Name: Section Registered In: Math 125 Exam 1 Version 1 February 21, 2006 60 points possible 1. (a) (3pts) Define what it means for a linear system to be inconsistent. Solution: A linear system is inconsistent if it does not have any solutions. (b) (2pts) Clearly explain why the system x + 2y + 3z = 2 2x + 3y + 5z = 3 3x + 5y + 8z = 6 is inconsistent. Solution: Forming the augmented matrix, we get 1 2 3 2 2 3 5 3 3 5 8 6 rref 1 0 1 0 0 1 1 0 0 0 0 1. The Best Answer: Using the consistency theorem, we see that this system is inconsistent since the row rank of the coefficient matrix is 2 and the row rank of the augmented matrix is 3. Since these row ranks are not equal, we know that there is no solution. A solution: In order for this system to be consistent, the last equation of the rref matrix requires that 0 = 1. Since this is clearly not possible, we know that the system is inconsistent.
2. (5pts) Clearly use a lines-of-constancy argument to determine at which points (A, B, C, or D) the function z = 4x y attains its maximum and minimum values over the following feasibility region. Again, clearly justify your conclusions. B 4 2 C -2 2 4 6-2 A -4 D Solution: To start, we take the objective function z = 4x y and solve for the y variable. This results in slope-intercept form y = 4x z. We graph a few lines by choosing varying values of z and examine how the lines move as we increase and decrease values of z. z = 0 : y = 4x z = 4 : y = 4x 4
Graphing these lines over the feasibility region, we observe the following: as z increases, 10 7.5 B 5 z 4 2.5 C A -2 2 4 6 z 0-2.5-5 D the y-intercept of the lines of constancy move downward. Due to the slope of the line, as z increases the last line of constancy that intersects the feasibility region will intersect it at the point A. Similarly, as z decreases, the last line of constancy that intersects the feasibility region will intersect the point D. Therefore, the maximum value of z is attained at the point A and the minimum value of z is attained at the point D.
3. (a) (3pts) Precisely state the Consistency Theorem. Solution: A linear system is consistent if and only if the row rank of the coefficient matrix equals the row rank of the augmented matrix. (b) (7pts) Find a row echelon form for the linear system 2 1 1 a 1 1 2 b 0 3 3 c and determine the conditions that a, b, and c must satisfy in order for the system to be consistent. Solution: R 1 R 2 1 1 2 b 2 1 1 a 0 3 3 c R 2 = R 2 2R 1 1 1 2 b 0 3 3 a 2b 0 3 3 c R 2 = 1 3 R 2 1 1 2 b 0 1 1 (a 2b)/3 0 3 3 c 1 1 2 b R 3 = R 3 3R 2 0 1 1 (a 2b)/3 0 0 0 a 2b + c By the consistency theorem, since the row rank of the coefficient matrix is 2, we require that the row rank of the augmented matrix be 2 as well. This will occur when a 2b + c = 0.
4. (10pts) A local store budgets no more than $1000 per month for newspaper and radio advertising. The News-Gazette charges $50 per ad and requires at least four ads per month. WHMS-FM 97.5 (home of the Illini Sports Network) charges $100 per minute of radio ad and requires a minimum of 5 minutes of advertising per month. It is estimated that each newpaper ad reaches 8,000 people and that each minute of radio advertising reaches 15,000 people. Set up but do not solve a linear program that enables the store to reach the maximum number of people. Solution: Let x = the number of print ads purchased from the News-Gazette and y = the number of minutes of advertising purchased from WHMS-FM. The objective here is to maximize the audience of the ads. P = 8000x + 15000y. The people function is Since there is only at most a $1000 to spend each month, we get the inequality 50x + 100y 1000. Similarly, due to the minimum ad purchase requirements by each company, we get the inequalities x 4 and y 5. The above discussion accurately describes the linear program. represent this in its mathematical form: One could choose to maximize P = 8000x + 15000y subject to 50x + 100y 1000 x 4 y 5 (Note: You did not have to order the linear program like this for credit.)
