Chapter 5 Systems and Matricies
5.1 Systems of Linear Equations
Linear Systems Any equation of the form a1x1 + a2x2 + + anxn = b, for all real numbers a1,a2,…,an (not all of which are 0) and b, is a linear equation or a first-degree equation in n unknowns. A set of equations is called a systems of equations. The solutions of a system of equations must satisfy every equation in the system. If all the equations in a system are linear, the system is a system of linear equations, or a linear system.
Linear Systems Consistent The graphs of two equations intersect a single point. The coordinates of this point give the only solution of the system.
Linear Systems Inconsistent The graphs are distinct parallel lines. The equations are independent. That is, there is no solution common to both equations.
Linear Systems Dependent The graphs are the same line. Any solution of one equation is also the solution of the other. Thus there are infinite number of solutions.
Substitution Method In a system of two equations with two variables, the substitution method involves using one equation to find an expression for one variable in terms of the other, then substituting into the other equation of the system. Example: Solve the system. 4x + 2y = 8 (1) 3x 7y = 11 (2)
Solution 4x + 2y = 8 (1) 3x 7y = 11 (2) Begin by solving one of the equations for one of the variables. 4x + 2y = 8 2y = 4x + 8 y = 2x + 4 (3) Now replace y with 2x + 4 in the second equation and solve for x. 3x 7(2x + 4) = 11 3x + 14x 28 = 11 17x = 17 x = 1 Replace x with 1 in equation (3) to obtain y = 2(1) + 4 = 2. The solution of the ordered pair is (1,2). Check the solution in both equations (1) and (2). 4x + 2y = 8 4(1) + 2(2) = 8 8 = 8 3x 7y = 11 3(1) 7(2) = 11 11 = 11
Elimination Method The elimination method uses multiplication and addition to eliminate a variable from one equation. To eliminate a variable, the coefficients of that variable in the two equations must be additive inverses. To achieve this, we use properties of algebra to change the system to an equivalent system, one with the same solution set.
Equivalent Systems Transformations of a Linear System Interchange any two equations of the system. Multiply or divide any equation of the system by a nonzero real number. Replace any equation of the system by the sum of that equation and a multiple of another equation in the system.
Example Solve the system using the elimination method. 6x + 2y = 4 If we multiply the first equation by 5 and the second equation by 3, we will be able to eliminate the x variable. 30x + 10y = 20 Substituting: 6x + 2y = 4 30x 21y = 24 6x + 2(4) = 4 11y = 44 6x 8 = 4 y = 4 6x = 12 The solution is (2, 4) x = 2
Solving an Applied Problem by Writing a System of Equations Step 1 Read the problem carefully until you understand what is given and what is to be found. Step 2 Assign variables to represent the unknown values, using diagrams or tables as needed. Write down what each variable represents. Step 3 Write a system of equations that relates the unknowns. Step 4 Solve the system of equations. Step 5 State the answer to the problem. Does it seem reasonable? Step 6 Check the answer in the words of the original problem.
Solving Systems with Three Unknowns (Variables) The graph of a linear equation in three unknowns requires a three-dimensional coordinate system. Some of the possible intersections of planes representing three equations are shown below.
Systems of Three Equations with Three Variables To solve a system with three unknowns, first eliminate a variable from any two of the equations. Then eliminate the same variable from a different pair of equations. Eliminate a second variable using the resulting two equations in two variables to get an equation with just one variable whose value you can now determine. Find the values of the remaining variables by substitution. Solutions of the systems are written as ordered triples. Example: Solve the system. 3x + 9y + 6z = 3 (1) 2x + y z = 2 (2) x + y + z = 2 (3)
Solution 3x + 9y + 6z = 3 (1) 2x + y z = 2 (2) x + y + z = 2 (3) Eliminate z by adding equations (2) and (3) to get 3x + 2y = 4 (4) To eliminate z from another pair of equations, multiply both sides of equations (2) by 6 and add the result to equation (1). 3x + 9y + 6z = 3 12x + 6y 6z = 12 (1) 15x + 15y = 15 (5) To eliminate x from the equations (4) and (5), multiply both sides of equation (4) by 5 and add the result to equation (5). Solve the new equation for y. 15x 10y = 20 15x + 15y = 15 5y = 5 y = 1
Solution continued Using y = 1, find x from equation (4) by substitution. 3x + 2(1) = 4 (4) x = 2 Substitute 2 for x and 1 for y in equation (3) to find z. 2 + (1 ) + z = 2 (3) z = 1 Verify that the ordered triple (2, 1, 1) satisfies all three equations in the original system. The solution set is {(2, 1, 1)}.
