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**Systems of equations and matricies**

Lecture 3 Systems of equations and matricies

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**Systems of Two Equations in Two Variables**

We are given the linear system ax + by = c dx + ey = f A solution is an ordered pair (x0, y0) that will satisfy each equation (make a true equation when substituted into that equation). The solution set is the set of all ordered pairs that satisfy both equations. In this section, we wish to find the solution set of a system of linear equations.

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**Solve by Graphing? Substitution?**

This is practical for two-variable systems, but very impractical once problem gets larger: Solve the following system: 3x + 5y = -9 x+ 4y = -10

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Solve by Substitution 3x + 5y = -9 x+ 4y = -10 Substitution: Solve equation 1 for x, replace for x in second equation. 3x =- 5y -9 , x=(-5/3)y-3 (-5/3)y-3+ 4y = -10 (-5/3)y+(12/3)y=-7 (7/3)y=-7 Y=-3 Plug back into either equation to solve for x. 3x+5(-3)=-93x=6x=2… OR x+4(-3)=-10x=2 Solution? (2,-3)

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**Method of Substitution**

The method of substitution is an algebraic one. It works well when the coefficients of x or y are either 1 or -1. For example, let’s solve the system 2x + 3 = y x + 2y = -4 using the method of substitution.

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**Method of Substitution (continued)**

The steps for this method are as follows: Solve one of the equations for either x or y. Substitute that result into the other equation to obtain an equation in a single variable (either x or y). Solve the equation for that variable. Substitute this value into any convenient equation to obtain the value of the remaining variable.

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**Example (continued) 2x+3 = y x +2y = -4 x + 2(2x + 3) = -4**

Substitute y from first equation into second equation Solve the resulting equation After we find x = -2, then from the first equation, we have 2(-2)+3 = y or y = -1. Our solution is (-2, -1)

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**Another Example Solve the system using substitution: 3x - 2y = 7**

y = 2x - 3

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**Another Example Solution**

Solve the system using substitution: 3x - 2y = 7 y = 2x - 3 Solution:

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Solve by Substitution Notice how quickly this becomes very cumbersome. We will first learn a method called Jordan elimination to learn how to manipulate matricies, but this method too is also very cumbersome. Learning to use some computer program is essential for larger problems. Solve still works in Mathematica, but it too becomes very cumbersome.

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**X1=2, X2=3, X3=1, as before… but what did I just do and how?**

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Terminology A consistent linear system is one that has one or more solutions. If a consistent system has exactly one solution, it is said to be independent. An independent system will occur when the two lines have different slopes. If a consistent system has more than one solution, it is said to be dependent. A dependent system will occur when the two lines have the same slope and the same y intercept. In other words, the graphs of the lines will coincide. There will be an infinite number of points of intersection.

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**Terminology (continued)**

An inconsistent linear system is one that has no solutions. This will occur when two lines have the same slope but different y intercepts. In this case, the lines will be parallel and will never intersect. Example: Determine if the system is consistent, independent, dependent or inconsistent: 2x – 5y = x + 10y = -1

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Solution of Example Solve each equation for y to obtain the slope intercept form of the equation: Since each equation has the same slope but different y intercepts, they will not intersect. This is an inconsistent system. Notice that Mathematica has problems as well. You must know what is going on…

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Note, when I use the methods as I showed in the last 3 equation 3 variable example using Mathematica, I get an error that means something… You will come back to this slide after we learn about matricies

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**Elimination by Addition**

The method of substitution is not preferable if none of the coefficients of x and y are 1 or -1. For example, substitution is not the preferred method for the system below: x – 7y = 3 -5x + 3y = 7 A better method is elimination by addition. The following operations can be used to produce equivalent systems: 1. Two equations can be interchanged. 2. An equation can be multiplied by a non-zero constant. 3. An equation can be multiplied by a non-zero constant and then added to another equation.

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**Elimination by Addition Example**

For our system, we will seek to eliminate the x variable. The coefficients are 2 and -5. Our goal is to obtain coefficients of x that are additive inverses of each other. We can accomplish this by multiplying the first equation by 5, and the second equation by 2. Next, we can add the two equations to eliminate the x-variable. Solve for y Substitute y value into original equation and solve for x Write solution as an ordered pair

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**Elimination by Addition Another Example**

Solve 2x - 5y = 6 -4x + 10y = -1

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**Elimination by Addition Another Example**

Solve 2x - 5y = 6 -4x + 10y = -1 Solution: 1. Eliminate x by multiplying equation 1 by Add two equations 3. Upon adding the equations, both variables are eliminated producing the false equation 0 = 11 4. Conclusion: If a false equation arises, the system is inconsistent and there is no solution. See how Mathematica responds… answer is empty set

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Application A man walks at a rate of 3 miles per hour and jogs at a rate of 5 miles per hour. He walks and jogs a total distance of 3.5 miles in 0.9 hours. How long does the man jog?