5. (a) (4pts) Define the elementary row operations. Solution: 1. interchange two rows 2. multiply a row by a nonzero constant 3. add a multiple of one row to another row (b) (2pts) What property of the elementary row operations allow one to use them to solve a system of equations? Solution: The importance of the elementary row operation is that the new system of equations (the augmented matrix after a row op.) has the exact same solution as the original system of equations. (c) (5pts) Use a calculator to determine the solution set for the linear system 2 3 1 2 3 4 4 4 1 4 11 4. 2 5 2 2 1 9 0 2 1 0 4 5 Solution: This matrix in reduced row echelon form is 1 0 1/4 1 0 25/4 0 1 1/2 0 0 9/2 0 0 0 0 1 1 0 0 0 0 0 0. First, we notice that the system is consistent since the row rank of the coefficient matrix and augmented matrix is 3. Since the third variable column (the x 3 -column) doesn t have a leading 1 in it, we define x 3 = s. Similarly, the fourth variable column also does not have a
leading 1 in it. (Remark: That is not a leading 1 at the top of the 4th column. Leading 1 s are for rows, not columns.) Therefore, we define x 4 = t The above matrix then yields the resulting system of equations: x 1 + 1 4 s + t = 25 4 x 2 + 1 2 s = 9 2 x 5 = 1 This results in the solution set ( 1 4 s t 25 4, 1 2 s 9, s, t, 1) for all s and t. 2
6. (a) (2pts) Write the equation of the plane 7x + 4y + 14z = 28 in intercept form. Solution: x 4 + y 7 + z 2 = 1. (b) (3pts) Sketch the first-octant portion of the plane by drawing lines between the intercepts. Solution: Recall the advantage of the equation being in intercept form is that it is significantly easier to find the intercepts of the plane. The x-intercept is (4, 0, 0), the y-intercept is (0, 7, 0) and the z-intercept is (0, 0, 2). Plotting these points on the axis in the first-octant and connecting the line segments between the intercepts, we find the following portion of the plane: 6 z 4 2 6 4 x 2 0 0 2 4 y 6 8
7. (a) (3pts) Define a homogeneous system of equations. Solution: A system of equation is homogeneous if every equation in the system equals zero. (b) (2pts) Explain why a homogeneous system is always consistent. Solution: A system is consistent if it has one or more solutions. Since the zero solution (or the trivial solution) always solves the homogeneous system, the system must be consistent.
8. (5pts) Graph the feasibility region defined by the following system of inequalities. 2x 3y 3 4x + 2y 6 4x + 6y 2 Solution: To graph the half-plane defined by 2x 3y 3, we start by graphing the associated line 2x 3y = 3. This line has the intercepts (0, 1) and ( 3/2, 0) and is represented by the red line below. Since the point (0, 0) is not on the line, we use it as the -5-4 -3-2 -1 1 2 x -1-2 -3 y 4 3 2 1-4 test point to determine which side of the half-plane the inequality defines. Notice that 2(0) 3(0) = 0 is not 3. Therefore, the origin is not in the half-plane and we shade the other region.
-5-4 -3-2 -1 1 2 x -1-2 -3 y 4 3 2 1-4 Similarly, graphing the line 4x+2y = 6 and noticing that the origin is in the half-plane, we get the following combined region: y 6 4 2-5 -4-3 -2-1 1 2 x -2-4
Lastly, graphing the line 4x + 6y = 2 and noticing that the origin again lies in the half-plane, results in the feasibility region for the entire system of inequalities: y 4 3 2 1-5 -4-3 -2-1 1 2 x -1-2 -3-4
9. (5pts) Determine if the following statement is true or false. For each part, you will receive 1 point for a correct answer, 1 point for an incorrect answer, and 0 points for no answer. The lowest possible score on this problem is zero. (a) Every elementary row operation is reversible. Solution: TRUE. Notice that R i R j reverses itself, R i = 1 k R i reverses R i = kr i for all k 0, R j = R j kr i reverses R j = R j + kr i. (b) If a 3 5 coefficient matrix for a system has three leading 1 s, then it is consistent. Solution: TRUE. Since there are 3 leading 1 s, there are 3 non-zero rows in the coefficient matrix. Hence, by definition, the coefficient matrix has the row rank 3. Since there are only three rows in the associated augmented matrix and all the leading 1 s are in the coefficient matrix, the augmented matrix also must have row rank 3. By the Consistency Theorem, the system is consistent. (c) solutions. Whenever a consistent system has free variables, the solution set contains many Solution: TRUE. Whenever there is a free variable, the solution set is parametric. Parametric solution sets contain an infinite number of points. (d) In some cases, by using different sequences of row operations, a matrix may be row reduced to more than one matrix in reduced row echelon form. Solution: FALSE. This was the point of Homework 2, Problem 5. There is only one reduced row echelon form for any given matrix. (e) Gaussian elimination applies only to augmented matrices for a linear system. Solution: FALSE. Gaussian elimination is a method for turning any matrix or table of num-
bers into row echelon form. When used on an augmented system, it helps one determine the solutions of the given system. But there is nothing in the definition of Gaussian elimination that requires that you only reduce an augmented matrix.
Name: Section Registered In: Math 125 Exam 1 Version 2 February 21, 2006 60 points possible 1. (a) (3pts) Define the row rank of a matrix. Solution: The row rank of a matrix is the number of non-zero rows the matrix has after it is in row echelon form. 1 2 1 2 2 2 1 2 3 2 (b) (2pts) Determine the row rank of the matrix. 1 2 3 4 5 4 5 4 1 6 1 0 0 25/6 17/12 0 1 0 7/3 2/3 Solution: The reduced row echelon form of this matrix is. Three 0 0 1 3/2 3/4 0 0 0 0 0 nonzero rows tell us that the row rank is 3.