Using Curve Fitting to Find an Equation Through Three Points Example: Find the equation of the parabola y = ax2 + bx + c that passes through (2,4), (1, 1), and (2,5). Solution: Since the three points lie on the graph of the equation y = ax2 + bx + c, they must satisfy the equation. Substituting each ordered pair into the equation gives three equations with three variables. 4 = a(2)2 + b (2) + c or 4 = 4a + 2b + c (1) 1 = a(1)2 + b(1) + c or 1 = a b + c (2) 5 = a(2)2 + b(2) + c or 5 = 4a 2b + c (3)
Using Curve Fitting to Find an Equation Through Three Points continued This system can be solved by the elimination method. First eliminate c using equations (1) and (2). 4 = 4a + 2b + c (1) 1 = a + b c 3 = 3a + 3b (4) Now, use equations (2) and (3) to eliminate the same variable (c). 1 = a b + c (2) 5 = 4a + 2b c 4 = 3a + b (5) Solving systems of equations (4) and (5) in two variables by eliminating a. 3 = 3a + 3b (4) 4 = 3a + b (5) 1 = 4b Find a by substituting for b in equation (4), which is equivalent to 1 = a + b. 1 = a + b 1 = a
Using Curve Fitting to Find an Equation Through Three Points continued Finally, find c by substituting a = and b = in equation (2). An equation of the parabola is
5.2 Matrix Solution of Linear Systems
Definitions A matrix is a rectangular array of numbers enclosed in brackets. Each number is called an element of the matrix. The size of a matrix is determined by the number of row (horizontal) and columns (vertical).
Definitions continued Linear System Augmented matrix columns rows
Matrix Row Transformations For any augmented matrix of a system of linear equations, the following row transformations will result in the matrix of an equivalent system. 1. Interchange any two rows. 2. Multiply or divide the elements of any row by a nonzero real number. 3. Replace any row of the matrix by the sum of the elements of that row and a multiple of the elements of another row.
Using the Gauss-Jordan Method to Put a Matrix into Diagonal Form Step 1 Obtain 1 as the first element of the first column. Step 2 Use the first row to transform the remaining entries in the first column to 0. Step 3 Obtain 1 as the second entry in the second column. Step 4 Use the second row to transform the remaining entries in the second column to 0. Step 5 Continue in this manner as far as possible.
Example Solve the system 5x 3y = 14 4x + y = 18 Write the augmented matrix. Work with the columns and rows so that it is transformed into the form
Example continued Multiply row 1 by 1/5 to get a 1 in the first position. Introduce 0 in the second row by multiplying each element of R1 by 4 and adding to R2. Obtain 1 in the second row, second column by multiplying each element of the second row by 5/17.
Example continued Finally, to get 0 in the first row, second column multiply each element of the second row by 3/5 and add to the first row. This last matrix corresponds to the system x = 4 y = 2 that has the solution set (4, 2). Check the solution in both equations of the original system.
Example Solve the system. 1 in the first row, first column. Introduce 0 in the second row of the first column by multiplying each element of row 1 by 2 and adding to row 2.
Example continued To change the third element in the first column to 0, multiply the first row by 1 and add. Use the same procedure to transform the second and third columns. For both of these columns, perform the additional step of getting 1 in the appropriate position of each column. Do this by multiplying the elements of the row by the reciprocal of the number in that position.