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Application A man walks at a rate of 3 miles per hour and jogs at a rate of 5 miles per hour. He walks and jogs a total distance of 3.5 miles in 0.9 hours. How long does the man jog? Solution: Let x represent the amount of time spent walking and y represent the amount of time spent jogging. Since the total time spent walking and jogging is 0.9 hours, we have the equation x + y = We are given the total distance traveled as 3.5 miles. Since Distance = Rate x Time, distance walking = 3x and distance jogging = 5y. Then total distance is 3x+5y= 3.5.

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**Application (continued)**

We can solve the system using substitution. 1. Solve the first equation for y 2. Substitute this expression into the second equation. 3. Solve second equation for x 4. Find the y value by substituting this x value back into the first equation. 5. Answer the question: Time spent jogging is 0.4 hours. Solution:

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Supply and Demand The quantity of a product that people are willing to buy during some period of time depends on its price. Generally, the higher the price, the less the quantity demanded; the lower the price, the greater the quantity demanded. Similarly, the quantity of a product that a supplier is willing to sell during some period of time also depends on the price. Generally, a supplier will be willing to supply more of a product at higher prices and less of a product at lower prices. The simplest supply and demand model is a linear model where the graphs of a demand equation and a supply equation are straight lines. You will see these quite a bit.

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**Supply and Demand (continued)**

In supply and demand problems we are usually interested in finding the price at which supply will equal demand. This is called the equilibrium price, and the quantity sold at that price is called the equilibrium quantity. If we graph the the supply equation and the demand equation on the same axis, the point where the two lines intersect is called the equilibrium point. Its horizontal coordinate is the value of the equilibrium quantity, and its vertical coordinate is the value of the equilibrium price.

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**Supply and Demand Example**

Example: Suppose that the supply equation for long-life light bulbs is given by p = 1.04 q , and that the demand equation for the bulbs is p = -0.81q + 7.5 where q is in thousands of cases. Find the equilibrium price and quantity, and graph the two equations in the same coordinate system.

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**Supply and Demand (Example continued)**

If we graph the two equations and find the intersection point, we see the graph below. Demand Curve Supply Curve Thus the equilibrium point is (7.854, 1.14), the equilibrium price is $1.14 per bulb, and the equilibrium quantity is 7,854 cases.

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**Now, Solve the Opening Example**

A restaurant serves two types of fish dinners- small for $5.99 each and large for $ One day, there were 134 total orders of fish, and the total receipts for these 134 orders was $ How many small dinners and how many large dinners were ordered?

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Solution A restaurant serves two types of fish dinners- small for $5.99 each and large for $ One day, there were 134 total orders of fish, and the total receipts for these 134 orders was $ How many small dinners and how many large dinners were ordered? Answer: 60 small orders and 74 large orders

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Matrix Methods It is impractical to solve more complicated linear systems by hand. Computers and calculators now have built in routines to solve larger and more complex systems. Matrices, in conjunction with graphing utilities and or computers are used for solving more complex systems. In this section, we will develop certain matrix methods for solving two by two systems.

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Matrices A matrix is a rectangular array of numbers written within brackets. Here is an example of a matrix which has three rows and three columns: The subscripts give the “address” of each entry of the matrix. For example the entry a23 is found in the second row and third column Since this matrix has 3 rows and 3 columns, the dimensions of the matrix are 3 x 3. Remember, columns hold up things… Each number in the matrix is called an element.

