 0\,x + 0\,y + 0\,z = 1 \), i.e. \( 0 = 1 \), which is false.
The system is inconsistent — no solution.
EXAMPLE 6 A System with Infinitely Many Solutions
Find the complete solution of \(\;\begin{cases} -3x - 5y + 36z = 10 \\ -x + 7z = 5 \\ x + y - 10z = -4 \end{cases}\)
Solution
Reduced row-echelon form gives a row of zeros (\(0 = 0\)), so the system reduces to \( x - 7z = -5 \) and \( y + 3z = -1 \).
Let \( z = t \): \( x = 7t - 5,\; y = -3t - 1,\; z = t \).
Solution: \((7t - 5,\, -3t - 1,\, t)\) for any real \(t\).
EXAMPLE 7 Infinitely Many Solutions (Two Parameters)
Find the complete solution of \(\;\begin{cases} x + 2y - 3z - 4w = 10 \\ x + 3y - 3z - 4w = 15 \\ 2x + 2y - 6z - 8w = 10 \end{cases}\)
Solution
Reduced row-echelon form leaves \( x - 3z - 4w = 0 \) and \( y = 5 \), with the third row \(0 = 0\) dropped.
Let \( z = s \) and \( w = t \): \( x = 3s + 4t,\; y = 5,\; z = s,\; w = t \), for any real \(s, t\).
EXAMPLE 8 Nutritional Analysis
A subject must consume exactly 500 mg potassium, 75 g protein, and 1150 units vitamin D daily from three foods (MiniCal, LiquiFast, SlimQuick) whose nutrient content per ounce is given. How many ounces of each?
Solution
Let \(x, y, z\) be ounces of MiniCal, LiquiFast, SlimQuick: \( \begin{cases} 50x + 75y + 10z = 500 \\ 5x + 10y + 3z = 75 \\ 90x + 100y + 50z = 1150 \end{cases} \).
Reduce (e.g. with rref): \( x = 5,\; y = 2,\; z = 10 \).
Answer: 5 oz of MiniCal, 2 oz of LiquiFast, and 10 oz of SlimQuick.
A linear system and its augmented matrix carry the same information; row operations on the matrix mirror equation operations on the system.
Figure 10.3.1
ADMIN NOTE
Matrices are just bookkeeping for the same elimination you did in 10.2 — you stop rewriting the variable letters every line. For hand work I prefer Gaussian (row-echelon + back-substitution) because it needs fewer operations; for a calculator or code, Gauss-Jordan (rref) is cleaner since the answer falls out directly. Watch the bottom row: \(0=c\,(c\neq0)\) ⇒ no solution; \(0=0\) ⇒ infinitely many. This same row-reduction is the workhorse behind solving the large stiffness/circuit systems you'll meet in mechanics and electrical courses.
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SECTION 10.4

The Algebra of Matrices

Equality of Matrices

Two matrices are equal when they have the same dimension and every pair of corresponding entries is equal.

Addition, Subtraction, and Scalar Multiplication

Matrices of the same dimension are added or subtracted entry by entry (otherwise the operation is undefined). To multiply a matrix by a number (a scalar), multiply every entry by that number — the scalar product.

Matrix Multiplication

The product \(AB\) is defined only when the number of columns of \(A\) equals the number of rows of \(B\): if \(A\) is \(m \times n\) and \(B\) is \(n \times k\), then \(AB\) is \(m \times k\). The entry \(c_{ij}\) of \(AB\) is the inner product of the \(i\)th row of \(A\) with the \(j\)th column of \(B\). Matrix multiplication is not commutative: in general \(AB \neq BA\).

Stochastic Matrix

A matrix whose columns each sum to \(1\). Stochastic matrices arise frequently in statistics and demographic modeling, where each column distributes proportions that must total \(100\%\).

Computer Graphics (Application)

Digital images are stored as matrices: each pixel gets a value (e.g. \(0\)–\(255\) on a gray scale). Operations on the matrix transform the image — adding a constant darkens it, subtracting lightens it, and raising the dark values while lowering the light ones increases contrast.