Example continued continued The solution set is (1, 3, 2).
Special Systems When a row consists entirely of 0’s, the equations are dependent and the system is equivalent.
Special Systems continued When we obtain a row whose only nonzero entry occurs in the last column, we have an inconsistent system of equations. For example, the last row corresponds to the false equation 0 = 9, so we know the original system has no solution.
5.3 Determinant Solution of Linear Systems
Determinants Every n n matrix A is associated with a real number called the determinant, of A, written |A|. The determinant of a 2 2 matrix is defined as follows. Note: Matrices are enclosed with square brackets, while determinants are denoted with vertical bars.
Example Let Find |A|.
Determinant of a 3 3 Matrix
Cofactor Let Mij be the minor for the element aij in an n n matrix. The cofactor if aij written Aij, is Find the cofactor of the element (2). The cofactor is
Finding the Determinant of a Matrix Multiply each element in any row or column of the matrix by its cofactor. The sum of these products give the value of the determinant.
Example Evaluate expanding by the third row.’
Example continued Now find the cofactor of each element of these minors. Find the determinant by multiplying each cofactor by its corresponding element in the matrix and finding the sum of these products.
Cramer’s Rule for Two Equations in Two Variables
Example Use Cramer’s Rule to solve the system. 7x + 3y = 15 Find D first, since if D = 0, Cramer’s Rule does not apply. The solution set is {(3, 2)}.
General Form of Cramer’s Rule
Example Use Cramer’s Rule to solve the system.
Example continued Thus: The solution set is
5.4 Partial Fractions
Decomposition of Rational Expressions The sums of rational expressions are found by combining two or more rational expressions into one rational expression. A special type of sum of rational expressions is called the partial fraction decomposition, each term in the sum is a partial fraction. Add rational expressions Partial fraction decomposition
Partial Fraction Decomposition
Partial Fraction Decomposition continued
Distinct Linear Factors Example: Find the partial fraction decomposition of Solution: The given fraction is not a proper fraction; the numerator has greater degree than the denominator. Perform the division.
Distinct Linear Factors continued The quotient is Now, work with the remainder fraction. Factor the denomina- tor as x3 4x = x(x + 2)(x 2). Since the factors are distinct linear factors, use Step 3(a) to write the decomposition as Where A, B, and C are constants that need to be found. Multiply both sides of the equation (1) by x(x + 2)(x 2) to obtain 5x 2 = A(x + 2)(x 2) + Bx(x 2) + Cx(x + 2). Equation (1) is an identity as is equation (2). Equation (1) holds for all values of x except 0, 2, and 2. However, equation (2) holds for all values of x.
Distinct Linear Factors continued Substituting 0 for x in equation (2), A = . When choosing x = 2, B = . When choosing x = 2, C = 1. The remainder rational expression can be written as the following sum of partial factions: And the given rational expression can be written as This can be checked by combining the terms on the right.
Distinct Linear and Quadratic Factors Find the partial fraction decomposition of Solution: The denominator has distinct linear and quadratic factors, where neither is repeated. Since x2 + 2 cannot be factored, it is irreducible. The partial fraction decomposition is Multiply both sides by (x + 1)(x2 + 2) to get x2 + 3x 1 = A(x2 + 2) + (Bx + C)(x + 1). (1)
Distinct Linear and Quadratic Factors continued First, substitute 1 for x to get (1)2 + 3(1) 1 = A[(1)2 + 2] + 0 3 = 3A A = 1. Replace A, with 1 in equation (1) and substitute any value for x. For instance, if x = 0, then 02 + 3(0) 1 = 1(02 + 2) + (B 0 + C)(0 + 1) 1 = 2 + C C = 1. Now, letting A = 1 and C = 1, substitution again in equation (1), using another number for x.
Distinct Linear and Quadratic Factors continued For x = 1, 3 = 3 + (B + 1)(2) 6 = 2B + 2 B = 2. Using A = 1, B = 2, and C = 1, the partial fraction decomposition is Again this work can be checked by combining terms on the right side.
