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Jim Jack (J²) MATH 1324 – Math for Business & Economics Admin Class Grading Attendance 4. Systems of Equations and Inequalities 4.1. Review: Systems of Linear Equations in Two Variables DFN: Given the linear system ax by h cx dy k a,b,c,d,h,k real constants, x0 , y0 is a solution of this system if each equation is satisfied by the pair. The set of all such ordered pairs is called the solution set. Graphing If 2 adult tickets and 1 child ticket cost $8, while 1 adult ticket and 3 child tickets cost $9, what is the price of each? 2x y 8 x 3y 9 x 2y 2 x y 5 x 2 y 4 2x 4 y 8 2x 4 y 8 x 2y 4 DFN: A system is consistent if at least one solution exists, otherwise, it is inconsistent. A consistent system is independent if there is a unique solution (exactly one solution), otherwise it is dependent. Two systems are equivalent if they have the same solution set. Thm 1: The linear system ax by h cx dy k must have either A: exactly one solution B: no solution C: infinitely many solutions. There are no other possibilities. Solve a system with a graphing calculator 5 x 2 y 15 2 x 3 y 16 Method of substitution 1. Solve one equation for one variable 2. Sub the result into the other equation, then solve for the other variable. 3. Solve equation from step 1 using step 2. 5x y 4 2x 3y 5 Elimination by addition Thm 2: A system of linear equations is transformed into an equivalent system if A: Two equations are interchanged B: An equation is multiplied by a non-zero constant (both sides). C: A constant multiple of one equation is added to another equation. 3x 2 y 8 2 x 5 y 1 2 x 6 y 3 x 3y 2 x 1 y 4 2 2x y 8 A woman wants to use milk and orange juice to increase the amount of calcium and vitamin A in her daily diet. An ounce of milk contains 37 mg of calcium and 57 g of vitamin A. An ounce of orange juice contains 5mg of calcium and 65 g of vitamin A. How many ounces of milk and OJ should she drink each day to provide exactly 500 mg calcium and 1200 g vitamin A? 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 demand. Similarly, the quantity of a product that a supplier is willing to sell during some period of time increases with the price. Suppose the data collected on the sale of cherries shows that: p 0.07q 0.76 supply p 0.2q 4 demand Supply/Demand @ $2.09/lb, $1.32/lb, equilibrium 4.2 Systems of Linear Equations and Augmented Matrices A matrix is a rectangular array of numbers. 4 5 12 0 1 8 1 4 5 7 0 2 3 10 9 6 0 1 a11 1 a12 4 a13 5 a21 7 a22 0 a23 2 The dimension of a matrix is m n if it has m rows and n columns. A square matrix is n n . a1 x b1 y c1 z d1 a1 b1 c1 d1 a b c d a2 x b2 y c2 z d 2 2 2 2 2 a3 x b3 y c3 z d 3 a3 b3 c3 d 3 square matrix m n row matrix m 1 column matrix n 1 principal diagonal i j coefficient, constant, augmented matrix 2x 3y 5 x 2 y 3 Thm 1 (like thm2 from last unit): An augmented matrix is transformed into a row- equivalent matrix by performing any of the following row operations. A: two rows are interchanged Ri R j B: A row is multiplied by a non-zero constant kRi Ri C: a constant multiple of one row is added to another row. kR j Ri Ri Gauss-Jordan elimination 1. Do row ops to get a one at the upper left. 2. Do row ops to get a zero below that. 3. Do row ops to get a one at bottom of second column. 4. Do row ops to get a zero above it. 3 x1 4 x2 1 x1 2 x2 7 2 x1 3 x2 6 3x1 4 x2 1 2 2 x1 x2 4 6 x1 3 x2 12 2 x1 6 x2 3 x1 3 x2 2 4.3 Gauss-Jordan Elimination 3 possibilities: a11 a12 k1 a21 a22 k 2 Reduced Row-Echelon Form 1 0 0 1 1 4 0 0 3 0 1 0 1 1 0 0 0 0 1 0 2 0 1 4 0 0 1 2 0 0 0 1 6 1 0 4 5 1 0 3 0 1 3 0 0 1 1 0 0 0 1 0 0 0 A matrix is said to be in reduced row echelon form if 1. Each row consisting entirely of zeros is below any row having at least one nonzero element. 2. The left-most non-zero element in any row is a one. 3. All other elements in the column containing the leftmost 1 of a given row are zeros. 4. The leftmost 1 in any row is to the right of the leftmost 1 in the row above. Main diagonal 1s, then maybe zeros First non-zero element is a 1 (leading 1) Leftmost leading 1 listed first Rows with only zeros at the bottom All elements below main diagonal are zeros 1 0 0 1 0 2 0 3 0 0 1 5 0 1 2 1 0 3 1 2 2 3 0 0 1 1 1 0 3 0 0 0 0 1 2 2x 2 y z 3 3x y z 7 x 3y 2z 0 Gauss-Jordan Elimination 1. Choose the leftmost nonzero column, and use row operations to get a 1 at the top. 2. Use multiples of the row containing the 1 at the top to get zeros in all the remaining places in the column containing the 1. 3. Repeat step 1 with the submatrix formed by mentally deleting the row used in step 2 and all rows above this row. 4. Repeat step 2 with the entire matrix including all the rows deleted mentally. Continue until matrix is in reduced row echelon form. 2 x 4 y z 4 4x 8 y 7z 2 2 x 4 y 3z 5 3x 6 y 9 z 15 2 x 4 y 6 z 10 2 x 3 y 4 z 6 x1 2 x2 4 x3 x4 x5 1 2 x1 4 x2 8 x3 3 x4 4 x5 2 x1 3 x2 7 x3 3 x5 2 A company that rents small moving trucks wants to purchase 25 trucks with a combined capacity of 28000 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 1400 cubic feet. How many trucks of each type should the company purchase? 4.4 Matrices: Basic Operations Two matrices are equal if they are the same size and all corresponding elements are equal ( aij bij i, j ) a b c u v w a u b v c w d e f x y z iff d x e y f z Addition The sum of two matrices of the same size is the matrix with elements that are the sum of the corresponding elements of the given matrices. a b w x a w b x c d y z c y d z 2 3 0 3 1 2 1 2 5 3 2 5 1 7 5 0 2 0 6 1 3 8 2 8 Properties of Matrices Addition is commutative A B B A Addition is associative A B C A B C A matrix with all zero elements is the zero matrix. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The negative of a matrix M, denoted M , is a matrix with elements that are the negatives of the elements in matrix M. a b a b If M , then M c d . c d We can then define subtraction as A B A B. 3 2 2 2 5 0 3 4 The product of a scalar (real number) k and a matrix M, denoted kM, is the matrix formed by multiplying each element of M by the number k. 3 1 0 2 2 1 3 0 1 2 Ms. Smith and Mr. Jones are salespeople in a new car agency that sells only two models. August was the last month for this year’s models, and next year’s models were introduced in September. Gross dollar sales for each month are given in the following matrices: August Sales September Sales Compact Luxury Compact Luxury Smith $54,000 $88,000 $228,000 $368,000 Jones $126,000 0 $304,000 $322,000 (Matrix A) (Matrix B) What were the combined sales in August and September for each salesperson by model? What was the increase in dollar sales from August to September? If both salespeople receive 5% commission on gross dollar sales, what is the commission each receives by model for September? Product of a row matrix with a column matrix The product of a 1 n matrix with an n 1 matrix is an 11 matrix given by: b1 b 2 a1 a2 a3 an b3 a1b1 a2b2 a3b3 anbn bn 5 2 3 0 2 2 A factory produces a slalom water ski that requires 3 labor hours in the assembly department and 1 labor hour in the finishing department. Assembly personnel receive $9/hour while finishing personnel receive $6/hour. What is the labor cost per ski? 9 3 1 6 Matrix product If A is an m p matrix and B is a p n matrix, the matrix product of A and B, denoted AB, is an m n matrix whose ij element is obtained by the product of the i row of A with the j column of B. The elements are aip bpj . If the number of columns of p 1, n A is not equal to the number of rows of B, the matrix product is not defined. m p p n 1 3 2 3 1 2 1 2 2 0 1 2 2 1 1 0 1 1 0 1 2 1 2 0 1 2 2 1 1 1 0 1 2 1 2 0 1 0 1 2 2 6 1 2 1 3 3 6 1 2 2 6 3 6 1 3 5 2 3 0 2 2 5 2 2 3 0 2 Notes: Matrix multiplication is not commutative. The principle of zero products does not hold. Combine the time and labor costs for slalom and trick skis from previous example. Wages Hours 12 13 5 1.5 7 8 3 1 Hours Wages 5 1.5 12 13 3 1 7 8 4.5 Inverse of a Square Matrix The identity element for multiplication for the set of all square matrices of order n is the square matrix of order n, denoted by I, with 1's along the principal diagonal and 0's elsewhere. 1 0 0 1 0 0 1 0 0 1 0 0 1 1 0 0 a b c 0 1 0 d e f 0 0 1 g h i a b c 1 0 0 d e f 0 1 0 g h i 0 0 1 1 0 a b c 0 1 d e f 1 0 0 a b c d e 0 1 0 f 0 0 1 IM MI M Multiplicative Inverse Let M be a square matrix of order n, and I be the identity matrix of order n. If a matrix M 1 M 1M MM 1 I then M 1 is called the multiplicative inverse of M or more simply the inverse of M. 2 3 a c 1 2 b d 1 1 1 a d g 0 2 1 b e h 2 3 0 c f i 1 1 1 1 0 0 0 2 1 0 1 0 2 3 0 0 0 1 Inverse of a Square Matrix M If M I can be transformed by row operations into I B , then the resulting matrix B is M 1 . However, if we obtain all 0's in any row left of the vertical line, then M 1 does not exist. Find the inverse of 4 1 6 2 Find the inverse of 2 4 3 6 blank A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 0 1 2 3 4 5 6 7 8 91 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 01234567 890123 456 19 5 3 18 5 20 0 3 15 4 5 4 3 19 3 5 0 15 5 1 1 5 18 20 3 4 0 91 24 66 21 80 25 9 3 72 19 20 5 Decode 46 84 85 28 47 46 4 5 10 30 48 72 29 57 38 38 57 95 1 1 1 5 2 1 using 2 1 2 A1= 2 1 0 2 3 1 4 1 1 4.6 Matrix Equations and Systems of Linear Equations ax b ax b c Basic Properties of Matrices Addition A B C A B C A B B A A0 0 A A A A A Multiplication ABC ABC AI IA A AA1 A1 A I Combined AB C AB AC B C A BA CA A B CA CB and AC BC Solving a matrix equation AX=B x y z 1 2y z 1 2x 3y 1 3 3 1 A1 2 2 1 4 5 2 x yz 3 2y z 1 2x 3y 4 x y z 5 2y z 2 2x 3y 3 If the number of equations in a system equals the number of variables, and the coefficient matrix has an inverse, then the system will have a unique solution that can be found using the inverse. Will not work if: 1) coefficient matrix is singular 2) number of variables does not equal the number of equations. An investment advisor currently has two types of investments available for clients: A conservative investment A that pays 10%/year, and a higher risk investment B that pays 20% per year. Clients may divide investments between the two to achieve any total return between 10% and 20%. Higher return is higher risk. How should each client listed invest to achieve the desired return? Client 1 2 3 k Investment $20000 $50000 $10000 k1 Rturn $2400 $7500 $1300 k2 12% 15% 13% x1 x2 k1 0.1x1 0.2 x2 k 2 4.7 Leontief Input-Output Analysis Wassily Leontief 1973 Nobel Prize 500 sectors of the economy interacting Two industry model Electric and Water Suppose production of $1 worth of electricity requires $0.30 worth of electricity and $0.10 worth of water. Production of $1 worth of water requires $0.20 worth of electricity and $0.40 worth of water. Outside demand is: d1 $12 million for electricity d 2 $8 million for water To produce this amount of water and electricity: Basic Input-Output problem Given the internal demands for each industry’s output, determine the output levels for the various industries that will meet the given final (outside) level demand as well as the internal demand. x1 0.3 x1 0.2 x2 d1 x2 0.1x1 0.4 x2 d 2 x1 0.3 0.2 x1 d1 x 0.1 0.4 x d 2 2 2 X MX D 0.7 0.2 1 .5 0 .5 and I M 0.25 1.75 1 I M 0.1 0.6 Solution to a Two-Industry Input-Output Problem Given two industries C1 and C2 , with C1 C2 x1 d1 M C1 a11 a12 , X , and D x2 d 2 C2 a21 a22 where aij is the input required of Ci to produce a dollar’s worth of output for C j , the solution to the input-output matrix equation X MX D is X I M D 1 assuming that I M has an inverse. Three-Industry model An economy is based on three sectors, agriculture, energy, and manufacturing. Production of a dollar’s worth of agriculture requires an input of $0.20 from the agriculture sector, and $0.40 from the energy sector. Production of a dollar’s worth of energy requires an input of $0.20 from the energy sector, and $0.40 from the manufacturing sector. Production of a dollar’s worth of manufacturing requires an input of $0.10 from the agriculture sector, $0.10 from the energy sector, and $0.30 from the manufacturing sector. Find the output from each sector that is needed to satisfy a final demand of $20billion for agriculture, $10billion from energy, and $30billion from manufacturing. A E M A E M I M I M 1 5. Linear Inequalities and Linear Programming 5.1. Linear Inequalities in Two Variables y 2x 3 2x 3y 5 y 2x 3 2x 3y 5 3 x 4 y 24 3 x 4 y 24 3 x 4 y 24 Graphing of Linear Inequalities ax by c The graph of the linear inequality Ax By C or Ax By C is either the upper half-plane or the lower half-plane determined by the line Ax By C . Procedure: 1) First graph Ax By C as a dashed line if equality is not included in the original statement or a solid line if equality is included. 2) Choose a test point anywhere in the plane not on the line and substitute into the inequality. 3) The graph of the original inequality includes the half-plane containing the test point if the inequality is satisfied by that point, or the other half-plane if the inequality is not satisfied. 2x 3y 6 y 3 2x 5 x 3y Interpreting a graph Ax By C A concert sales promoter wants to book a rock group for a stadium concert. A ticket for the playing field costs $125, while a ticket in the stands costs $175. The group wants a total ticket sales of at least $700,000. How many tickets of each type must be sold? 5.2 Systems of Linear Inequalities x y 6 2x y 0 Solve Systems of Linear Inequalities; Corner Pts 2 x y 22 x y 13 2 x 5 y 50 x0 y0 A patient in a hospital requires at least 84 units of drug A and 120 units of drug B. Each gram of M contains 10 units of drug A and 8 units of drug B. Each gram of N contains 2 units of drug A and 4 units of drug B. How many grams of M and N can be mixed to meet the minimum daily rqmts? M N MDR A 10 2 84 B 8 4 120 5.3 Linear Programming in Two Dimensions: Geometric Approach Objective – find values of decision variables that produce optimal value Objective function maximized or minimized to make that decision Constraints – limits imposed in the decision Feasible solutions (feasible region) – solution set for system of linear equations. Optimal solution – the best feasible solution Iso-profit line – lines of constant profit A tent manufacturer makes a standard model and an expedition model. Each standard tent requires 1 labor-hour from the cutting department and 3 labor- hours from the assembly department. Each expedition tent requires 2 labor-hours from the cutting department and 4 labor-hours from the assembly department. The maximum labor hours available per day in the cutting and assembly departments are 32 and 84 respectively. If the company makes a $50 profit on each standard tent, and $80 on each expedition tent, how many tents should be made daily to maximize profit? Labor hours req per tent Std Exp Max Cut 1 2 32 Assy 3 4 84 Profit 50 80 Maximize P 50x1 80x2 x1 2 x2 32 3x1 4 x2 84 Subject to x1 0 x2 0 Fundamental Theorem of Linear Programming If an optimal value for a linear programming problem exists, then it occurs at one or more corner points of the region of feasible solutions. Procedure 1. Identify the decision variables. 2. Summarize variable information in a table. 3. Determine objective and linear obj function. 4. Write problem constraints (& non-neg). 5. Graphically determine feasible region. 6. Determine the corner points of feasible region. 7. Determine optimal solution by evaluating objective function at corner points. 8. Interpret results in terms of original problem. Existence of feasible solutions 1. bounded region 2. unbounded region 3. empty feasible region Minimize and maximize z 3x1 x2 , subject to x1 0 , x2 0 , 2 x1 x2 20, 10x1 x2 36 , and 2 x1 5 x2 36 Solving a Linear Programming Problem 1. Read problem carefully. 2. Use a table to write the objective function and all of the constraints. 3. Sketch a graph of the region of feasible solutions. Identify vertices. 4. Evaluate the objective function at each one of the vertices. Choose the min/max. Minimize and maximize z 10x1 20 x2 , subject to x1 0 , x2 0 , 6 x1 2 x2 36, 2 x1 4 x2 32, and x2 20 Minimize and maximize P 2 x1 3x2 , subject to x1 0 , x2 0 , x1 x2 8, x1 2 x2 8, and 2 x1 x2 10 A patient in a hospital requires at least 84 units of drug A and 120 units of drug B. Each gram of M contains 10 units of drug A and 8 units of drug B. Each gram of N contains 2 units of drug A and 4 units of drug B. Now suppose that both M and N contain an undesirable drug D, 3 units/gram in M and 1 unit/gram in N. How many grams of M and N can be mixed to meet the minimum daily requirements, while minimizing intake of drug D? M N MDR A 10 2 84 B 8 4 120 D 3 1 6. Linear Programming: The Simplex Method 6.1 Geometric Introduction of the Simplex Method Maximize P 50x1 80x2 x1 2 x2 32 3x1 4 x2 84 Subject to x1 0 x2 0 Bounded and unbounded variables Standard Form A linear programming problem is said to be a standard maximization problem in standard form if its mathematical model is of the following form: Maximize the objective function P c1 x1 c2 x2 cn xn subject to constraints of the following form: a1 x1 a2 x2 an xn b, b 0 with non-negative constraints x1, x2 , xn 0 Slack variables x1 2 x2 32 x1 2 x2 s1 32 3 x1 4 x2 84 3 x1 4 x2 s2 84 Basic/Non-basic Variables Basic Solutions Basic Feasible Solutions Given a system of linear equations associated with a linear programming problem: The variables are divided into two (mutually exclusive) groups: Basic variables are selected arbitrarily with the restriction that there are as many basic variables as equations. The rest of the variables are nonbasic variables. A solution found by setting nonbasic variables to zero and solving for the basics is a basic solution. If a basic solution has no negative values, it is a basic feasible solution. Suppose there is a system of three problem constraint equations with eight variables associated with a standard maximization problem. Basic/Nonbasic variables? Setting nonbasic variables to zero Basic solution has all non-negative elements Feasible? Fundamental Theorem of Linear Programming If an optimal value of the objective function in a linear programming problem exists, then it occurs at one or more of the basic feasible solutions. 6.2 Simplex Method: Maximization with problem constraints of the form Maximize P 50x1 80x2 x1 2 x2 32 3x1 4 x2 84 Subject to x1 0 x2 0 x1 2 x2 s1 32 3 x1 4 x2 s2 84 50 x1 80 x2 P0 x1, x2 , s1 , s2 0 Basic Solutions and Basic Feasible Solutions (bfs) 1. Let P always be a basic variable. 2. Note that a basic solution of this system is still basic after P is deleted. 3. If a basic solution to the above is a bfs without P, it is also a bfs with P. 4. A bfs can contain a negative number only if it is the value of P. Fundamental Theorem of Linear Programming If an optimal value of the objective function in a linear programming problem exists, then it occurs at one or more of the basic feasible solutions of the initial system. x1 2 x2 s1 32 3 x1 4 x2 s2 84 50 x1 80 x2 P0 Initial basic feasible solution, pick slacks and P Simplex Tableau x1 x2 s1 s2 P s1 1 2 1 0 0 32 s2 3 4 0 1 0 84 P 50 80 0 0 1 0 Pivot Operation x1 x2 s1 s2 P s1 1 2 1 0 0 32 s2 3 4 0 1 0 84 P 50 80 0 0 1 0 Pivot Element 1.Select most negative indicator in bottom row. This is the pivot column. 2.Divide each positive element in the pivot column into the corresponding element in the last column. (if no positive elements in pivot column, stop) The pivot row is the one that generates the smallest quotient. 3.The pivot element is at the intersection of pivot row and column. Pivot operation 1. Use division to put a 1 in the pivot element. 2. Use row operations to zero all other elements in that column . x1 x2 s1 s2 P s1 1 2 1 0 0 32 s2 3 4 0 1 0 84 P 50 80 0 0 1 0 Maximize P 10 x1 5 x2 4 x1 x2 28 2 x1 3 x2 24 Subject to x1 0 x2 0 Maximize P 6 x1 3x2 2 x1 3x2 9 x1 3x2 12 Subject to x1 0 x2 0 A farmer owns a 100-acre farm and plans to plant at most 3 crops. The seed for crops A, B, and C cost $40, $20, and $30 per acre respectively. A max of $3200 can be spent on seed. These crops require 1, 2, and 1 workday per acre, and there are a max of 160 workdays available. If the farmer can realize profit of $100, $300, and $200 per acre respectively, how many acres of each crop should be planted to max profit? Objective: Constraints: 6.3 Systems of Linear Equations in Three Variables (end) 2x 3y 4z 4 y 2z 0 3,2,1 x 5 y 6z 7 x 4 y 2 z 15 1,3,2 4x y z 1 1,10,13 6 x 2 y 3z 6 Solving with elimination and substitution 1. Eliminate one variable from two of the equations. 2. Apply the techniques from 6.1 and 6.2 to solve the resulting two equations. 3. Substitute back in to find the third variable. x y 2z 6 2 x y 2 z 3 x 2 y 3z 7 One thousand tickets were sold for a play, which generated $3800. The prices of the tickets were $3 for children, $4 for students, and $5 for adults. There were 100 fewer student tickets sold than adult tickets. Find the number of each ticket sold. Three students buy lunch in the cafeteria. One student buys 2 hamburgers, 1 order of fries, and a soda for $9. Another student buys 1 hamburger, 2 orders of fries, and a soda for $8. A third student buys 3 hamburgers, 3 fries, and 2 sodas for $18. If possible, find the cost of each item. x y z 2 x 2 y 2 z 3 y z 1 6.4 Solutions of Linear Systems Using Matrices Carl Friedrich Gauss (1777-1855) Gaussian Elimination with Backward Substitution Matrix – a rectangular array of numbers The dimension of a matrix is m n if it has m rows and n columns. A square matrix is n n . a1 x b1 y c1 z d1 a1 b1 c1 d1 a2 x b2 y c2 z d 2 a1 b1 c1 d1 a3 x b3 y c3 z d 3 a1 b1 c1 d1 3x 4 y 6 5 x y 5 2 x 5 y 6 z 3 3x 7 y 3z 8 x 7y 5 1 0 2 3 2 2 10 3 1 2 3 5 1 2 3 4 0 1 6 7 0 0 1 8 Row-Echelon Form 1 3 0 1 1 3 1 5 1 2 0 0 1 6 1 0 1 1 3 0 1 4 0 0 1 2 0 0 1 0 1 3 1 5 1 3 5 0 0 1 3 0 0 1 0 0 0 0 Main diagonal 1s, then maybe zeros First non-zero element is a 1 (leading 1) Leftmost leading 1 listed first Rows with only zeros at the bottom All elements below main diagonal are zeros Solve the system 1 1 3 12 0 1 2 4 0 0 1 3 1 1 5 5 0 1 3 3 0 0 0 0 Gaussian Elimination Matrix Row Transformations 1. Any two rows may be interchanged 2. The elements of any row may be multiplied by a non-zero constant 3. Any row may be changed by adding or subtracting a multiple of another row. x y z 1 x y z 5 y 2z 5 2x 4 y 4z 4 x 3y z 4 x 3 y 2 z 1 Reduced Row-Echelon Form Row-Echelon Form with elements above and below the leading 1 are zeros 1 0 0 1 0 1 0 1 0 3 0 1 0 0 1 0 0 0 1 2 0 0 1 1 0 0 3 1 0 4 8 0 1 0 1 0 1 1 2 0 0 1 1 0 0 0 0 Solve the system 1 0 0 3 0 1 0 1 0 0 1 2 1 0 6 0 1 5 Transform to Reduced Row-Echelon Form 2 x y 2 z 10 x 2z 5 x 2 y 2z 1 6.5 Properties and Applications of Matrices a11 a12 a13 a14 a11 a12 a13 a11 a12 a11 a12 a 21 a 22 a 23 a 24 a11 a12 a13 a 21 a 22 a 23 a 21 a 22 a 21 a 22 a 31 a 32 a 33 a 34 a21 a22 a23 a31 a32 a33 a 31 a 32 a 41 a 42 a 43 a 44 An element of matrix A is designated aij ( arc ) Two matrices A and B are equal if all corresponding elements are equal ( aij bij i, j ) 3 3 7 3 x 7 Let A= 1 6 2 B= 1 6 2 4 2 5 4 5 2 a12 b32 a13 What value of x would make A=B? a31b13 a32b23 a33b33 Sum, Difference, Scalar Multiplication The sum of two m n matrices A and B is the m n matrix A B , where each element is the sum of corresponding elements in A and B. aij bij i, j 1 i m,1 j n 2 2 2 1 1 1 0 2 0 1 1 1 0 2 0 1 1 1 1 2 1 1 1 1 2 2 2 1 1 1 1 2 1 1 1 1 The difference of two m n matrices A and B is the m n matrix A B , where each element is the sum of corresponding elements in A and B. aij bij i, j 1 i m,1 j n 3 3 3 1 1 1 1 3 1 1 1 1 1 3 1 1 1 1 7 8 1 5 2 10 A= B= 0 1 6 3 2 4 A+B= B+A= A – B= 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 The product of a scalar (real number) k and m n matrix A is the m n matrix kA , where each element is the sum of corresponding elements in A and B. kaij i, j 1 i m,1 j n 2 7 11 If A= 1 3 5, find 4 A 0 9 12 1 1 1 3 4 2 0 1 A= B= C= 0 7 D= 9 7 3 5 2 3 4 2 1 8 A 3B AC 2C 3D Matrix Products A B $ Student1 10 7 CollegeA 60 Student2 11 4 CollegeB 80 Cost for Student 1 Cost for Student 2 10 7 60 AB= 11 4 80 The product of an m n matrix with an n k matrix is an m k matrix. The elements are aij b jk . j 1, n m n n k 1 1 1 1 2 1 1 2 3 A= 0 3 B= C= D= 0 3 2 2 4 5 6 4 2 3 4 5 AB= CA= DC= CD= 1 0 7 4 6 7 A= 3 2 1 B= 8 9 10 5 2 5 0 1 3 AB= Graphing Calculator 2 A 3B 3 6.6 Inverses of Matrices A1 like f 1 Translation 1 0 h x A0 1 k Xy AX 0 0 1 1 Translate the point 1,2, 3 units right, down 4. 1 0 3 A1 0 1 4 0 0 1 AA1 A1 A The n n identity matrix, denoted by I n , has only 1s on the main diagonal and 0s elsewhere. Let A be an n n matrix. If there exists an n n matrix, denoted A1, that satisfies A1 A I n and AA1 I n then A1is the inverse of A. If A1 exists, A is invertible or non-singular If A is not invertible, then A is singular. 5 3 2 3 Let A B B A1? 3 2 3 5 90° clockwise rotation 0 1 0 0 1 0 Let A 1 0 0 and A1 1 0 0 0 0 1 0 0 1 Rotate the point 2,0 clockwise 90° about origin What will A1 do? Find A1. Representing linear systems with matrix equations 3x 2 y 4 z 5 2 x y 3z 9 x 5 y 2z 5 3x 4 y 7 x 6 y 3 x 5y 2 3x 2 y z 7 4 x 5 y 6 z 10 Solving linear systems with inverses AX B x 4y 3 9 4 2x 9 y 5 2 1 x 3y z 6 1.25 2 .25 2 y z 2 .25 1 .25 x y 3z 4 .5 1 .5 Modeling blood pressure P A,W a bA cW P A W 113 39 142 138 53 181 152 65 191 Finding Inverses Symbolically 1 1 4 Find A if A 2 9 1 0 1 Find A1 if A 2 1 3 1 1 1