Sum, Difference, and Scalar Product
\[ A + B = [\,a_{ij} + b_{ij}\,], \qquad A - B = [\,a_{ij} - b_{ij}\,], \qquad cA = [\,c\,a_{ij}\,] \]
Inner Product of a Row and a Column
\[ \begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} = a_1 b_1 + a_2 b_2 + \cdots + a_n b_n \]
Matrix Product
\[ C = AB, \qquad c_{ij} = \sum_{r=1}^{n} a_{ir}\,b_{rj} \quad (\text{inner product of row } i \text{ of } A \text{ and column } j \text{ of } B) \]
Properties of Addition and Scalar Multiplication
\[ A + B = B + A, \quad (A+B)+C = A+(B+C) \] \[ c(dA) = (cd)A, \quad (c+d)A = cA + dA, \quad c(A+B) = cA + cB \]
Properties of Matrix Multiplication
\[ A(BC) = (AB)C, \quad A(B+C) = AB + AC, \quad (B+C)A = BA + CA \]
EXAMPLE 1 Equal Matrices
Find \(a, b, c, d\) if \(\;\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 5 & 2 \end{bmatrix}\)
Solution
Corresponding entries must match: \( a = 1,\; b = 3,\; c = 5,\; d = 2 \).
EXAMPLE 2 Algebraic Operations on Matrices
With \(A = \begin{bmatrix} 2 & -3 \\ 0 & 5 \\ 7 & -\tfrac12 \end{bmatrix}\), \(B = \begin{bmatrix} 1 & 0 \\ -3 & 1 \\ 2 & 2 \end{bmatrix}\), \(C = \begin{bmatrix} 7 & -3 & 0 \\ 0 & 1 & 5 \end{bmatrix}\), \(D = \begin{bmatrix} 6 & 0 & -6 \\ 8 & 1 & 9 \end{bmatrix}\), carry out each operation or explain why it cannot be done: (a) \(A + B\); (b) \(C - D\); (c) \(C + A\); (d) \(5A\).
Solution
(a) \(A + B = \begin{bmatrix} 3 & -3 \\ -3 & 6 \\ 9 & \tfrac32 \end{bmatrix}\) (add entrywise).
(b) \(C - D = \begin{bmatrix} 1 & -3 & 6 \\ -8 & 0 & -4 \end{bmatrix}\).
(c) \(C + A\) is undefined — different dimensions (\(2 \times 3\) vs \(3 \times 2\)).
(d) \(5A = \begin{bmatrix} 10 & -15 \\ 0 & 25 \\ 35 & -\tfrac52 \end{bmatrix}\).
EXAMPLE 3 Solving a Matrix Equation
Solve \(2X - A = B\) for \(X\), where \(A = \begin{bmatrix} 2 & 3 \\ -5 & 1 \end{bmatrix}\), \(B = \begin{bmatrix} 4 & -1 \\ 1 & 3 \end{bmatrix}\).
Solution
\( 2X = B + A \Rightarrow X = \tfrac12(B + A) \).
\( B + A = \begin{bmatrix} 6 & 2 \\ -4 & 4 \end{bmatrix} \Rightarrow X = \begin{bmatrix} 3 & 1 \\ -2 & 2 \end{bmatrix} \).
EXAMPLE 4 Multiplying Matrices
With \(A = \begin{bmatrix} 1 & 3 \\ -1 & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} -1 & 5 & 2 \\ 0 & 4 & 7 \end{bmatrix}\), compute \(AB\) and \(BA\) if possible.
Solution
\(A\) is \(2\times2\), \(B\) is \(2\times3\), so \(AB\) is defined (\(2\times3\)): \( AB = \begin{bmatrix} -1 & 17 & 23 \\ 1 & -5 & -2 \end{bmatrix} \).
\(BA\) is not defined: \(B\) is \(2\times3\) and \(A\) is \(2\times2\) — inner dimensions \(3\) and \(2\) don't match.
EXAMPLE 5 Matrix Multiplication Is Not Commutative
With \(A = \begin{bmatrix} 5 & 7 \\ -3 & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 2 \\ 9 & -1 \end{bmatrix}\), compute \(AB\) and \(BA\).
Solution
\( AB = \begin{bmatrix} 68 & 3 \\ -3 & -6 \end{bmatrix} \), \( BA = \begin{bmatrix} -1 & 7 \\ 48 & 63 \end{bmatrix} \).
\( AB \neq BA \) — they don't even share an entry, confirming multiplication is not commutative.
EXAMPLE 6 Writing a Linear System as a Matrix Equation
Show that the matrix equation below is equivalent to the system from §10.3 Example 2.
Solution
\[ \begin{bmatrix} 1 & -1 & 3 \\ 1 & 2 & -2 \\ 3 & -1 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ 10 \\ 14 \end{bmatrix} \]
Performing the multiplication and equating entries reproduces \( \begin{cases} x - y + 3z = 4 \\ x + 2y - 2z = 10 \\ 3x - y + 5z = 14 \end{cases} \).
EXAMPLE 7 Representing Demographic Data by Matrices
A matrix \(A\) gives the proportion of voters in each age group registered to each party; matrix \(B\) gives voters by age and sex. (a) Compute \(AB\). (b) How many males are registered Democrats? (c) How many females are registered Republicans?
Solution
(a) \( AB = \begin{bmatrix} 13{,}500 & 16{,}500 \\ 9{,}000 & 10{,}950 \\ 4{,}500 & 5{,}550 \end{bmatrix} \) (rows: Democrat / Republican / Independent; columns: Male / Female).
(b) Entry \(c_{11} = 13{,}500\) — male Democrats.
(c) Entry \(c_{22} = 10{,}950\) — female Republicans. (The columns of \(A\) sum to \(1\), so \(A\) is a stochastic matrix.)
An eight-level gray scale: each cell of an image is matched to a value 0–7 to build a matrix representation.
Figure 10.4.1
Digitizing an image: overlaying a grid, assigning each cell a gray value, storing the values in a matrix, and reconstructing the digital image.
Figure 10.4.2
ADMIN NOTE
The single most common mistake here is dimension matching for products: write the dimensions side by side, \( (m\times n)(n\times k)\) — the two inner numbers must agree, and the outer numbers give the result's size. Remember \(AB\neq BA\) in general; order matters, which is exactly why we'll later be careful about multiplying matrix equations "on the left" by an inverse. The image-as-matrix idea is your first taste of why matrices dominate graphics, robotics, and signal processing.
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SECTION 10.5

Inverses of Matrices and Matrix Equations

Identity Matrix

The identity matrix \(I_n\) is the \(n \times n\) matrix with \(1\)s on the main diagonal and \(0\)s elsewhere. It behaves like the number \(1\): \( A\,I_n = A \) and \( I_n\,B = B \) whenever the products are defined. A square matrix has equal numbers of rows and columns; its main diagonal runs from top-left to bottom-right.

Inverse of a Matrix

For a square \(n \times n\) matrix \(A\), the inverse \(A^{-1}\) (when it exists) satisfies \( AA^{-1} = A^{-1}A = I_n \). It plays the role of a reciprocal. A matrix with no inverse is called singular. For a \(2 \times 2\) matrix, the quantity \(ad - bc\) is the determinant; if it is \(0\), no inverse exists.

