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## Precalculus – MAT 129

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**Precalculus – MAT 129**Instructor: Rachel Graham Location: BETTS Rm. 107 Time: 8 – 11:20 a.m. MWF**Chapter Seven**Linear Systems and Matrices**Ch. 7 Overview**• Solving Systems of Equations • Systems of Linear Equations in Two Variables • Multivariable Linear Systems • Matrices and Systems of Equations**Ch. 7 Overview (cont.)**• Operations with Matrices • The Inverse of a Square Matrix • The Determinant of a Square Matrix • Applications of Matrices and Determinants**7.1 – Solving Systems of Equations**• The Method of Substitution**7.1 – The Method of Substitution**• Systems of equations • thenumber of equations must equal the number of unknown variables • A solution of a system is an ordered pair that satisfies each equation**7.1 – The Method of Substitution**• Solve one of the equations for one variable. • Substitute into the other equation. • Solve the equation you substituted into (it will now only be in one variable) • Back-substitute into the first equation. • Check the solution(s).**Example 1.7.1**Pg. 481 #5 Solve the system by the method of substitution. 2x + y = 6 -x + y = 0**Solution Example 1.7.1**Pg. 481 #5 Following the steps: • Solving for y in the second equation gives y=x. • Substituting in the first equation gives 2x + x = 6 3x = 6 x = 2. • Back substituting gives y = 2. • Check (2,2) in both equations.**Things that can happen**• No-Solution Case • The graphs don’t touch. • Two-Solution Case • The graphs touch in two places.**Example 2.7.1**Solve the system by the method of substitution. 6x + 5y = -3 -x – (5/6)y = -7**Solution Example 2.7.1**Solving for y in the first equation yields: y = -(3/5) – (6/5)x Substituting for y in the second equation: -x – (5/6)(-3/5 – (6/5)x ) ½ = -7. This means that there is no solution.**Example 3.7.1**Pg. 482 #27 Solve the system by the method of substitution. x3 – y = 0 x – y = 0**Solution Example 3.7.1**Pg. 482 #27 Solving for y in the second equation yields: y = x Substituting for y in the first equation: x3 – x = 0 x(x2 – 1) = 0 x = 1, -1, and 0. This means that there are multiple solutions.**Example 4.7.1**Pg. 478 Example 5 Solve the system of equations graphically. y = ln(x) x + y = 1**Solution Example 4.7.1**Looking at the graph of these two equations shows us that the point of intersection is (1,0). Make sure to still check graphical solutions in the equations!**Activities (479)**1. Solve the system by substitution. 3x + 2y = 14 and x – 2y = 10. 2. Find all points of intersection. 4x – y – 5 = 0 and 4x2 – 8x + y + 5 = 0 3. Solve the system graphically. 3x + 2y = 6 and y = ln(x-1)**7.2 – Systems of Linear Equations in Two Variables**• The Method of Elimination • Graphical Interpretation of Two-Variable Systems • Application**7.2 – The Method of Elimination**• Find coefficients that when added will cancel each other out. • Remember you can multiply to make this happen! • Add the equations to eliminate one variable; solve the resulting equation. • Back-substitute into the first equation. • Check your solutions in both original equations.**Example 1.7.2**Pg. 486 Example 2 Solve the system of linear equations. 5x + 3y = 9 2x - 4y = 14**Solution Example 1.7.2**Pg. 486 Example 2 (3, -2)**Example 2.7.2**Pg. 491 #11 Solve the system of linear equations. 3r + 2s = 10 2r + 5s = 3**Solution Example 2.7.2**Pg. 491 #11 Multiplying: -2*(3r + 2s = 10) and 3*(2r + 5s = 3) Adding: 11s = -11 s = -1**Solution Example 2.7.2 (cont.)