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**Matrix Solution of Linear Systems**

When solving systems of linear equations, we can represent a linear system of equations by an augmented matrix, a matrix which stores the coefficients and constants of the linear system and then manipulate the augmented matrix to obtain the solution of the system. This gets cumbersome quick, but for now, it’s a useful way to get used to seeing matricies. Example: x + 3y = 5 2x – y = 3 The augmented matrix associated with the above system is

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Generalization Linear system: Associated augmented matrix:

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**Operations that Produce Row-Equivalent Matrices**

1. Two rows are interchanged: 2. A row is multiplied by a nonzero constant: 3. A constant multiple of one row is added to another row:

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**Augmented Matrix Method Example 1**

Solve x + 3y = 5 2x – y = 3 1. Augmented system 2. Eliminate 2 in 2nd row by row operation 3. Divide row two by -7 to obtain a coefficient of Eliminate the 3 in first row, second position. 5. Read solution from matrix R2

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**Augmented Matrix Method Example 2**

Solve x + 2y = 4 x + (1/2)y = 4 Eliminate fraction in second equation by multiplying by 2 Write system as augmented matrix. Multiply row 1 by -2 and add to row 2 Divide row 2 by -3 Multiply row 2 by -2 and add to row 1. Read solution : x = 4, y = 0 (4,0)

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**Augmented Matrix Method Example 3**

Solve 10x - 2y = 6 -5x + y = Represent as augmented matrix. 2. Divide row 1 by 2 3. Add row 1 to row 2 and replace row 2 by sum 4. Since 0 = 0 is always true, we have a dependent system. The two equations are identical, and there are infinitely many solutions. See how Mathematica tells you this…

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**Augmented Matrix Method Example 4**

Solve Rewrite second equation Add first row to second row The last row is the equivalent of 0x + 0y = -5 Since we have an impossible equation, there is no solution. The two lines are parallel and do not intersect. Here Mathematica gives the empty set.

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**Possible Final Matrix Forms for a Linear System in Two Variables**

Form 1: Unique Solution (Consistent and Independent) Form 2: Infinitely Many Solutions (Consistent and Dependent) Form 3: No Solution (Inconsistent)

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**Gauss-Jordan Elimination**

Any linear system must have exactly one solution, no solution, or an infinite number of solutions. Previously we considered the 2x2 case, in which the term consistent is used to describe a system with a unique solution, inconsistent is used to describe a system with no solution, and dependent is used for a system with an infinite number of solutions. In this section we will consider larger systems with more variables and more equations, but the same three terms are used to describe them.

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**Matrix Representations of Consistent, Inconsistent and Dependent Systems**

The following matrix representations of three linear equations in three unknowns illustrate the three different cases: From this matrix representation, you can determine that x = 3 y = 4 z = 5 Case I: consistent

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**Matrix Representations (continued)**

Case 2: inconsistent From the second row of the matrix, we find that 0x + 0y +0z =6 or 0 = 6, an impossible equation. From this, we conclude that there are no solutions to the linear system.

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**Matrix Representations (continued)**

Case 3: dependent When there are fewer non-zero rows of a system than there are variables, there will be infinitely many solutions, and therefore the system is called dependent.

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**Reduced Row Echelon Form**

A matrix is said to be in reduced row echelon form or, more simply, in reduced form, if Each row consisting entirely of zeros is below any row having at least one non-zero element. The leftmost nonzero element in each row is 1. All other elements in the column containing the leftmost 1 of a given row are zeros. The leftmost 1 in any row is to the right of the leftmost 1 in the row above.

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**Examples of Reduced Row Echelon Form**

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**Solving a System Using Gauss-Jordan Elimination**

Example: Solve x + y – z = -2 2x – y + z = 5 -x + 2y + 2z = 1

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**Solving a System Using Gauss-Jordan Elimination**

Example: Solve x + y – z = -2 2x – y + z = 5 -x + 2y + 2z = 1 Solution: We begin by writing the system as an augmented matrix

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Example (continued) We already have a 1 in the diagonal position of first column. Now we want 0’s below the 1. The first 0 can be obtained by multiplying row 1 by -2 and adding the results to row 2: Row 1 is unchanged (-2) times Row 1 is added to Row 2 Row 3 is unchanged

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Example (continued) The second 0 can be obtained by adding row 1 to row 3: Row 1 is unchanged Row 2 is unchanged Row 1 is added to Row 3

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Example (continued) Moving to the second column, we want a 1 in the diagonal position (where there was a –3). We get this by dividing every element in row 2 by -3: Row 1 is unchanged Row 2 is divided by –3 Row 3 is unchanged

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Example (continued) To obtain a 0 below the 1 , we multiply row 2 by -3 and add it to the third row: Row 1 is unchanged Row 2 is unchanged (-3) times row 2 is added to row 3

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Example (continued) To obtain a 1 in the third position of the third row, we divide that row by 4. Rows 1 and 2 do not change.

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Example (continued) We can now work upwards to get zeros in the third column, above the 1 in the third row. Add R3 to R2 and replace R2 with that sum Add R3 to R1 and replace R1 with the sum. Row 3 will not be changed. All that remains to obtain reduced row echelon form is to eliminate the 1 in the first row, 2nd position.

