Linear Systems Two or more unknown quantities that are related to each other can be described by a “system of equations”. If the equations are linear, we have a linear system. A solution to the system is a set of values for these quantities that makes all of the equations true. We will solve linear systems using graphical and algebraic methods.

Graphical Methods Example: Solve the system: -3x + 2y = 8 (equation 1) Graph the two equations. Their point of intersection is the solution. Check the solution.

Algebraic Methods Example: Solve the system: -3x + 2y = 8 (equation 1) Solve for y in one equation in terms of x. Substitute your expression for y (from step 1) into the other equation and solve for x. Knowing the value of x, substitute that value into either equation and solve for y.

More Algebraic Methods Example: Solve the system: -3x + 2y = 8 (equation 1) x + 2y = -8 (equation 2) Often we can solve a system faster by forming an “equivalent system” (having the same solution) and eliminating one of the variables by adding the equations. Try multiplying both sides of equation 2 by -1 and adding the two equations.

Classifying a system of Linear Equations. Consistent System: System has at least one solution. Linearly Independent system: Has one solution. Consists of the equations of two lines that intersect at one point. Linearly dependent system: Has infinite number of solutions. The equations describe lines that coincide. Inconsistent System: System has NO solutions. The equations describe lines that do not intersect; that is, they are parallel.

How can we tell if a system is independent, dependent, or inconsistent? Put each line in slope-intercept form If slopes and intercepts are the same, the lines are the same (linearly dependent). If slope is the same, but intercepts are different, the lines are parallel and have no solution (inconsistent). If slopes are different, lines are linearly independent and have one solution.

Classify the following systems of equations. 4y - 2x = 6, 8y = 4x - 12 -3x + y = 4, x - 1 3 y=1 2x + 3y = 1, 4x + y = -3 y = 2x – 3, 6x – 3y = 9

How can we solve a system of inequalities by graphing? Solve: 2x – y ≥ -3, y ≥ - 1 2 x +1 (Hint: First put both inequalities in slope-intercept form) Solve: x + 2y ≤ 4, y ≥ -x – 1 Page 153, Problem 21 Page 153, Problem 31 Page 154, Problem 48 Page 154, Problem 51 Homework 12.

Linear Programming A method for optimizing some quantity (say costs or profits) subject to given constraints. Objective function: The quantity we are trying to optimize. (ex: minimize costs, maximize profits) Constraints: Restrictions on quantities forming a set of inequalities. Feasible Region: The region in space that satisfies all of the constraints. Vertex Principle: If there is a maximum or minimum value of the objective function, it occurs at one or more vertices of the feasible region.

EXAMPLE 1: You want to spend no more than $40 for at most 15 tomato plants. You want to maximize the number of pounds of tomatoes you’ll get. Roma plants will yield 8 lbs per plant and cost $2 each. Cherry plants will yield 10 lbs per plant and cost $3 each. How many of each plant should you buy? Write the objective function as an equation. Write the constraints as inequalities. Graph the constraints and shade in the feasible region. Test the vertices of the feasible region for an optimal value of the objective function.

Example 2 You’re making T-Shirts and Sweatshirts to sell at a festival. You have only 20 hours to do the job. You can only spend up to $600 for supplies. You want to have at least 50 items to sell. Also: A T-shirt costs $4 and 10 minutes to make with a profit of $6. A sweatshirt costs $20 and 30 minutes to make with a profit of $20. How many of each should you make to maximize your profit? What is your maximum profit?

Example 2 continued…. You can organize the information in a table: Write the objective function (Profit): P = Write the constraints using the table. Graph the constraints (Note: Allow max value of x at 150) Shade the feasible region Test the vertices constraint T-Shirts (x) Sweatshirts (y) Total Minutes 10x 30y 1200 Number x y 50 Cost ($) 4x 20y 600

Methods of Solving Simultaneous Equations We’ve seen two methods of solving simultaneous equations: Multiplying entire equations by numbers and adding them to eliminate one variable. (“Elimination”) Solving one equation for one variable (say, y) in terms of the other (say, x), then substituting for y in the other equation and solving for x. (“Substitution”)

Systems with Three Variables In two variable systems, we’ve visualized solutions graphically as the intersection of two lines, each with the form: Ax + By = C. In three variable systems, we can visualize solutions graphically as the intersection of three planes, each with the form: Ax + By + Cz = D

Solving Systems Using Elimination 2x –y + z = 4, x + 3y –z = 11, 4x + y –z = 14 x – y + z = -1, x + y + 3z = -3, 2x –y + 2z = 0

Solving Systems Using Substitution 2x + 3y – 2z = -1, x + 5y = 9, 4z – 5x = 4 x – 2y + z = -4, -4x + y -2z = 1, 2x + 2y – z = 10

Some Situations Inventory Management: You have a clothing store and a budget of $6000 to use to buy 200 shirts. You can buy T-shirts for $12 each, polo shirts for $24 each, and rugby shirts for $36 each. You wish to have twice as many rugby shirts as polo shirts. Geometry Application: In triangle PQR, the measure of angle Q is three times the measure of angle P. The measure of angle R is 20⁰ more than the measure of angle P. Find the measure of each angle. What are the unknowns? What system of equations represent the situation?

Homework 14: Pages 171-172: 10, 21,30, 32 Show all work.

Solving Linear Systems Using Matrices A matrix is an array of numbers that can be used to represent a system of linear equations. Below is a 2X3 (“2 by 3”) matrix: 2 4 1 6 5 3 has 2 rows and 3 columns. The above matrix can be used to represent the following system of linear equations 2x + 4y = 1 6x + 5y = 3

Structure of Matrices We refer to the numbers in a matrix as “elements” and they are referenced by indices: aij where i = row and j= column. Labeling of elements in a 3X3 matrix: 𝑎 11 𝑎 12 𝑎 13 𝑎 21 𝑎 22 𝒂 𝟐𝟑 𝑎 31 𝑎 32 𝑎 33 The matrix element 𝑎 23 (in boldface) is in the second row and third column.

Representing Systems of Equations with Matrices Example: x – 3y + z = 6 x + 3z = 12 y = -5x + 1 Rewrite system in the same variable order, lining up the coefficients of like variables: x – 3y + z = 6 x + 3z = 12 5x + y = 1 Write the matrix using only coefficients and constants and use a vertical line to separate coefficients from constants: 1 −3 1 1 0 3 5 1 0 | | | 6 12 1

Represent the following systems using matrices -4x – 2y = 7 3x + y = -5 4x – y + 2z = 1 y + 5z = 20 2x = -y + 7 What system is represented by: 5 2 0 1 | 7 | 9 ?

An Especially Convenient Matrix What system is represented by: 1 0 0 0 1 0 0 0 1 | 𝑎 | 𝑏 | 𝑐 ? Solving systems using matrices is a convenient way to solve using elimination. We perform “row operations” that leave the system unchanged until we get the matrix in the above form, called “Reduced Row Echelon Form” (RREF).

Row Operations We can perform the following operations on a matrix without changing the solution of the system. (Remember: Each row corresponds to an equation.): Switch any two rows. (That’s just changing the order in which we look at the equations.) Multiply a row by a constant. (That’s just multiplying both sides of an equation by the same number.) Add one row to another. (That’s just adding the right hand sides and left hand sides of two equations). Any combination of the above. We have done all of these things before, just not in matrix form.

Using Row Operations to reduce a Matrix to RREF. Find the solution to: x+ 4y = -1 and 2x + 5y = 4 Write in matrix form as: 1 4 2 5 | −1 | 4 Get bottom row reduced first: Multiply top row by -2 and add to bottom row : 1 4 0 −3 | −1 | 6 Multiply bottom row by (-1/3): 1 4 0 1 | −1 | −2 Multiply bottom row by -4 and add to top row: 1 0 0 1 | 7 | −2 So: x =7 and y = -2.

Gauss-Jordan Elimination: Systematic Method to reduce a Matrix to RREF Think of the elements to the left of the constants as a “square matrix” (i.e. 2 by 2, 3 by 3, …. n by n). You want to get all the elements along the diagonal (called the “diagonal elements”) equal to 1 and all the others; zero. In general, get the elements below the diagonal of the leftmost column zeroed starting from the bottom using row operations, until you reach the diagonal element. Repeat for each column, working your way to the right. At the end of this step, all elements below the diagonal are zero. Then multiply the bottom most row by the reciprocal of its diagonal element to turn it into 1. Now work your way up the rows: For each row, zero the off diagonal elements using multiples of the lower rows , then multiply by the reciprocal of the diagonal element.

Solve the following Systems Using Gaussian Elimination A) x + 3y = 22 2x – y = 2 4 𝑥 + 1 𝑦 =1 Hint: Let 𝑚= 1 𝑥 and 𝑛= 1 𝑦 and rewrite system solving for n and m 8 𝑥 + 4 𝑦 =3 C) 2x + 3y + z = 13 5x – 2y – 4z = 7 4x + 5y + 3z = 25

More About Gauss-Jordan Elimination Provides a more concise, less cluttered representation of elimination. Easy to program in a computer in case you have thousands of equations and unknowns. Has further applications (especially in physics) beyond solving systems of linear equations. More complex programming methods are possible that are faster than Gauss-Jordan Elimination. If you stop row operations after the lower off diagonal elements are zero and the diagonal elements are 1, your matrix will be in “row echelon form” (“REF”) and this process is called “Gaussian Elimination”. If you translate the result back to a system of equations, it can be easily solved using substitution.

Homework 15 Homework 15: Pages 179-181: 8-11, 28, 30, 32, 41 46. (You must use matrix methods MANUALLY WITHOUT a graphing calculator to solve 28, 32, 41, 46).