Finding the Inverse of an n × n Matrix

Form the \(n \times 2n\) matrix \([\,A \mid I_n\,]\) and use elementary row operations to reduce the left half to the identity. The right half is then transformed automatically into \(A^{-1}\). If a row of zeros appears on the left, \(A\) has no inverse.

Matrix Equations

A linear system \(AX = B\) (with \(A\) the coefficient matrix, \(X\) the variable matrix, \(B\) the constants) is solved by multiplying on the left by \(A^{-1}\): \( X = A^{-1}B \), provided \(A^{-1}\) exists. This is especially efficient when several systems share the same coefficient matrix — \(A^{-1}\) is found once and reused.

Inverse Defining Property
\[ AA^{-1} = A^{-1}A = I_n \]
Inverse of a 2 × 2 Matrix
\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \;\Rightarrow\; A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}, \qquad ad - bc \neq 0 \]
Solving a Matrix Equation
\[ AX = B \;\Longrightarrow\; X = A^{-1}B \]
EXAMPLE 1 Identity Matrices
Show that multiplying a matrix by an identity matrix of suitable size leaves it unchanged.
Solution
\( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 3 & -5 & 6 \\ 1 & -2 & 7 \end{bmatrix} = \begin{bmatrix} 3 & -5 & 6 \\ 1 & -2 & 7 \end{bmatrix} \). The identity acts like the number \(1\).
EXAMPLE 2 Verifying That a Matrix Is an Inverse
Verify that \(B = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}\) is the inverse of \(A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}\).
Solution
Compute \( AB = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and \( BA = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). Both equal \(I_2\), so \(B = A^{-1}\).
EXAMPLE 3 Inverse of a 2 × 2 Matrix
Find \(A^{-1}\) for \(A = \begin{bmatrix} 4 & 5 \\ 2 & 3 \end{bmatrix}\).
Solution
\( ad - bc = 4\cdot3 - 5\cdot2 = 2 \).
\( A^{-1} = \tfrac12\begin{bmatrix} 3 & -5 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} \tfrac32 & -\tfrac52 \\ -1 & 2 \end{bmatrix} \).
EXAMPLE 4 Inverse of a 3 × 3 Matrix
Find \(A^{-1}\) for \(A = \begin{bmatrix} 1 & -2 & -4 \\ 2 & -3 & -6 \\ -3 & 6 & 15 \end{bmatrix}\).
Solution
Augment with \(I_3\) and row-reduce the left half to the identity. The right half becomes:
\[ A^{-1} = \begin{bmatrix} 3 & -2 & 0 \\ 4 & -1 & -\tfrac23 \\ 1 & 0 & \tfrac13 \end{bmatrix} \]
Verify \( AA^{-1} = A^{-1}A = I_3 \).
EXAMPLE 5 A Matrix That Does Not Have an Inverse
Find the inverse of \(\begin{bmatrix} 2 & -3 & -7 \\ 1 & 2 & 7 \\ 1 & 1 & 4 \end{bmatrix}\).
Solution
Row-reduction produces a row whose left portion is all zeros, so the left half cannot become the identity.
The matrix has no inverse — it is singular.
EXAMPLE 6 Solving a System Using a Matrix Inverse
Solve \(\;\begin{cases} -2x - 5y = 15 \\ 3x + 6y = 36 \end{cases}\) using a matrix inverse.
Solution
Write as \( AX = B \) with \( A = \begin{bmatrix} -2 & -5 \\ 3 & 6 \end{bmatrix},\; B = \begin{bmatrix} 15 \\ 36 \end{bmatrix} \).
\( \det A = (-2)(6) - (-5)(3) = 3 \Rightarrow A^{-1} = \tfrac13\begin{bmatrix} 6 & 5 \\ -3 & -2 \end{bmatrix} \).
\( X = A^{-1}B = \begin{bmatrix} 30 \\ -9 \end{bmatrix} \Rightarrow x = 30,\; y = -9 \).
EXAMPLE 7 Modeling Nutritional Requirements
A pet-store owner mixes three foods to meet daily protein/fat/carbohydrate targets for hamsters (340, 280, 440 mg) and gerbils (480, 360, 680 mg). The nutrient matrix \(A\) is the same for both, so solve \(AX = B\) and \(AY = C\).
Solution
With \( A = \begin{bmatrix} 10 & 0 & 20 \\ 10 & 20 & 10 \\ 5 & 10 & 30 \end{bmatrix} \), compute \(A^{-1}\) once, then \( X = A^{-1}B \) and \( Y = A^{-1}C \).
\( X = \begin{bmatrix} 10 \\ 3 \\ 12 \end{bmatrix},\quad Y = \begin{bmatrix} 8 \\ 4 \\ 20 \end{bmatrix} \).
Answer: each hamster gets 10 g / 3 g / 12 g of the three foods; each gerbil gets 8 g / 4 g / 20 g.
The main diagonal of a square matrix; the identity matrix carries 1s here and 0s elsewhere.
Figure 10.5.1
ADMIN NOTE
The \(2\times2\) inverse formula is worth memorizing: swap \(a\) and \(d\), negate \(b\) and \(c\), divide by \(ad-bc\). If \(ad-bc=0\), stop — the matrix is singular and the system has no unique solution. The big payoff of \(X=A^{-1}B\) is reuse: when you must solve the same physical structure under many different load vectors \(B\), invert \(A\) once and just re-multiply. (In production code we factor rather than literally invert, but the idea is the same.)
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SECTION 10.6

Determinants and Cramer's Rule

Determinant

A single number assigned to every square matrix, written \(\det(A)\) or \(|A|\). For a \(1\times1\) matrix \([a]\), \(\det = a\). The determinant is central to deciding invertibility and to Cramer's Rule.

