Systems of Equations and Inequalities Chapter 7 Systems of Equations and Inequalities Copyright © 2014, 2010 Pearson Education, Inc.

Section 7.2 Systems of Linear Equations in Three Variables Solve a system of linear equations in three variables. Classify systems as consistent and inconsistent. Solve nonsquare systems. Interpret linear systems in three variables geometrically. Use linear systems in applications.

Definitions A linear equation in the variables x1, x2, …, xn is an equation that can be written in the form. where b and the coefficients a1, a2, …, an, are real numbers. The subscript n may be any positive integer.

Definitions A system of linear equations (or a linear system) in three variables is a collection of two or more linear equations involving the same variables. For example, is a system of three linear equations in three variables x, y, and z.

Definitions An ordered triple (a, b, c) is a solution of a system of three equations in three variables x, y, and z if each equation in the system is a true statement when a, b, and c are substituted for x, y, and z respectively.

Example: Verifying a Solution Determine whether the ordered triple (2, –1, 3) is a solution of the given linear system Replace x with 2 and y with –1, and z by 3 in all three equations.

Example: Verifying a Solution (cont) (2, –1, 3) satisfies all three equations, so it is a solution of the system.

INCONSISTENT SYSTEM If in the process of converting a linear system to triangular form, an equation of the form 0 = a occurs, where a ≠ 0, then the system has no solution and is inconsistent.

Example: Attempting to Solve a Linear System with No Solution Solve the system of equations. Steps 1-2 To eliminate x from equation (2), add –2 times equation (1) to equation (2).

Example: Attempting to Solve a Linear System with No Solution (cont) Next add –3 times equation (1) to equation (3) to eliminate x from equation (3). We now have the following system:

Example: Attempting to Solve a Linear System with No Solution (cont) Step 3 Multiply equation (4) by to obtain To eliminate y from equation (5), add –1 times equation (6) to equation (5).

Example: Attempting to Solve a Linear System with No Solution (cont) We now have the system in triangular form: This system is equivalent to the original system. Since equation (7) is false, we conclude that the solution set of the system is and the system is inconsistent.

DEPENDENT EQUATIONS If in the process of converting a linear system to triangular form, an equation of the form 0 = a (a ≠ 0) does not occur, but an equation of the form 0 = 0 does occur, then the system of equations has infinitely many solutions and the equations are dependent.

Example: Solving a System with Infinitely Many Solutions Solve the system of equations. Steps 1-2 Eliminate x from equation (2) by adding –3 times equation (1) to equation (2).

Example: Solving a System with Infinitely Many Solutions (cont) Eliminate x from equation (3) by adding –4 times equation (1) to equation (3). We now have the equivalent system.

Example: Solving a System with Infinitely Many Solutions (cont) Step 3 To eliminate y from equation (5), add –1 times equation (4) to equation (5). We finally have the system in triangular form.

Example: Solving a System with Infinitely Many Solutions (cont) The equation 0 = 0 may be interpreted as 0z = 0, which is true for every value of z. Solving equation (4) for y, we have y = 3z – 2. Substituting into equation (1) and solving for x. Thus every triple (x, y, z) = (2z + 5, 3z – 2, z) is a solution of the system for each value of z.

NONSQUARE SYSTEMS Sometimes in a linear system, the number of equations is not the same as the number of variables. Such systems are called nonsquare linear systems.

Example: Solving a Nonsquare Linear System Solve the system of equations. Steps 1-2 Eliminate x from equation (2) by adding –2 times equation (1) to equation (2).

Example: Solving a Nonsquare Linear System (cont) We now have the equivalent system. Steps 3-4 Since there is no third equation, steps 3 and 4 are not needed. Step 5 Solve equation (3) for y in terms of z to obtain y = 10z – 8.

Example: Solving a Nonsquare Linear System (cont) Step 6 Back-substitute y = 10z – 8 into equation (1) and solve for x. z determines both the x and y values. The system has infinitely many solutions, one for each real number z. The solution set is {(−53z + 47), (10z – 8, z)}.