**Substituting: 3r + 2(-1) = 10 3r = 12 r = 4 Checking r = 4 and s = -1 into both original solutions: 12 – 2 = 10 √ 8 – 5 = 3 √**7.2 – Graphical Interpretation of Two-Variable Systems**There are three kinds of systems that you will encounter: • One solution • Infinitely many solutions (identical lines) • Equations with at least one solution are called consistent. • No solution (parallel lines) • These equations are inconsistent.**Example 3.7.2**Pg. 487 Example 3 The graphs of the three different types of systems.**Example 4.7.2**Pg. 491 #21 Solve the system of linear equations by elimination. 4x + 3y = 3 3x + 11y = 13**Solution Example 4.7.2**Pg. 491 #21 Multiplying: -11*(4x + 3y = 3) and 3*(3x + 11y = 13) Adding: -35x = 6 x = -(6/35)**Solution Example 4.7.2 (cont.)**Substituting: y = 43/35 Checking the solution shows that they work in both equations.**7.3 – Multivariable Linear Systems**• Row-Echelon Form and Back Substitution • Gaussian Elimination • Nonsquare Systems • Graphical Interpretation of Three-Variable Systems • Partial Fraction Decomposition and Other Applications**7.3 – Row-Echelon Form and Back Substitution**Row-echelon form is a “stair-step” pattern and all leading coefficients are 1. See Example 1 on pg. 495**Example 1.7.3**Pg. 505 #5 Use back-substitution to solve the system of linear equations. 2x – y + 5z = 16 y + 2z = 2 z = 2**Solution Example 1.7.3**Pg. 505 #5 Substitute z=6 into the second equation: y + 2(6) = 4 y = -8 Then substitute both z and y into the first: 2x + 8 + 30 = 24 2x = -14 x = -7**Solution Example 1.7.3 (cont.)**Pg. 505 #5 So the ordered triple that is the solution of this system of equations is: (-7, -8, 6) Check these solutions!**7.3 – Gaussian Elimination**Elementary Row Operations • Interchange two equations. • Multiply one of the equations by a nonzero constant. • Add a multiple of one equation to another equation.**7.3 – Gaussian Elimination**Use elementary row operations to get the equations in row-echelon form.**Example 2.7.3**Pg. 496 Example 2 Read the row operations that were used in the red boxes to the right of the equations.**Example 3.7.3**Pg. 506 #15 Solve the system of equations and check any solution. x + y + z = 6 2x – y + z = 3 3x - z = 0 Do on the board.**Solution Example 3.7.3**Pg. 506 #15 So the ordered triple that is the solution of this system of equations is: (1, 2, 3) Don’t forget to check these solutions.**Example 4.7.3**Pg. 497-498 Examples 3 and 4 Recall these special cases. The same rules apply as when we are in two variables.**7.3 – Nonsquare Systems**In a non-square system of equations, the number of equations differs from the number of variables.**Example 5.7.3**Pg. 499 Example 5 Pay close attention these are pretty hard.**Example 6.7.3**Pg. 506 #37 Solve the system of linear equations and check your solutions! x – 2y + 5z = 2 4x – z = 0**Solution Example 6.7.3**Pg. 506 #37 So the ordered triple that is the solution of this system of equations is: (2a, 21a – 1, 8a) Don’t forget to check these solutions!**7.3 – Graphs in Three Dimensions**Note the pictures of what you are finding for solutions on the bottom of pg. 500.**7.3 – Partial Fractions**• Distinct Linear Factors • Repeated Linear Factors**Example 7.7.3**Pg. 502 Example 6**Example 8.7.3**Now look at Example 7 on pg. 503 Note: When you get double roots you have to include the single factor and the squared factor.**Activities (499 & 502)**1. Solve the system: x – y + z = 4 x +3y – 2z = -3 3x + 2y + z = 5 2. Solve the system: x – 2y – z = -5 2x + y + z = 5 3 and 4 are partial fractions that I will put on the board.**7.4 – Matrices and Systems of Equations**• Matrices • Elementary Row Operations • Gaussian Elimination with Back Substitution • Gauss-Jordan Elimination