Minors and Cofactors

The minor \(M_{ij}\) of entry \(a_{ij}\) is the determinant of the matrix left after deleting row \(i\) and column \(j\). The cofactor is \( A_{ij} = (-1)^{i+j}M_{ij} \) — the minor with a sign from the checkerboard pattern \(\begin{smallmatrix}+&-&+\\-&+&-\\+&-&+\end{smallmatrix}\).

Expanding a Determinant

The determinant of an \(n\times n\) matrix is obtained by multiplying each entry of the first row by its cofactor and summing. In fact you may expand along any row or column and get the same value — choose one with many zeros to save work.

Invertibility Criterion

A square matrix \(A\) has an inverse if and only if \(\det(A) \neq 0\). This is the reason for the name "determinant" — it determines whether the inverse exists.

Row and Column Transformations

Adding a multiple of one row to another row (or one column to another column) does not change the determinant. Use this to introduce zeros before expanding, greatly reducing the arithmetic.

Cramer's Rule

A method that expresses the solution of a square linear system \(DX = B\) (with \(\det(D) \neq 0\)) as ratios of determinants. Convenient for small systems but impractical for large ones, and inapplicable when \(\det(D) = 0\).

Determinant of a 2 × 2 Matrix
\[ \det\!\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \]
Cofactor
\[ A_{ij} = (-1)^{i+j} M_{ij} \]
Determinant by First-Row Expansion
\[ \det(A) = a_{11}A_{11} + a_{12}A_{12} + \cdots + a_{1n}A_{1n} \]
Cramer's Rule (Two Variables)
For \(\begin{cases} ax + by = r \\ cx + dy = s \end{cases}\): \[ x = \frac{\begin{vmatrix} r & b \\ s & d \end{vmatrix}}{\begin{vmatrix} a & b \\ c & d \end{vmatrix}}, \qquad y = \frac{\begin{vmatrix} a & r \\ c & s \end{vmatrix}}{\begin{vmatrix} a & b \\ c & d \end{vmatrix}}, \qquad \begin{vmatrix} a & b \\ c & d \end{vmatrix} \neq 0 \]
Cramer's Rule (n Variables)
\[ x_i = \frac{|D_{x_i}|}{|D|}, \qquad i = 1, 2, \ldots, n, \qquad |D| \neq 0 \]

where \(D_{x_i}\) is \(D\) with its \(i\)th column replaced by the constants \(B\).

Area of a Triangle
For vertices \((a_1,b_1),\,(a_2,b_2),\,(a_3,b_3)\): \[ \mathcal{A} = \pm\,\frac{1}{2}\begin{vmatrix} a_1 & b_1 & 1 \\ a_2 & b_2 & 1 \\ a_3 & b_3 & 1 \end{vmatrix} \]

choosing the sign that makes the area positive.

EXAMPLE 1 Determinant of a 2 × 2 Matrix
Evaluate \(|A|\) for \(A = \begin{bmatrix} 6 & -3 \\ 2 & 3 \end{bmatrix}\).
Solution
\( |A| = 6\cdot3 - (-3)\cdot2 = 18 + 6 = 24 \).
EXAMPLE 2 Determinant of a 3 × 3 Matrix
Evaluate \(\det(A)\) for \(A = \begin{bmatrix} 2 & 3 & -1 \\ 0 & 2 & 4 \\ -2 & 5 & 6 \end{bmatrix}\).
Solution
Expand by the first row: \( 2\begin{vmatrix} 2 & 4 \\ 5 & 6 \end{vmatrix} - 3\begin{vmatrix} 0 & 4 \\ -2 & 6 \end{vmatrix} + (-1)\begin{vmatrix} 0 & 2 \\ -2 & 5 \end{vmatrix} \).
\( = 2(-8) - 3(8) - 1(4) = -16 - 24 - 4 = -44 \).
EXAMPLE 3 Expanding About a Row and a Column
For the matrix of Example 2, evaluate the determinant by (a) the second row, (b) the third column, and verify both give \(-44\).
Solution
(a) Second row: \( 0 - 2\begin{vmatrix} 2 & -1 \\ -2 & 6 \end{vmatrix} + 4\begin{vmatrix} 2 & 3 \\ -2 & 5 \end{vmatrix} = 0 - 20 - 64 \cdot(\ldots) = -44 \).
(b) Third column also yields \(-44\). Expansion along any row or column gives the same value.
EXAMPLE 4 Using the Determinant to Show a Matrix Is Not Invertible
Show that \(A = \begin{bmatrix} 1 & 2 & 0 & 4 \\ 0 & 0 & 0 & 3 \\ -5 & 6 & 2 & -6 \\ -2 & 4 & 0 & 9 \end{bmatrix}\) has no inverse.
Solution
Expand by the second row (mostly zeros); only the cofactor \(A_{24}\) survives, and expanding that \(3\times3\) along its third column gives \(0\).
\( \det(A) = 0 \), so by the Invertibility Criterion \(A\) has no inverse.
EXAMPLE 5 Row and Column Transformations
Find \(\det(A)\) for \(A = \begin{bmatrix} 8 & 2 & 1 & 4 \\ 3 & 5 & -3 & 11 \\ 24 & 6 & 1 & -12 \\ 2 & 2 & 7 & -1 \end{bmatrix}\); does it have an inverse?
Solution
Add \(-3\) times row 1 to row 3 to create zeros, then expand; introduce more zeros with a column transformation.
The work reduces to \( \det(A) = -4(-25)\bigl[2(-1) - (-4)(2)\bigr] = 600 \).
Since \( \det(A) = 600 \neq 0 \), \(A\) does have an inverse.
EXAMPLE 6 Cramer's Rule with Two Variables
Solve \(\;\begin{cases} 2x + 6y = -1 \\ x + 8y = 2 \end{cases}\) using Cramer's Rule.
Solution
\( |D| = \begin{vmatrix} 2 & 6 \\ 1 & 8 \end{vmatrix} = 10,\;\; |D_x| = \begin{vmatrix} -1 & 6 \\ 2 & 8 \end{vmatrix} = -20,\;\; |D_y| = \begin{vmatrix} 2 & -1 \\ 1 & 2 \end{vmatrix} = 5 \).
\( x = \dfrac{-20}{10} = -2,\quad y = \dfrac{5}{10} = \tfrac12 \).
EXAMPLE 7 Cramer's Rule with Three Variables
Solve \(\;\begin{cases} 2x - 3y + 4z = 1 \\ x + 6z = 0 \\ 3x - 2y = 5 \end{cases}\) using Cramer's Rule.
Solution
\( |D| = 38,\; |D_x| = -78,\; |D_y| = -22,\; |D_z| = 13 \) (replace each column of \(D\) by the constants in turn).
\( x = \dfrac{-78}{38} = -\dfrac{39}{19},\quad y = \dfrac{-22}{38} = -\dfrac{11}{19},\quad z = \dfrac{13}{38} \).
EXAMPLE 8 Area of a Triangle
Find the area of the triangle with vertices \((-1, 2),\,(3, 6),\,(1, 4)\).
Solution
\( \mathcal{A} = \pm\tfrac12\begin{vmatrix} -1 & 2 & 1 \\ 3 & 6 & 1 \\ 1 & 4 & 1 \end{vmatrix} = \pm\tfrac12(-12) \).
Choose the sign that makes the area positive: \( \mathcal{A} = 6 \).
Evaluating a 2×2 determinant: subtract the up-diagonal product from the down-diagonal product, \(ad - bc\).
Figure 10.6.1
A triangle whose area is computed from the determinant of its vertex coordinates.
Figure 10.6.2
ADMIN NOTE
Two practical habits: (1) always expand along the row or column with the most zeros — it can turn a brutal \(4\times4\) into a single \(3\times3\); (2) use row/column transformations to manufacture those zeros first, since they don't change the determinant. Treat Cramer's Rule as a neat closed-form tool for \(2\times2\) and \(3\times3\) hand problems only; for anything bigger, Gaussian elimination or a solver is far faster. And keep \(\det A \neq 0 \Leftrightarrow A\) invertible in your back pocket — it answers "does a unique solution exist?" without solving anything.
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SECTION 10.7

Partial Fractions

Partial Fraction Decomposition

The reverse of combining fractions over a common denominator: a rational function \( r(x) = \dfrac{P(x)}{Q(x)} \) (with \(\deg P < \deg Q\)) is written as a sum of simpler fractions called partial fractions. Every polynomial with real coefficients factors completely into linear factors \(ax + b\) and irreducible quadratic factors \(ax^2 + bx + c\), and each factor contributes a term to the decomposition.

Case 1 — Distinct Linear Factors

When \(Q(x)\) is a product of distinct linear factors, each factor \(a_i x + b_i\) contributes one term \(\dfrac{A_i}{a_i x + b_i}\).

Case 2 — Repeated Linear Factors

When a linear factor \(ax + b\) is repeated \(k\) times, it contributes \(k\) terms, one for each power up to \(k\): \(\dfrac{A_1}{ax+b} + \dfrac{A_2}{(ax+b)^2} + \cdots + \dfrac{A_k}{(ax+b)^k}\).

Case 3 — Distinct Irreducible Quadratic Factors

Each irreducible quadratic factor \(ax^2 + bx + c\) contributes a term with a linear numerator: \(\dfrac{Ax + B}{ax^2 + bx + c}\).

Case 4 — Repeated Irreducible Quadratic Factors

An irreducible quadratic raised to the \(k\)th power contributes \(k\) terms, one for each power, each with a linear numerator.

Improper Rational Functions

These techniques require \(\deg P < \deg Q\). If the numerator's degree is not smaller, first use long division to write \(r(x)\) as a polynomial plus a proper remainder fraction, then decompose the remainder.

Case 1 — Distinct Linear Factors
\[ \frac{P(x)}{(a_1x+b_1)(a_2x+b_2)\cdots(a_nx+b_n)} = \frac{A_1}{a_1x+b_1} + \frac{A_2}{a_2x+b_2} + \cdots + \frac{A_n}{a_nx+b_n} \]
Case 2 — Repeated Linear Factor \((ax+b)^k\)
\[ \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \cdots + \frac{A_k}{(ax+b)^k} \]
Case 3 — Irreducible Quadratic Factor
\[ \frac{Ax + B}{ax^2 + bx + c} \]
Case 4 — Repeated Irreducible Quadratic \((ax^2+bx+c)^k\)
\[ \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + \cdots + \frac{A_kx+B_k}{(ax^2+bx+c)^k} \]
EXAMPLE 1 Distinct Linear Factors
Find the partial fraction decomposition of \(\dfrac{5x + 7}{x^3 + 2x^2 - x - 2}\).
Solution
Factor the denominator: \( x^3 + 2x^2 - x - 2 = (x-1)(x+1)(x+2) \).
Set \( \dfrac{5x+7}{(x-1)(x+1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{x+2} \), clear denominators, and equate coefficients to get \( \begin{cases} A + B + C = 0 \\ 3A + B = 5 \\ 2A - 2B - C = 7 \end{cases} \).
Solve: \( A = 2,\; B = -1,\; C = -1 \). Thus \( \dfrac{5x+7}{x^3+2x^2-x-2} = \dfrac{2}{x-1} - \dfrac{1}{x+1} - \dfrac{1}{x+2} \).
EXAMPLE 2 Repeated Linear Factors
Find the partial fraction decomposition of \(\dfrac{x^2 + 1}{x(x-1)^3}\).
Solution
Set \( \dfrac{x^2+1}{x(x-1)^3} = \dfrac{A}{x} + \dfrac{B}{x-1} + \dfrac{C}{(x-1)^2} + \dfrac{D}{(x-1)^3} \).
Clearing and equating coefficients gives \( A = 1,\; B = -1,\; C = 0,\; D = 2 \).
Thus \( \dfrac{x^2+1}{x(x-1)^3} = \dfrac{1}{x} - \dfrac{1}{x-1} + \dfrac{2}{(x-1)^3} \).
EXAMPLE 3 Distinct Quadratic Factors
Find the partial fraction decomposition of \(\dfrac{2x^2 - x + 4}{x^3 + 4x}\).
Solution
Factor: \( x^3 + 4x = x(x^2 + 4) \), so set \( \dfrac{2x^2 - x + 4}{x(x^2+4)} = \dfrac{A}{x} + \dfrac{Bx + C}{x^2 + 4} \).
Equating coefficients gives \( A = 1,\; B = 1,\; C = -1 \).
Thus \( \dfrac{2x^2 - x + 4}{x^3 + 4x} = \dfrac{1}{x} + \dfrac{x - 1}{x^2 + 4} \).
EXAMPLE 4 Repeated Quadratic Factors (Form Only)
Write the form of the partial fraction decomposition of \(\dfrac{x^5 - 3x^2 + 12x - 1}{x^3(x^2 + x + 1)(x^2 + 2)^3}\).
Solution
\[ \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx+E}{x^2+x+1} + \frac{Fx+G}{x^2+2} + \frac{Hx+I}{(x^2+2)^2} + \frac{Jx+K}{(x^2+2)^3} \]
(Finding all 11 constants would require solving an 11-equation linear system.)
EXAMPLE 5 Long Division Before Decomposition
Find the partial fraction decomposition of \(\dfrac{2x^4 + 4x^3 - 2x^2 + x + 7}{x^3 + 2x^2 - x - 2}\).
Solution
Numerator degree \(\geq\) denominator degree, so divide first: the quotient is \(2x\) with remainder \(5x + 7\), giving \( 2x + \dfrac{5x+7}{x^3 + 2x^2 - x - 2} \).
Decompose the remainder as in Example 1: \( 2x + \dfrac{2}{x-1} - \dfrac{1}{x+1} - \dfrac{1}{x+2} \).
Partial fraction decomposition reverses the process of adding fractions over a common denominator.
Figure 10.7.1
ADMIN NOTE
This section is mostly a setup skill for calculus (integration) and for Laplace transforms in systems/controls — that's why we keep it. The reliable recipe: factor \(Q(x)\) completely, write the correct template for each factor type (linear → constant; quadratic → linear numerator; repeated → one term per power), clear denominators, then either substitute clever \(x\)-values (the roots) or equate coefficients to solve for the unknowns. Never skip the degree check — if the top degree isn't smaller, long-divide first.
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SECTION 10.8

Systems of Nonlinear Equations

Nonlinear Systems

Systems in which the equations are not all linear. The same tools from §10.1 — substitution and elimination — still apply, and the graphical method is often especially useful because the graphs (circles, parabolas, etc.) can intersect in several points.

Graphical Interpretation

The solutions of a nonlinear system are the intersection points of the graphs. Because curves can cross more than once, a nonlinear system may have several solutions — and curves that appear close may not actually intersect, so checking algebraically (or zooming carefully) matters.

Strategy (no single closed formula)

Solve by the same methods as linear systems:

• Substitution — solve one equation for a variable, substitute into the other.
• Elimination — scale and add equations to cancel a term.
• Graphical — graph each curve and locate intersection points.

EXAMPLE 1 Substitution Method
Find all solutions of \(\;\begin{cases} x^2 + y^2 = 100 \\ 3x - y = 10 \end{cases}\)
Solution
Solve Equation 2 for \(y\): \( y = 3x - 10 \).
Substitute: \( x^2 + (3x - 10)^2 = 100 \Rightarrow 10x^2 - 60x = 0 \Rightarrow 10x(x - 6) = 0 \Rightarrow x = 0 \text{ or } x = 6 \).
Back-substitute: \( x = 0 \Rightarrow y = -10 \); \( x = 6 \Rightarrow y = 8 \).
Solutions: \((0,\,-10)\) and \((6,\,8)\) — a circle meeting a line at two points.
EXAMPLE 2 Elimination Method
Find all solutions of \(\;\begin{cases} 3x^2 + 2y = 26 \\ 5x^2 + 7y = 3 \end{cases}\)
Solution
Multiply Equation 1 by \(5\), Equation 2 by \(-3\), and add to eliminate \(x^2\): \( -11y = 121 \Rightarrow y = -11 \).
Back-substitute into \( 3x^2 + 2y = 26 \): \( 3x^2 - 22 = 26 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4 \).
Solutions: \((4,\,-11)\) and \((-4,\,-11)\).
EXAMPLE 3 Graphical Method
Find all solutions of \(\;\begin{cases} x^2 - y = 2 \\ 2x - y = -1 \end{cases}\)
Solution
Solve each for \(y\): \( y = x^2 - 2 \) and \( y = 2x + 1 \); graph both.
The graphs intersect at \((-1,\,-1)\) and \((3,\,7)\).
EXAMPLE 4 Solving a System Graphically (Rounded)
Find all solutions, rounded to one decimal place, of \(\;\begin{cases} x^2 + y^2 = 12 \\ y = 2x^2 - 5x \end{cases}\)
Solution
The first graph is a circle (graph \(y = \pm\sqrt{12 - x^2}\)); the second is a parabola.
Intersections occur near \((-0.559,\,3.419)\) and \((2.847,\,1.974)\); a near-miss in Quadrant IV is not an actual intersection.
Solutions (to one decimal): \((-0.6,\,3.4)\) and \((2.8,\,2.0)\).
A nonlinear system as a circle meeting a line; the two intersection points are the solutions.
Figure 10.8.1
Curves may approach without intersecting — zooming in (or checking algebraically) distinguishes a true solution from a near-miss.
Figure 10.8.2
ADMIN NOTE
When one equation contains only \(x^2\) (or \(y^2\)) terms, elimination across the two equations is usually cleanest — you knock out the squared variable in one move, as in Example 2. After substitution, expect a quadratic (or higher) in one variable, so don't be surprised by two or more solution pairs, and don't drop the \(\pm\) when you take a square root. Finally, trust algebra over eyeballs: a graph can make two curves look tangent when they actually miss, or vice versa.
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SECTION 10.9

Systems of Inequalities

Graph of an Inequality

The graph of an inequality is a region of the plane whose boundary is the graph of the corresponding equation. Use a dashed curve for strict inequalities (\(<\) or \(>\)) and a solid curve for \(\leq\) or \(\geq\). A test point not on the boundary tells you which side to shade: if it satisfies the inequality, shade that side.

System of Inequalities

The solution set of a system of inequalities is the set of all points satisfying every inequality — the intersection of the individual solution regions. Graph each inequality on the same axes, shade the overlap, and label the vertices (corner points), found by solving the relevant boundary equations simultaneously. A vertex is part of the solution only if it satisfies all inequalities (open circle if not).

Linear Inequalities

An inequality is linear if it can be written as \(ax + by \leq c\) (or with \(<, >, \geq\)). Systems of linear inequalities bound regions whose edges are line segments.

Bounded vs. Unbounded Regions

A region is bounded if it can be enclosed in a (large enough) circle; otherwise it is unbounded.

Feasible Region (Application)

In applied problems the variables face constraints (limited labor, materials, pollution limits). Expressed as a system of inequalities, the solution set is called the feasible region: its points represent the possible (feasible) combinations of the quantities being studied.

Forms of a Linear Inequality
\[ ax + by \leq c, \quad ax + by \geq c, \quad ax + by < c, \quad ax + by > c \]
Graphing an Inequality (procedure)

1. Graph the boundary equation (dashed for \(<,>\); solid for \(\leq,\geq\)).
2. Test a point off the boundary; shade the side that satisfies the inequality.

EXAMPLE 1 Graphs of Inequalities
Graph each inequality: (a) \(x^2 + y^2 < 25\); (b) \(x + 2y \geq 5\).
Solution
(a) Boundary is the circle of radius \(5\) (dashed, since strict). Test \((0,0)\): \(0 < 25\) ✓ — shade inside the circle.
(b) Boundary is the line \(x + 2y = 5\) (solid). Test \((0,0)\): \(0 \geq 5\) ✗ — shade the side above the line (containing, e.g., \((5,5)\)).
EXAMPLE 2 A System of Two Inequalities
Graph the solution of \(\;\begin{cases} x^2 + y^2 < 25 \\ x + 2y \geq 5 \end{cases}\) and label its vertices.
Solution
The solution is where the inside of the circle overlaps the region above the line.
Vertices (intersection of the two boundaries) found by substitution \(x = 5 - 2y\): \( 5y(4 - y) = 0 \Rightarrow y = 0 \text{ or } 4 \), giving \((5, 0)\) and \((-3, 4)\).
The vertices lie on the dashed circle, so they are not part of the solution set (open circles).
EXAMPLE 3 A System of Four Linear Inequalities
Graph the solution of \(\;\begin{cases} x + 3y \leq 12 \\ x + y \leq 8 \\ x \geq 0 \\ y \geq 0 \end{cases}\) and label its vertices.
Solution
Each inequality is satisfied below/left of its line (test \((0,0)\)); \(x \geq 0,\, y \geq 0\) restrict to the first quadrant.
Vertices: \((0,0)\), \((8,0)\), \((0,4)\), and the intersection of \(x + 3y = 12\) with \(x + y = 8\), which is \((6, 2)\).
All four vertices are part of the (bounded) solution set.
EXAMPLE 4 A System of Linear Inequalities
Graph the solution of \(\;\begin{cases} 10x + 20y \geq 60 \\ 30x + 20y \geq 100 \\ 10x + 40y \geq 80 \\ x \geq 0,\; y \geq 0 \end{cases}\) (and the variant with the first inequality reversed).
Solution
Rewrite as \( y \geq 3 - \tfrac12 x,\; y \geq 5 - \tfrac32 x,\; y \geq 2 - \tfrac14 x \) in the first quadrant; shade the common region.
The resulting region is unbounded with vertices such as \((0,5),\,(2,2),\,(4,1),\,(8,0)\). Reversing the first inequality (\(\leq\)) yields a different, bounded overlap.
EXAMPLE 5 A System with a Triangular Solution
Graph the solution of \(\;\begin{cases} x + 2y \leq 8 \\ x - 2y \leq 4 \\ -3x + 2y \leq 8 \end{cases}\) and label the vertices.
Solution
Rewrite each as \( y \leq -\tfrac12 x + 4,\; y \geq \tfrac12 x - 2,\; y \leq \tfrac32 x + 4 \); the overlap is a triangle.
Vertices (from intersecting pairs of boundary lines): \((2, 3)\), \((4, 2)\), and \((-... )\) — the labeled corners \((2,3),\,(4,2),\,(6,5)\) of the bounded triangular region.
EXAMPLE 6 Restricting Pollutant Outputs (Feasible Region)
A factory emits, per barrel: pesticide A — 0.25 kg CO and 0.60 kg SO₂; pesticide B — 0.50 kg CO and 0.20 kg SO₂. Limits: 75 kg CO and 90 kg SO₂ per day. (a) Write and graph the feasible region. (b) Is 100 barrels of A and 80 of B legal? (c) Is 60 barrels of A and 160 of B legal?
Solution
Let \(x, y\) be barrels of A, B. Constraints: \( \begin{cases} 0.25x + 0.50y \leq 75 \\ 0.60x + 0.20y \leq 90 \\ x \geq 0,\; y \geq 0 \end{cases} \), which simplify to \( x + 2y \leq 300 \) and \( 3x + y \leq 450 \).
(b) \((100, 80)\) lies inside the feasible region, so this plan is legal.
(c) Test \((60, 160)\) in the constraints to decide legality (check both pollution limits).
Solid boundary curves for \(\geq,\leq\); dashed boundary curves for \(>,<\). Shading shows which side satisfies the inequality.
Figure 10.9.1
A bounded region fits inside a circle; an unbounded region does not. Corner points are the vertices of the solution set.
Figure 10.9.2
ADMIN NOTE
Two habits keep this clean: dashed vs. solid (strict vs. inclusive) and a single test point — \((0,0)\) is ideal whenever the boundary doesn't pass through the origin. The vertices are the payoff: in the next section's linear programming, the optimum always sits at a vertex, so finding and labeling them here is the whole point. Note that a vertex on a dashed boundary is a "corner" of the region but not actually in the solution set.
--- ## Focus on Modeling: Linear Programming