Repeated Quadratic Factors Find the partial fraction decomposition of Solution: This expression has both a linear factor and a repeated quadratic factor. By the previously stated steps, 3(a) and 4(b), Multiplying both sides by (x2 + 1)2(x 1) leads to 2x = (Ax + B)(x2 + 1)(x 1)+ (Cx + D)(x 1)+ E(x2 + 1)2. (1)
Repeated Quadratic Factors continued If x = 1, then equation (1) reduces to 2 = 4E, or E = . Substituting for E in equation (1) and combining terms on the right gives To get additional equations involving the unknowns, equate the coefficients of like powers of x on the sides of equation (2). Setting corresponding coefficients of x4 equal, From the corresponding coefficients of x3, 0 = A + B. Since A = , B = . Using the coefficients of x2, 0 = A B + C + 1.
Repeated Quadratic Factors continued Since A = and B = , C = 1. Finally, from the coefficients of x, 2 = A + B + D C. Substituting for A, B, and C gives D = 1. With the given fraction has the partial decomposition
Decomposition into Partial Fractions Method 1 For Linear Factors Multiply both sides of the rational expression by the common denominator. Substitute the zero of each factor in the resulting equation. For repeated linear factors, substitute as many other numbers as necessary to find all the constants in the numerators. The number of substitutions required will equal the number of constants A, B,….
Decomposition into Partial Fractions continued Method 2 For Quadratic Factors Multiply both sides of a rational expression by the common denominator. Collect like terms on the right side of the resulting equation. Equate the coefficients of the like terms to get a system of equations. Solve the system to find the constants in the numerators.
5.5 Nonlinear Systems of Equations
Nonlinear System A system of equations in which at least one equation is not linear is called a nonlinear system. Solve the system When one of the equations in a nonlinear system is linear, it is usually best to begin by solving the linear equation for one of the variables.
Nonlinear System continued Solve (2) for y. y = 2 x Substitute this result for y in equation (1).
Nonlinear System continued Substitute for x in equation (2) gives y = 2 and y = 3. The solution set of the given system is {(4, 2), (1, 3)}.
Example Find y by substituting back into equation (1). Solve the system. Add the two equations to eliminate y. Find y by substituting back into equation (1).
Example continued The solutions of the given system are:
Example Solve the system. Solution
Example continued Substitute y into either equation (1) or (2).
Example continued Substitute the x-values into y = 4/x to find y values. The solution set of the system is
Example Solve the system. Solve equation (2) for |x| gives |x| = 3 y. Since |x| > 0 for all x, 3 y 0 and thus y 3. In equation (1), the first term is x2, which is the same as |x|2.
Example continued If y = 0, then If y = 3, then Solve for corresponding x values, use either equation (1) or (2). If y = 0, then If y = 3, then
Example continued The solution set, {(3, 0), (3, 0), (0, 3)}, includes the points of intersection shown in the figure at the right. Be sure to check the solutions in the original system.
Application Find two numbers whose sum is 17 and whose product is 42. Step 1: Read the problem. Step 2: Assign variables. Let x represent one number and y represent the other number. Step 3: Write a system of equations. The sum of the two numbers is 17: x + y = 17 The product of the two numbers is 42: xy = 42
Application continued Step 4 Solve the system. Solve equation (1) for y. y = 17 x, and substitute into equation (2). Step 5 State the answer. The two numbers whose sum is 17 and whose product is 42 are 3 and 14. Step 6 Check the answers.
5.6 Systems of Inequalities and Linear Programming
Definitions A line divides a plane into three sets of points: the points of the line itself and the points belonging to the two regions determined by the line. Each of these two regions is called a half-plane. The line is called the boundary.