GEOMETRIC INTERPRETATION The graph of a linear equation in three variables, such as ax + by + cz = d (where a, b, and c are not all zero), is a plane in three-dimensional space. Following are the possible situations for a system of three linear equations in three variables.

GEOMETRIC INTERPRETATION a. Three planes intersect in a single point. The system has only one solution. b. Three planes intersect in one line. The system has infinitely many solutions.

GEOMETRIC INTERPRETATION Three planes coincide with each other. The system has infinitely many solutions. d. There are three parallel planes. The system has no solution.

GEOMETRIC INTERPRETATION e. Two parallel planes are intersected by a third plane. The system has no solution. f. Three planes have no point in common. The system has no solution.

Section 7.4 Determinants and Cramer’s Rule Calculate the determinant of a 2 × 2 matrix. Find minors and cofactors. Evaluate the determinant of an n × n matrix. Apply Cramer’s Rule.

DETERMINANT OF A 2 × 2 MATRIX The determinant of the matrix denoted by det(A), |A| or is called a 2 by 2 determinant and is defined by

Example: Calculating the Determinant of a 2 × 2 Matrix Evaluate each determinant.

Example: Calculating the Determinant of a 2 × 2 Matrix (cont) Evaluate each determinant.

MINORS AND COFACTORS IN AN n × n MATRIX Let A be an n × n square matrix. The minor Mij of the element aij is the determinant of the (n – 1) × (n – 1) matrix obtained by deleting the ith row and the jth column of A.

Example: Finding Minors and Cofactors For the matrix find a. the minors M11, M23, and M32 b. the cofactors A11, A23, and A32 a. (i) To find M11 delete the first row and first column of the matrix A.

Example: Finding Minors and Cofactors (cont) Now find the determinant of the resulting matrix. (ii) To find M23 delete the second row and third column of the matrix A.

Example: Finding Minors and Cofactors (cont) Now find the determinant of the resulting matrix. (iii) To find M32 delete the third row and second column of the matrix A.

Example: Finding Minors and Cofactors (cont) Now find the determinant of the resulting matrix. b. To find the cofactors, use the formula Aij = (−1)i+jMij

n × n DETERMINANT Let A be a square matrix of order n ≥ 3. The determinant of A is the sum of the entries in any row of A (or column of A), multiplied by their respective cofactors.

Example: Evaluating a 3 by 3 Determinant Find the determinant of Expanding by the first row, we have

Example: Evaluating a 3 by 3 Determinant (cont) Thus

CRAMER’S RULE FOR SOLVING TWO EQUATIONS IN TWO VARIABLES The system of two equations in two variables has a unique solution (x, y) given by provided that D ≠ 0, where

Example: Using Cramer’s Rule Use Cramer’s rule to solve the system: Step 1: Form the determinant D of the coefficient matrix. Since D = 7 ≠ 0, the system has a unique solution.

Example: Using Cramer’s Rule(cont) Step 2 Replace the column of coefficients of x in D by the constant terms to obtain Step 3 Replace the column of coefficients of y in D by the constant terms to obtain

Example: Using Cramer’s Rule(cont) Step 4 By Cramer’s rule, The solution of the system is x = –3 and y = 4. The solution set is {(–3, 4)}. Check: You should check the solution in the original system.

Section 7.5 Partial-Fraction Decomposition Become familiar with partial-fraction decomposition. Decompose P(x)/Q(x) when Q(x) has only distinct linear factors. Decompose P(x)/Q(x) when Q(x) has repeated linear factors. Decompose P(x)/Q(x) when Q(x) has distinct irreducible quadratic factors.

PARTIAL FRACTIONS Each of the two fractions on the right is called a partial fraction. Their sum is called the partial-fraction decomposition of the rational expression on the left.

PARTIAL FRACTIONS A rational expression is called improper if the degree P(x) ≥ degree Q(x) and is called proper if the degree P(x) < Q(x). We restrict our discussion of the decomposition of into partial fractions to cases involving proper rational expressions.