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Example (continued) To get a zero in the first row and second position, we multiply row 2 by -1 and add the result to row 1 and replace row 1 by that result. Rows 2 and 3 remain unaffected.

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Final Result We can now “read” our solution from this last matrix. We have x = 1, y = -1 z = 2. Written as an ordered triple, we have (1, -1, 2). This is a consistent system with a unique solution.

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**Example 2 Example: Solve the system 3x – 4y + 4z = 7 x – y – 2z = 2**

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**Example 2 Example: Solve the system 3x – 4y + 4z = 7 x – y – 2z = 2**

Solution: Begin by representing the system as an augmented matrix:

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Example 2 (continued) Since the first number in the second row is a 1, we interchange rows 1 and 2 and leave row 3 unchanged:

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Example 2 (continued) In this step, we will get zeros in the entries beneath the 1 in the first column: Multiply row 1 by -3 , add to row 2 and replace row 2: -3*R1+R2 → R2. Multiply row 1 by -2, add to row 3 and replace row 3: -2*R1+R3 → R3.

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Final Result To get a zero in the third row, second entry we multiply row 2 by -1 and add the result to R3 and replace R3 by that sum: Notice this operations “wipes out” row 3 so row 3 consists entirely of zeros. Any time you have fewer non-zero rows than variables you will have a dependent system.

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**Representation of a Solution of a Dependent System**

We can interpret the second row of this matrix as –y + 10z = 1, or z – 1 = y So, if we let z = t (arbitrary real number,) then in terms of t, y = 10t - 1. Next we can express the variable x in terms of t as follows: From the first row of the matrix, we have x – y -2z = 2. If z = t and y = 10t – 1, we have x – (10t-1) - 2t = 2 or x = 12t+1. Our general solution can now be expressed in terms of t: (12t+1,10t-1,t), where t is an arbitrary real number.

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**Procedure for Gauss-Jordan Elimination**

Step 1. Choose the leftmost nonzero column and use appropriate row operations to get a 1 at the top. Step 2. Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1. Step 3. Repeat step 1 with the submatrix formed by (mentally) deleting the row used in step 2 and all rows above this row. Step 4. Repeat step 2 with the entire matrix, including the rows deleted mentally. Continue this process until the entire matrix is in reduced form. Note: If at any point in this process we obtain a row with all zeros to the left of the vertical line and a nonzero number to the right, we can stop because we will have a contradiction.

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Applications Example: Purchasing. A company that rents small moving trucks wants to purchase 25 trucks with a combined capacity of 28,000 cubic feet. Three different types of trucks are available: a 10-foot truck with a capacity of 350 cubic feet, a 14-foot truck with a capacity of 700 cubic feet, and a 24-foot truck with a capacity of 1,400 cubic feet. How many of each type of truck should the company purchase?

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Solution Step 1. The question in this example indicates that the relevant variables are the number of each type of truck. x = number of 10-foot trucks y = number of 14-foot trucks z = number of 24-foot trucks We form the mathematical model: x + y + z = (Total number of trucks) 350x + 700y + 1,400z = 28,000 (Total capacity) With two equations and three unknowns… we won’t get a solution to our problem, but we can find a general solution

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Solution (continued) Step 2. Now we form the augmented coefficient matrix of the system and solve by using Gauss-Jordan elimination: (1/350)R R2 -R1 + R R2 -R2 + R R1 Matrix is in reduced form. x - 2z = -30 or x = 2z -30, y + 3z = 55 or y = -3z + 55.

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**Solution (continued) Let z = t. Then for t any real number**

x = 2t – 30 y = -3t z = t is a solution to our mathematical model. Step 3. We must interpret this solution in terms of the original problem. Since the variables x, y, and z represent numbers of trucks, they must be nonnegative. And since we can’t purchase a fractional number of trucks, each must be a nonnegative whole number.

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Solution (continued) Since t = z, it follows that t must also be a nonnegative whole number. The first and second equations in the model place additional restrictions on the values t can assume: x = 2t - 30 > 0 implies that t > 15 y = -3t + 55 > 0 implies that t < 55/3 Thus the only possible values of t that will produce meaningful solutions to the original problem are 15, 16, 17, and 18. A table is a convenient way to display these solutions. 10-ft truck 14-ft. truck 24-ft truck t x y z 15 10 16 2 7 17 4 18 6 1

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