Solutions to ExercisesCollege of the RedwoodsMathematics DepartmentMath 45 — Linear AlgebraExam #3Linear AlgebraDavid ArnoldCopyrightc 2000 [email protected] Revision Date: November 27, 2001 Version 1.002Essay QuestionsRead Carefully! You have the weekend to complete the exam. The exam is due, on my desk, at thebeginning of class on Wednesday.This exam is open notes, open b ook. You may use a calculator or computer to check your work whereappropriate. You must answer all of the exercises on your own. You are not allowed to work in groups onthe exam. You are not allowed to enlist the aid of a tutor or friend to help with the exam. You are notallowed to read the exercises in the exam, then seek help on similar questions. Once you open the exam andread the questions, you may not seek any outside help of any kind. From the moment you open the exam,you must do everything by yourself.Place the solution to each exercise on a separate sheet of paper. On a good sheet of paper, write out(longhand) and sign the following honor pledge.I promise that all work found herein is my own. I have received no help from tutors, colleagues,or other teachers. I have honored all of the examination constraints listed in the directions.Arrange the problems in order, place these examination pages on top of your solutions, then place the honorpledge on top of the examination as a cover sheet. Staple. Good luck!Exercise 1. Let S be the space of solutions to the equationx1− x2+ x3− 2x4=0.(a) Find a basis for the space S.(b) Use the Gram-Schmidt orthogonalization process and the vectors found in part (a) to find an orthonormalbasis for the space S.(c) Use the basis found in part (b) to find the vector in S that is closest to the vector b =(1, 1, 1, 1)T.Exercise 2. Consider the matrixA =1 −12311175.(a) Use Gaussian elimination to find the determinant of A.(b) Use the permutation definition to find the determinant of A.(c) Use the cofactor expansion to find the determinant of A.Exercise 3. Let A and B be 3 × 3 matrices with |A| =4and|B| = −3.(a) Evaluate |ATA|.(b) Evaluate |AB−1|.(c) Evaluate |−3B|.(d) If P is an invertible 3 × 3 matrix, evaluate det(PAP−1).Exercise 4. An n × n matrix A is said to be idempotent if A2= A. Show that if λ is an eigenvalue of anidempotent matrix A, then λ must either be 0 or 1.Exercise 5. Find the eigenvalues and eigenvectors ofA =7 −105 −8.All work is to be done by hand, pencil and paper calculations only.3Exercise 6. Consider the matrixA =5 −34020−88−7.(a) Using pencil and paper calculations only, find the characteristic polynomial for matrix A.(b) Use a graphing calculator to draw the characteristic polynomial and find its roots. These are theeigenvalues. Make a copy of the graph and its roots on your examination paper.(c) Using pencil and paper calculations only, compute the eigenvectors of matrix A.Exercise 7. Consider the matrixA =.7 .2.3 .8.(a) Diagonalize the matrix A. That is, find an invertible S and a diagonal matrix D so that A = SDS−1.(b) Use the result from part (a) to find Ak. Write a 2 × 2 matrix with correct entries for this result.(c) Use the result from part (c) to findlimk→∞Ak.Solutions to Exercises 4Solutions to ExercisesExercise 1(a) Let S be the space of solutions to the equationx1− x2+ x3− 2x4=0.In matrix from,1 −11−2x1x2x3x4=0.Thus, we find a basis for the null space ofA =1 −11−2.Note that column 2 is a multiple of column 1.col 2 = −1 · col 1Thus,1 · col 1 + 1 · col 2 + 0 · col 3 + 0 · col 4 = 0and1100is a basis element for N(A). Similarly,col 3 = 1 · col 1.Thus,−1 · col 1 + 0 · col 2 + 1 · col 3 + 0 · col 4 = 0and−1010is a second basis element for N(A). Finally,col 4 = −2 · col 1Thus,2 · col 1 + 0 · col 2 + 0 · col 3 + 1 · col 4 = 0and2001is a final basis element for N(A). Thus,B =1100−10102001is a basis for the null space of A or the solution space of x1− x2+ x3− 2x4=0. Solutions to Exercises 5Exercise 1(b) Leta =1100,b=−1010, and c =2001.LetA = a =1100.LetB = b −b · AA · AA=−1010−−121100=−1010+121100=−1/21/210To ease computation, letB =2B =−1120.Note that this new value of B is still orthogonal to A.Next,C = c −c · AA · AA −c · BB · BB=2001−221100−−26−1120=2001−1100+13−1120=2/3−2/32/31Solutions to Exercises 6Again, to ease computation, letC =3C =2−221.Check that C is orthogonal to both A and B.A · C =1100·2−223=0 and B · C =−11202−223=0.Divide each vector by its length to get an orthonormal basis.qA=AA=1100√2=1/√21/√200qB=BB=−1120√6=−1/61/√62/√60qC=CC=2−223√21=2/√21−2/√212/√213/√21Exercise 1(c) To find the vector in S that is closest to b =1111T, project b onto S. To find theprojection matrix,P = Q(QTQ)−1QTP = QIQTP = QQTSetQ =[qA,qB,qC]=1/√2 −1/√62/√211/√21/√6 −2/√2102/√62/√21003/√21.Solutions to Exercises 7Then,P = QQT=1/√2 −1/√62/√211/√21/√6 −2/√2102/√62/√21003/√211/√21/√20 0−1/√61/√62/√602/√21 −2/√21 2/√21 3/√21=1761−12161−2−11622 −22 3Thus, the vector in S closest to b =1111TisP = QQTb=1761−12161−2−11622 −22 31111=178685=8/76/78/75/7.Exercise 2(a) IfA =1 −12311175,subtract 3 times row 1 from row 2. Subtract 1 times row 1 from