Linear Programming

A modeling technique for finding the optimal allocation of resources — maximizing profit or minimizing cost subject to constraints. It is the most important practical application of systems of linear inequalities. (Kantorovich and Koopmans won the 1975 Nobel Prize in economics for developing it.)

Objective Function, Constraints, Feasible Region

The objective function is the linear expression to be maximized or minimized. The constraints are inequalities limiting the variables; their solution set is the feasible region. The key fact: the objective function attains its maximum and minimum at a vertex of the feasible region — so only the vertices need to be checked.

Guidelines for Linear Programming

(1) Choose the variables — name the quantities \(x\) and \(y\). (2) Find the objective function — write the expression to optimize. (3) Graph the feasible region — express constraints as inequalities and graph their solution. (4) Find the maximum or minimum — evaluate the objective function at each vertex.

Objective Function (general form)
\[ P = ax + by \quad (\text{maximize or minimize}) \]
Optimization Principle

A linear objective function attains its maximum and minimum values at the vertices of the feasible region — so evaluate it only at the corner points.

EXAMPLE 1 Manufacturing for Maximum Profit
A shoemaker makes oxfords and loafers. Cutting takes 15 min/pair for each; sewing takes 10 min/pair (oxfords) and 20 min/pair (loafers). Each machine runs 8 h/day. Profit is $15/oxford and $20/loafer. How many of each maximize profit?
Solution
Variables: \(x =\) pairs of oxfords, \(y =\) pairs of loafers. Objective: \( P = 15x + 20y \).
Constraints (hours): \( \tfrac14 x + \tfrac14 y \leq 8,\; \tfrac16 x + \tfrac13 y \leq 8,\; x \geq 0,\; y \geq 0 \), simplifying to \( x + y \leq 32 \) and \( x + 2y \leq 48 \).
Vertices of the feasible region: \((0,0),\,(0,24),\,(16,16),\,(32,0)\). Evaluate \(P\): \(0,\,480,\,560,\,480\).
Maximum: at \((16, 16)\), \(P = \$560\) — make 16 pairs of oxfords and 16 pairs of loafers daily.
EXAMPLE 2 A Shipping Problem (Minimum Cost)
Warehouses in Millville (15 cars) and Trenton (10 cars) supply dealerships in Camden (sells 10) and Atlantic City (sells 12). Shipping costs: $50, $40 (from Millville to Camden, A.C.) and $60, $55 (from Trenton). Minimize total shipping cost.
Solution
Variables: \(x =\) cars Millville→Camden, \(y =\) cars Millville→Atlantic City. Then Trenton ships \(10 - x\) and \(12 - y\).
Objective: \( C = 50x + 40y + 60(10 - x) + 55(12 - y) = 1260 - 10x - 15y \).
Constraints: \( x + y \leq 15,\; x + y \geq 12,\; 0 \leq x \leq 10,\; 0 \leq y \leq 12 \). Vertices: \((0,12),\,(3,12),\,(10,5),\,(10,2)\).
Evaluate \(C\): \(1080,\,1050,\,1085,\,1130\). Minimum: at \((3, 12)\), \(C = \$1050\) — ship 3 Millville→Camden, 12 Millville→A.C., 7 Trenton→Camden, 0 Trenton→A.C.
The feasible region for the shoe-manufacturing problem; the maximum profit occurs at the vertex (16, 16).
Figure FOM.1
Flow diagram for the shipping problem: routes, quantities available/required, and per-car shipping costs.
Figure FOM.2
ADMIN NOTE
This is where everything in the chapter pays off. The whole method reduces to one fact: a linear objective is optimized at a corner of the feasible region, so you never search the interior — just list the vertices and plug them in. Set it up in four disciplined steps (variables → objective → constraints/region → check vertices). Counter-intuitive results are common: in Example 1, loafers are more profitable per pair, yet the optimum is an equal split, because the sewing machine is the bottleneck. Real engineering and operations problems have hundreds of variables and use the matrix-based simplex/interior-point methods, but the geometric intuition you build here is exactly the same.
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