CASE 1: THE DENOMINATOR IS THE PRODUCT OF DISTINCT (NONREPEATED) LINEAR FACTORS Suppose Q(x) can be factored as with no factor repeated. The partial-fraction decomposition of is of the form where A1, A2, …, An, are constants to be determined.

Partial-Fraction Decomposition OBJECTIVE Find the partial-fraction decomposition of a rational expression. Step 1 Write the form of the partial-fraction decomposition with the unknown constants A, B, C,… in the numerators of the decomposition. EXAMPLE Find the partial fraction decomposition of 1.

Partial-Fraction Decomposition Step 2 Multiply both sides of the equation in Step 1 by the original denominator. Use the distributive property and eliminate common factors. Simplify. 2. Multiply both sides by (x – 3)(x + 4) and simplify to obtain 3x + 26 = (x + 4)A + (x – 3)B.

Partial-Fraction Decomposition Step 3 Write both sides of the equation in Step 2 in descending powers of x and equate the coefficients of like powers of x. 3x + 26 = (A + B)x + 4A – 3B

Partial-Fraction Decomposition Step 4 Solve the linear system resulting from Step 3 for the constants A, B, C,… Solving the system of equations in Step 3, we obtain A = 5 and B = –2.

Partial-Fraction Decomposition Step 5 Substitute the values you found for A, B, C,… into the equation in Step 1 and write the partial-fraction decomposition.

Example: Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors Find the partial-fraction decomposition of the expression. Factor the denominator. Step 1 Write the partial-fraction.

Example: Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors (cont) Alternative Solution (Multiply both sides by the denominator ) Substitute x = 2 in equation (1) to cause the terms containing A and C to be 0.

Substitute x = –2 in equation (1) to get Example: Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors (cont) Substitute x = –2 in equation (1) to get

Example: Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors (cont) Substitute x = 0 in equation (1) to get Thus, A = 1, B = 2 and C = –3 and the partial-fraction decomposition is given by

CASE 2: THE DENOMINATOR HAS A REPEATED LINEAR FACTOR Let (x – a)m be the linear factor (x – a) that is repeated m times in Q(x). Then the portion of the partial-fraction decomposition of that corresponds to the factor (x – a)m is where A1, A2, …, An, are constants.

Example: Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors Find the partial-fraction decomposition of the expression. Step 1 (x – 1) is repeated twice, (x + 3) is nonrepeating, the partial-fraction decomposition has the form

Example: Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors (cont) Step 2-4 Multiply by original denominator then use the distributive property to get Substitute x = 1 in the last equation to obtain

Example: Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors (cont) Substitute x = –3 to get A

Example: Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors (cont) To obtain the value of B, we replace x with any convenient number, say, 0. We have Now substitute A = 1/16 and C = 5/4

Example: Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors (cont) Substitute in the decomposition in Step 1 to obtain the partial-fraction decomposition. and

Example: Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors (cont) or

CASE 3: THE DENOMINATOR HAS A DISTINCT(NONREPEATED) IRREDUCIBLE QUADRATIC FACTOR Suppose ax2 + b + c is an irreducible quadratic factor of Q(x). Then the portion of that corresponds to ax2 + bx + c has the form the partial-fraction decomposition of

Example: Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor Find the partial-fraction decomposition of Step 1 (x – 4) is linear, (x2 + 1) is irreducible; so, the partial-fraction decomposition has the form

Example: Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor (cont) Step 2 Multiply both sides of the decomposition in Step 1 by the original denominator, and simplify to obtain Substitute x = 4 to obtain 17 = 17A, or A = 1. Step 3 Collect like terms, write both sides of the equation in descending powers of x.

Example: Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor (cont) Equate coefficients of the like powers of x to obtain Step 4 Substitute A = 1 in equation (1) to obtain 1 + B = 3, or B = 2. Substitute A = 1 in equation (3) to obtain 1 – 4C = 1, or C = 0.

Example: Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor (cont) Step 5 Substitute A = 1, B = 2, and C = 0 into the decomposition in Step 1 to get: