6-1 Study Guide and Intervention Multivariable Linear Systems and Row Operations

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1 6-1 Study Guide and Intervention Multivariable Linear Systems and Row Operations Gaussian Elimination You can solve a system of linear equations using matrices. Solving a system by transforming it into an equivalent system is called Gaussian elimination. First, create the augmented matrix. Then use elementary row operations to transform the matrix so that it is in row-echelon form. Then write the corresponding system of equations and use substitution to solve the system.. Example: Solve the system of equations using Gaussian elimination with matrices. x 2y + z = 1 2x + y 3z = 7 3x y + 2z = 0 Step 1 Write the augmented matrix [ ] Step 2 Apply elementary row operations to obtain a row echelon form of the matrix. a. b. c. 1 R 2 2R 1 [ ] [ ] R 5 2 [ ] R 3 3R d. e. [ ] R 3 5R [ ] 1 R Step 3 Write the corresponding system of equations and use substitution to solve the system. x 2y + z = 1 y z = 1 z = 2 The solution of the system is x = 1, y = 1, and z = 2 or ( 1, 1, 2). Solve each system of equations using Gaussian elimination with matrices. 1. 2x y + z = x y = x y z = 11 x + 2y z = 3 3x + 2y z = 8 x z = 2 3x + y 2z = 1 x 4y + z = 10 2y + 4z = 0 Chapter 6 5 Glencoe Precalculus

2 6-1 Study Guide and Intervention (continued) Multivariable Linear Systems and Row Operations Gauss-Jordan Elimination If you continue to apply elementary row operations to the rowechelon form of any augmented matrix, you can obtain a matrix in which every column has one element equal to 1 and the remaining elements equal to 0. This is called the reduced rowechelon form of the matrix. Solving a system by transforming an augmented matrix so that it is in reduced row echelon form is called Gauss Jordan elimination [ a b] c Example : Solve the system of equations. x 2y + z = 15 2x y + 2z = 1 x + y = 9 Write the augmented matrix. Apply elementary row operations to obtain a row echelon form. Then apply elementary row operations to obtain zeros above the leading 1s in each row. Augmented Matrix [ ] 2R 1 + R 2 [ ] [ ] R 1 + R R 2 6R 3 [ ] R 2 + 2R 3 [ ] Row echelon form [ ] R 2 + R R 1 + 2R [ ] [ ] R Reduced row-echelon form R 1 R [ ] The solution of the system is x = 4, y = 5, and z = 1 or (4, 5, 1). Solve each system of equations using Gaussian or Gauss Jordan elimination. 1. 3x 2y + z = x 4z = x y z = 1 4x + z = 31 2y + 3z = 2 x + 3y 2z = 24 2x 5y = 24 2x 5y = 6 4x + 2y + z = 2 Chapter 6 6 Glencoe Precalculus

3 6-2 Study Guide and Intervention Matrix Multiplication, Inverses, and Determinants Multiply Matrices To multiply matrix A by matrix B, the number of columns in A must be equal to the number of rows in B. If A has dimensions m r and B has dimensions r n, their product, AB, is an m n matrix. If the number of columns in A does not equal the number of rows in B, the matrices cannot be multiplied. a b f + bg af + bh [ ] [e ] =[ae c d g h ce + dg cf + dh ] 2 3 Example: Use matrices A = [ ] and B =[ 1 ]to find AB, if possible AB = [ 1 3 ] [ ] A is a 2 2 matrix and B is a 2 3 matrix. Because the number of columns for A is equal to the number of rows for B, the product AB exists. To find the first entry in AB, write the sum of the products of the entries in row 1 of A and in column 1 of B. [ 1 3 ] [ ]= [4( 1) + ( 2)( 2)] Follow this same procedure to find the entry for row 1, column 2 of AB. [ 1 3 ] [ ]= [4( 1) + ( 2)( 2)4(2) + ( 2)(4)] Continue multiplying each row by each column to find the sum for each entry. [ 1 3 ] [ ( 2)( 2) 4(2) + ( 2)(4) 4(3) + ( 2)( 1) ]= [4( 1) ( 1)( 1) + 3( 2) ( 1)(2) + 3(4) ( 1)(3) + 3( 1) ] Then simplify each sum. [ 1 3 ] [ ]= [ ] Find AB and BA, if possible A = [ ], B = [ 1 ] 2. A = [ 1 ], B = [ ] Chapter 6 11 Glencoe Precalculus

4 6-2 Study Guide and Intervention(continued) Matrix Multiplication, Inverses, and Determinants Inverses and Determinants The identity matrix is an n n matrix consisting of all 1s on its main diagonal, from upper left to lower right, and 0s for all other elements. LetI n be the identity matrix of order n and let A be an n n matrix. If there exists a matrix B such that AB = BA = I n, then B is called the inverse of A and is written as A 1. If a matrixhas an inverse, it is invertible. The determinant of a 2 2 matrix can be used to determine whether or not a matrix is invertible. If A =[ a b c d ], det(a) = ad cb. If ad cb 0, then A 1 = 1 [ d b ad cb c a ]. 7 4 Determine whether A = [ 5 3 ] and B = [3 4 ]are inverse matrices. 5 7 If A and B are inverse matrices, then AB = BA = I. 7 4 AB = [ 5 3 ] [3 4 + ( 4)(5) 7(4) + ( 4)(7) 0 ] = [7(3) ] or [ (3) + 3(5) 5(4) + 3(7) 0 1 ] BA = [ ] [ ( 5) 3( 4) + 4(3) 0 ] = [3(7) ] or [ (7) + 7( 5) 5( 4) + 7(3) 0 1 ] Because AB = BA = I, B = A 1 and A = B 1. Find the determinant of A = [ ].Then find A 1, if it exists. det(a) = A 1 = 1 [ ] = 2( 6) 3( 2) or 6 = [ ] Since det(a) 0, A is invertible. Determine whether A and B are inverse matrices. Explain your reasoning. 1. A=[ ], B = [ 1 5 ] 2.A = [3 2 5 ], B = [ ] Find the determinant of A = [ 10 2 ]. Then find A 1, if it exists. 4. Find the determinant of A = [ ]. Then find A 1, if it exists. Chapter 6 12 Glencoe Precalculus

5 6-3 Study Guide and Intervention Solving Linear Systems Using Inverses and Cramer s Rule Use Inverse Matrices A square system has the same number of equations as variables. If a square matrix has an inverse, the system has one unique solution. Example : Use an inverse matrix to solve each system of equations, if possible. a. 3x 7y = 16 x + 2y = 8 Write the system in matrix form. A X = B 3 7 [ 1 2 ] [x y ] = [ 16 8 ] Use the formula for an inverse of a 2 2 matrix to find the inverse A 1. A 1 1 = [ d b ad cb c a ] = 1 [2 7 (2)(3) ( 7)( 1) 1 3 ] b. 2x + y + z = 0 x + 2z = 9 x 2y 9z = 31 Write the system in matrix form. A X = B x 0 [ 1 0 2] [ y] = [ 9] z 31 Use a graphing calculator to find A 1. Multiply A 1 by B to solve the system. X = A 1 B 2 7 = [ 1 3 ] [ 16 8 ] = [ 24 8 ] So, the solution of the system is ( 24, 8). Multiply A 1 by B to solve the system. X = A 1 B = [ ] [ 9] = [ 2] So, the solution of the system is (1, 2, 4). Use an inverse matrix to solve each system of equations, if possible. 1. 2x + 5y = x y + 2z = 5 3x y = 10 x z = 4 3x + 2y + z = x + y = 7 4. x + y z = 5 2x 5y = 43 2x 3y + 2z = 20 y + 4z = 18 Chapter 6 16 Glencoe Precalculus

6 6-3 Study Guide and Intervention (continued) Solving Linear Systems Using Inverses and Cramer s Rule Use Cramer s Rule Another method, known as Cramer s Rule, can be used to solve a square system of equations. Let A be the coefficient matrix of a system of n linear equations in n variables given by AX = B. If det(a) 0, then the unique solution of the system is given by x 1 = A 1 A, x 2 = A 2 A, x 3 = A 3 A, x n = A n A, where Ai is the matrix obtained by replacing the ith column of A with the column of constants B. If det(a) = 0, then AX = B has either no solution or infinitely many solutions. Example: Use Cramer s Rule to find the solution of the system of linear equations, if a unique solution exists. 2x 1 + x 2 = 7 5x 1 2x 2 = 17 The coefficient matrix is A = [ 2 1 ]. Calculate the determinant of A. 5 2 A = 2 1 = ( 2) ( 2) 5(1) or Because the determinant of A does not equal zero, you can apply Cramer s Rule. x 1 = A 1 A x 2 = A 2 A = ( 2) 17(1) = = 3 or = (17) 5( 7) = = 1 or Therefore, the solution is x 1 = 3 and x 2 = 1 or (3, 1). Use Cramer s Rule to find the solution of each system of linear equations, if a unique solution exists. 1. x 2y = x 3y = 18 2x 5y = 8 x + 4y = x + y = x 4y = 2 x + 2y = 14 x + 3y = 3 Chapter 6 17 Glencoe Precalculus

7 6-4 Study Guide and Intervention Partial Fractions Linear Factors The function g(x) shown below can be written as the sum of two fractions with denominators that are linear factors of the original denominator. g(x) = 3x 1 = x 2 3x +1 x 1 2x 1 Each fraction in the sum is a partial fraction. The sum of these partial fractions make up the partial fraction decomposition of the original rational function. If the denominator of a rational expression contains a repeated linear factor, the partial fraction decomposition must include a partial fraction with its own constant numerator for each power of this factor. Example : Find the partial fraction decomposition of x + 11 x 2 3x 4. Rewrite the equation as partial fractions with constant numerators, A and B, and denominators that are the linear factors of the original denominator. x +11 = A + B x 2 3x 4 x 4 x +1 Form a partial fraction decomposition. x + 11 = A(x + 1) + B(x 4) Multiply each side by the LCD, x 2 3x 4. x + 11 = Ax + A + Bx 4B 1x + 11 = (A + B)x + (A + ( 4B)) Distributive Property Group like terms. Equate the coefficients on the left and right side of the equation to obtain a system of two equations. To solve the system, write it in matrix form CX = D and solve for X. A + B = 1 A + ( 4B) = 11 C X = D [ ] [A B ] = [ 1 11 ] Solving for X yields A = 3 and B = 2. Therefore x +11 = x 2 3x 4 x 4 x +1 Find the partial fraction decomposition of each rational expression. 1. 5x 34 x 2 x x + 13 x 2 5x x x2 x 1 2x(x 1) 2 x 2 (x 1) Chapter 6 22 Glencoe Precalculus

8 6-4 Study Guide and Intervention (continued) Partial Fractions Irreducible Quadratic Factors Not all rational expressions can be written as the sum of partial fractions using only linear factors in the denominator. If the denominator of a rational expression contains an irreducible quadratic factor, the partial fraction decomposition must include a partial fraction with a linear numerator of the form Bx + C for each power of this factor. Example : Find the partial fraction decomposition of 4x4 2x 3 13x 2 + 7x + 9 x(x 2 3) 2 This expression is proper. The denominator has one linear factor and one irreducible factor of multiplicity 2. 4x 4 2x 3 13x 2 + 7x + 9 x(x 2 3) 2 = A x Bx + C Dx + E + + x 2 3 (x 2 3) 2 4x 4 2x 3 13x 2 + 7x + 9 = A(x 2 3) 2 + (Bx + C)x(x 2 3) + (Dx + E)x 4x 4 2x 3 13x 2 + 7x + 9 = Ax 4 + Bx 4 + Cx 3 6Ax 2 3Bx 2 + Dx 2 3Cx + Ex + 9A 4x 4 2x 3 13x 2 + 7x + 9 = (A + B) x 4 + Cx 3 + ( 6A 3B + D) x 2 + ( 3C + E)x + 9A Write and solve the system of equations obtained by equating coefficients. A + B = 4 A = 1 C = 2 B = 3 6A 3B + D = 13 C = 2 3C + E = 7 D = 2 9A = 9 E = 1 Therefore, 4x4 2x 3 13x 2 + 7x + 9 x(x 2 3) 2 = 1 x + 3x + 2 x x + 1 (x 2 3) 2 Find the partial fraction decomposition of each rational expression x 3 + 5x 1 x x x x3 2x 2 8x + 5 (x 2 3) 2 3x 2 x 2 3 x 1 + (x 2 3) 2 2x 4. x3 + 2x (x 2 + 1) 2 (x 2 + 1) x3 + x x x 2 +1 (x 2 +1) 2 x x x (x 2 +1) 2 Chapter 6 23 Glencoe Precalculus

9 6-5 Study Guide and Intervention Linear Optimization Linear Programming Linear programming is a process for finding a minimum or maximum value for a specific quantity. The following steps can be used to solve a linear programming problem. Step 1 Write an objective function and a list of constraints to model the situation. Step 2 Graph the region corresponding to the solution of the system of constraints. Step 3 Find the coordinates of the vertices of the region formed. Step 4 Evaluate the objective function at each vertex to find the minimum or maximum. Example: A leather company wants to add belts and wallets to its product line. Belts require 2 hours of cutting time and 6 hours of sewing time. Wallets require 3 hours of cutting time and 3 hours of sewing time. The cutting machine is available 12 hours a week and the sewing machine is available 18 hours per week. Belts will net $18 in profit and wallets will net $12. How much of each product should be produced to achieve maximum profit? Let x represent the number of belts and y represent the number of wallets. The objective function is then given by f(x, y) = 18x + 12y. Write the constraints. x 0; y 0 Numbers of items cannot be negative. 2x + 3y 12 Cutting time 6x + 3y 18 Sewing time Graph the system. The solution is the shaded region, including its boundary segments. Find the coordinates of the four vertices by solving the system of boundary equations for each point of intersection. The coordinates are (0, 0), (0, 4), (1.5, 3), and (3, 0). Evaluate the objective function for each ordered pair. Point f(x, y) = 18x + 12y Result (0, 0) f(0, 0) = 18(0) + 12(0) 0 (0, 4) f(0, 4) = 18(0) + 12(4) 48 (1.5, 3) f(1.5, 3) = 18(1.5) + 12(3) 63 Maximum (3, 0) f(3, 0) = 18(3) + 12(0) 54 Since f is greatest at (1.5, 3), the company will maximize profit if it makes and sells 1.5 belts for every 3 wallets. Find the maximum and minimum values of the objective function f(x, y) and for what values of x and y they occur, subject to the given constraints. 1. f(x, y) = 3x 2y 2. f(x, y) = x + 2y 2x + y 10 x + y 4 x + 2y 8 x + 3y 6 x 0 x 0 y 0 y 0 Chapter 6 27 Glencoe Precalculus

10 6-5 Study Guide and Intervention (continued) Linear Optimization No or Multiple Optimal Solutions Linear programming models can have one, multiple, or no optimal solutions. If the graph of the objective function f to be optimized is coincident with one side of the region of feasible solutions, f has multiple optimal solutions. If the region does not form a polygon, but instead is unbounded, f may have no minimum value or maximum value. Example: Find the maximum value of the objective function f(x, y) = 6x + 3y and for what values of x and y it occurs, subject to the following constraints. 2x + y 8 y 4 x 3 x 0 y 0 Graph the region bounded by the given constraints. Find the value of the objective function f(x, y) = 6x + 3y at each vertex. f(0, 0) = 6(0) + 3(0) or 0 f(0, 4) = 6(0) + 3(4) or 12 f(2, 4) = 6(2) + 3(4) or 24 f(3, 2) = 6(3) + 3(2) or 24 f(3, 0) = 6(3) + 3(0) or 18 Because f(x, y) = 24 at (2, 4) and (3, 2), the problem has multiple optimal solutions. An equation of the line through these two vertices is y = 2x + 8. Therefore, f has a maximum value of 24 at every point on y = 2x + 8 for 2 x 3. Find the maximum and minimum values of the objective function f(x, y) and for what values of x and y they occur, subject to the given constraints. 1. f(x, y) = 2x + y 2. f(x, y) = 3x + 6y 2x + y 11 x - y 4 y 5 x + 2y 20 y 0 x 0 x 5 x 8 x 0 y 0 Chapter 6 28 Glencoe Precalculus