Graphing Inequalities For a function f, the graph of y < f(x) consists of all the points that are below the graph of y = f(x); the graph of y > f(x) consists of all the points that are above the graph of y = f(x). If the inequality is not or cannot be solved for y, choose a test point not on the boundary. If the test point satisfies the inequality, the graph includes all points on the same side of the boundary as the test point. Otherwise, the graph includes all points on the other side of the boundary.
Example Graph x + 2y < 6. Solution The boundary is a straight line x + 2y = 6. Since the points on the line do not satisfy the inequality the line is dashed. To decide which half-plane represents the solution set, solve for y.
Example continued Since y is less than, the graph of the solution set is the half-plane below the boundary. As a check, choose a point not on the boundary line and substitute into the inequality. Since the point (0, 0) is below the boundary line, the points that satisfy the inequality must be contained in the region.
Systems of Inequalities The solution set of a system of inequalities is the intersection of the solution sets of its members. Graph all solution sets on the same coordinate axes and identify, by shading, the region common to all graphs.
Example Graph the solution set of the system. Solve each equation for y. Graph the solution set for each equation.
Example continued Be sure to use the correct lines (dashed or solid) for each graph.
Example Graph the solution set of the system. If you graph each equation and shade appropriately you can see that the three inequalities have NO points in common.
Fundamental Theorem of Linear Programming If an optimal value for a linear programming problem exists, it occurs at the vertex of the region of feasible solutions.
Solving a Linear Programming Problem Step 1 Write the objective function and all necessary constraints. Step 2 Graph the region of feasible solutions. Step 3 Identify all vertices or corner points. Step 4 Find the value of the objective function at each vertex. Step 5 The solution is given by the vertex producing the optimal value of the objective function.
Example A tray of banana muffins requires 4 cups of milk and 3 cups of wheat flour. A tray of pumpkin muffins requires 2 cups of milk and 3 cups of wheat flour. There are 16 cups of milk and 15 cups of wheat flour available, and the baker makes $3 per tray profit on banana muffins and $2 per tray profit on pumpkin muffins. How many trays of each should the baker make in order to maximize profits?
Example continued Step 1 We let x = the number of banana muffins and y = the number of pumpkin muffins. Then the profit P is given by the function P = 3x + 2y We know that x muffins require 4 cups of milk and y muffins require 2 cups of milk. Since there are no more than 16 cups of milk, we have one constraint. 4x + 2y 16 Similarly, the muffins require 3 and 3 cups of wheat flour. There are no more than 15 cups of flour available, so we have a second constraint. 3x + 3y 15
Example continued We also know x 0 and y 0 because the baker cannot make a negative number of either muffin. Thus we want to maximize the objective function P = 3x + 2y subject to the constraints 4x + 2y 16, 3x + 3y 15, x 0, y 0 We graph the system of inequalities and determine the vertices. Next, we evaluate the objective function P at each vertex.
Example continued P = 3(3) + 2(2) = 13 (3, 2) P = 3(0) + 2(5) = 10 (0, 5) P = 3(4) + 2(0) = 12 (4, 0) P = 3(0) + 2(0) = 0 (0, 0) Profit P = 3x+ 2y Vertices Maximum The baker will make a maximum profit when 3 tray’s of banana muffins and 2 trays pumpkin muffins are produced.
5.7 Properties of Matrices
Basic Definitions It is customary to use capital letters to name matrices. Also, subscript notation is often used to name elements of a matrix, as shown. A n n matrix is a square matrix. A matrix with just one row is a row matrix. A matrix with just one column is a column matrix. Two matrices are equal if they are the same size and if corresponding elements, position by position, are equal.
Example Find the values of the variables which makes the statement true. From the definition of equality, the only way that the statement can be true is if a = 3, b = 4, x = 2 and y = 7.
Addition and Subtraction of Matrices To add two matrices of the same size, add corresponding elements. Only matrices of the same size can be added. If A and B are two matrices of the same size, then A B = A + (B).
Examples Add and subtract the following. Add Subtract
Examples continued Add or subtract, if possible. a) b) The matrices have different sizes so they cannot be added or subtracted.
Properties of Scalar Multiplication If A and B are matrices of the same size and c and d are scalars, then (c + d)A = cA + dA c(A)d = cd(A) c(A + B) = cA + cB (cd)A = c(dA) Example: Find the product.
Matrix Multiplication If the number of columns of an m n matrix A is the same as the number of rows of an n p matrix B (i.e., both n). The element cij of the product matrix C = AB is found as follows: Matrix AB will be an m p matrix.
Example Suppose A is a 3 4 matrix, while B is a 4 2 matrix. a) Can the product AB be calculated? b) If AB can be calculated, what size is it? c) Can BA be calculated? d) If BA can be calculated, what size is it?
Solutions a) AB can be calculated, because the number of columns of A is equal to the number of rows of B. 3 4 4 2 b) The product is a 3 2 matrix. c) BA cannot be calculated, the number of columns and rows do not match. d) Since BA cannot be calculated we cannot determine the size. must match Size of AB
Multiply the Matrices For find each of the following. a) AB b) BA c) AC
Solution AB A is a 2 3 matrix and B is a 3 2 matrix, so AB will be a 2 2 matrix.
Solution BA B is a 3 2 matrix and A is a 2 3 matrix, so BA will be a 3 3 matrix.
Solution AC The product AC is not defined because the number of columns of A, 3, is not equal to the number of rows of C, 2. Note that AB BA. Multiplication of matrices is generally not commutative.
Properties of Matrix Multiplication If A, B, and C are matrices such that all of the following products and sums exist, then (AB)C = A(BC) A(B + C) = AB + AC (B + C)A = BA + CA.
5.8 Matrix Inverses
Identity Matrices By the identity property for real numbers, a 1 = a and 1 a = a for any real number a. If there is to be a multiplicative identity matrix I, such that AI = A and IA = A for any matrix A, then A and I must be square matrices of the same size.
Identity Matrices 2 2 Identity Matrix If I2 represents the 2 2 identity matrix, then
Identity Matrices
Stating and Verifying the 3 3 Identity Matrix Example: Let Give the 3 3 identity matrix I and show that AI = A. Solution: By the definition of matrix multiplication,
Multiplicative Inverses If A is an n n matrix, then its multiplicative inverse, written A1, must satisfy both AA1 = In and A1A = In This means that only a square matrix can have a multiplicative inverse. Caution: Although for any nonzero real number a, if A is matrix, .
Finding an Inverse Matrix To obtain A1 for any n n matrix A for which A1 exists, follow these steps. Step 1 Form the augmented matrix where In is the n n identity matrix.. Step 2 Perform row transformations on to obtain a matrix of the Step 3 Matrix B is A1.
Example Find A1 if A = Solution: Use row transformations as follows. Step 1 Write the augmented matrix Step 2 Since 1 is already in the upper left hand corner, we begin by using row transformation that will result in 0 for the first element in the second row.
Example continued R2 2R1 R2 R3 R1 R3 R3 + R1 The following is done to obtain 0 as the first element in the third row. The following is done to obtain 1 as the third element in the third row. The following is done to obtain 0 for the third element in the first row. R2 2R1 R2 R3 R1 R3 R3 + R1
Example continued The following is done to obtain 0 for the third element in the second row. Step 3 The last transformation shows that the inverse is R2 R3
Solution of the Matrix Equation AX = B If A is an n n matrix with inverse A1, X is an n 1 matrix of variables, and B is an n 1 matrix, then the matrix equation AX = B has the solution GX = A1 B.
Solving Systems of Equations Using Matrix Inverses Example: 3x + 4y = 5 5x + 7y = 9 Solution: To represent the system as a matrix equation, use one matrix for the coefficients, one for the variables, and one for the constants, as follows The system can then be written in matrix form as the equation AX = B, since
Solving Systems of Equations Using Matrix Inverses continued To solve the system, first find A1. Then find A1 B. Since X = A1 B, The final matrix shows the solution set of the system is {(1, 2)}