2. (a) (3pts) Explain the difference between row echelon form and reduced row echelon form. Solution: A matrix in reduced row echelon form is a matrix that is in row echelon form but also has the addition property that above every leading 1 are zeros. (b) (2pts) Determine whether the given matrix is in reduced row echelon form, row echelon form, or neither. Clearly explain your answer. 0 0 0 0 0 0 0 1 2 3 0 0 0 1 0 0 0 0 0 0 Solution: NEITHER. This matrix is not in row echelon form (hence, can not be in reduced row echelon form) since, by definition of row echelon form, all rows of all zeros must be on the bottom of the matrix.
maximize z = 12x 15y subject to 3x y 4 3. Consider the linear program: 2x + 3y 6 x 0, y 0 (a) (6pts) Graph the feasibility region and find the corner points. Solution: Begin by graphing the line 3x y = 4, which has the intercepts (0, 4) and (4/3, 0). (Graphically, this is the red-line.) Using the point (0, 0) as a test point, we have the question Is 3(0) (0) = 0 4? No, it isn t. Therefore, the half-plane defined by 3x y 4 is the region possessing and below the red line. 3 2 y 1-2 -1 1 2 3 4 x -1-2 -3-4
Similarly, graphing 2x + 3y = 6 (using the intercepts (0, 2) and (3, 0)) we draw the blue line and find that the half plane defined by the in 2x + 3y 6 is the region including and below the line. Combining the two half planes and the fact that x and y must be non-negative, we find the following feasibility region: 2 1-2 -1 1 2 3 4 x -1-2 -3 y 3-4 As to the corner points, there are three for this region. The first is the intercept (4/3, 0). The second is the intercept (3, 0). The third occurs at the intersection of the two lines: (18/11, 10/11). (b) (4pts) Solve the linear program. Solution: Since we already have the corner points of the polygonal feasibility region, the easiest way to solve the linear program is to evaluate z at the corner points. z(3, 0) z = 12(3) 15(0) = 36 ( ) 4 z 3, 0 z = 16 ( 18 z 11, 10 ) z = 6 11 The conclusion is that the maximum z value is 36 and occurs at the point (3, 0).
4. (10pts) The Annual Springfield Dirt Bike Competition is coming up, and participants are looking for bikes. Apu has 18 wheels, 15 seats, and 14 exhaust pipes in his supply room. He can use these parts to assemble two different types of bikes: the Rider, or the Rover. The Rider consists of 2 wheels, 1 seat, and 2 exhaust pipes. The Rover has 3 wheels, 3 seats, and 1 exhaust pipe. Suppose Apu makes a profit of $15 for each Rider and $30 for each Rover. Assuming he can sell all that he manufactures, set up but do not solve a linear program that will maximize profits. Solution: Let x = the number of Riders and y = the number of Rovers manufactured by Apu. The objective here is to maximize profits. The profit function is P = 15x + 30y. Since there are only 18 wheels to use in manufacturing, we get the inequality 2x+3y 18. Since there are only 15 seats to use in manufacturing, we get the inequality x + 3y 15. Lastly, since there are only 14 exhaust pipes to use in manufacturing, we get the inequality 2x + y 14. We also know that our variables need to be non-negative. (It doesn t make sense to manufacture a negative number of bikes.) So we get the final inequalities, x 0 and y 0. The above discussion accurately describes the linear program. represent this in its mathematical form: One could choose to maximize P = 15x + 30y subject to 2x + 3y 18 x + 3y 15 2x + y 14 x 0, y 0 (Note: You did not have to order the linear program like this for credit.)
5. (a) (3pts) Explain the difference between Gaussian Elimination and Gauss-Jordan Elimination. Solution: Gaussian elimination is a technique that moves any matrix into row echelon form. It places the leading 1 s where they belong and annihilates all entries below the leading 1 s. Gauss-Jordan moves a matrix directly into reduced row echelon form by placing non-zero entries where the leading 1 s need to be and pivoting on them. (b) (7pts) Use Gaussian Elimination with back substitution to solve the following linear system. x + 2y 3z = a 2x + 3y + 3z = 1 5x + 9y 9z = 7 Solution: augmented 1 2 3 a 2 3 3 1 5 9 9 7 R 2 = R 2 2R 1 R 3 = R 3 5R 1 1 2 3 a 0 1 9 1 2a 0 1 6 7 5a R 2 = R 2 R 3 = R 3 + 5R 2 1 2 3 a 0 1 9 1 + 2a 0 0 3 6 3a R 3 = 1 3 R 3 1 2 3 a 0 1 9 1 + 2a 0 0 1 2 + a
Now that we are in row echelon form, we know we have completed Gaussian elimination. In order to begin back substitution we write the system of equations associated with the ref matrix. x + 2y 3z = a y 9z = 1 + 2a z = 2 + a We have already solved for z. To solve for y, substitute z = 2 + a into the second equation. y 9( 2 + a) = 1 + 2a y + 18 9a = 1 + 2a y = 19 + 11a Now, solve for x. x + 2( 19 + 11a) 3( 2 + a) = a x 38 + 22a + 6 3a = a x 32 + 19a = a x = 32 18a So the solution is x = 32 18a y = 19 + 11a z = 2 + a or the solution point (32 18a, 19 + 11a, 2 + a).
6. (a) (3pts) Find the point(s) of intersection, if any, of the following planes. 3x + 3y + 12z = 6 x + y + 4z = 2 2x + 5y + 20z = 10 x + 2y + 8z = 4 Solution: We need to form and move the augmented matrix into reduced row echelon form. 3 3 12 6 1 1 4 2. 2 5 20 10 1 2 8 4 Using the calculator to move this to reduced row echelon form, we get the matrix 1 0 0 0 0 1 4 2. 0 0 0 0 0 0 0 0 First we note that the system is consistent since the row rank of the coefficient matrix equals the row rank of the augmented matrix. Since the x 3 columns has no leading 1 in it, it becomes a parameter. Let x 3 = t. Then the reduced row echelon form corresponds to the system of equations x 1 = 0 x 2 + 4t = 2 Solving these equations for the variables, we get the solution set (0, 4t + 2, t) for all t. (b) (2pts) Identify the intersection of the planes as the empty set, a point, a line, or a plane. Justify your answer. Solution: Since the solution set has one parameters in it, we know that the solution set must define a line.
7. In class (and in the homework), we have shown that a two-dimensional linear system has either no solutions, exactly one solution or an infinite number of solutions. Recall that an arbitrary two-dimensional linear system has the form ax + by = k cx + dy = l where the values of a, b, c, d, k and l are any constants. (a) (3 pts.) Give an example of an augmented matrix that corresponds to a two-dimensional linear system with no solution and, by drawing a sketch, graphically display what is happening to cause the system to not have a solution. Solution: Any system that represents two distinct parallel lines is a solution to this problem. Lesson 2.3, Example 2(b) is an example. (b) (2 pts.) Give an example of an augmented matrix that corresponds to a twodimensional linear system with exactly one solution and, by drawing a sketch, graphically demonstrate what is happening in a linear system with a unique solution. Solution: Any system that represents any two lines of differing slopes is a solution to this problem. Lesson 2.3, Example 2(a) is an example.
8. (5pts) Use Gauss-Jordan Elimination to perform a pivot about the position indicated by the asterisk : 0 2 1 a 1 6 10 b 3 6 3 9 0 3 1 d Solution: R 3 = 1 3 R 3 0 2 1 a 1 6 10 b 1 2 1 3 0 3 1 d R 1 = R 1 + R 3 R 2 = R 2 + 10R 3 R 4 = R 4 R 3 1 4 0 a 3 9 26 0 b 30 1 2 1 3 1 1 0 d + 3
9. (5pts) Determine if the following statement is true or false. For each part, you will receive 1 point for a correct answer, 1 point for an incorrect answer, and 0 points for no answer. The lowest possible score on this problem is zero. (a) Elementary row operations on an augmented matrix never change the solution set of the associated linear system. Solution: TRUE. That is the exact reason that we are able to use them to solve systems of equations. (b) Gauss-Jordan elimination applies only to augmented matrices for a linear system. Solution: FALSE. Gauss-Jordan elimination is a method for turning any matrix or table of numbers into reduced row echelon form. When used on an augmented system, it helps one determine the solutions of the given system. But there is nothing in the definition of Gauss-Jordan elimination that requires that you only reduce an augmented matrix. (c) The row echelon form of a matrix is unique. Solution: FALSE. This was the point of Homework 2, Problem 5. Although there is only one reduced row echelon form for any given matrix. There can be an infinite number or row echelon forms associated with a matrix. It depends on the sequence of row operations used to move the matrix to row echelon form. (d) If a 3 5 coefficient matrix for a system has two leading 1 s, then it is inconsistent. Solution: FALSE. Knowing that the coefficient matrix of has a bottom row of all zeros, tells us nothing about the row rank of the augmented matrix. If there is a leading one after the augment, then the system is inconsistent. If there is a zero after the augment, then the system is consistent. There is not enough information here to be able to state with certainty that the system is inconsistent.
(e) A homogeneous system is always consistent. Solution: TRUE. A system is consistent if it has one or more solutions. Since the zero solution (or the trivial solution) always solves the homogeneous system, the system must be consistent.