blueSolutions to True or False Questions Solutions to ExercisesCollege of the RedwoodsMathematics DepartmentMath 45 — Linear AlgebraQuiz #5Linear AlgebraDavid ArnoldCopyrightc 2000 [email protected] Revision Date: October 9, 2001 Version 1.002True or False QuestionsDirections: Circle true o r false.1. The set S = {(x1,x2): x1= x2} is a subspace of R2.(a) True (b) False2. The set S = {(x1,x2,x3): x1+2x2+3x3=6} is a subspace ofR3ft.(a) True (b) FalseEssay QuestionsDirections: Answer each of the following. You must follow directionsexplicitly and show all work to receive full credit. No calculatorsallowed!Exercise 1. State the nullspace of matrixA =120−12001 1 −3000 0 03Exercise 2. Find the rank of of matrix A and justify your conclusion.A =12−1120135217Exercise 3. The following matrix is the reduced row echelon form ofa linear system of equations.110−2001 1000 0Write down the “complete” solution of the system without doing anywork.Solutions to True or False Questions 4Solutions to True or False QuestionsSolution to Question 1: Draw the graph of s{(x1,x2):x1= x2}in the plane.x1x2x1= x2Solutions to True or False Questions 5Take two points on the line; add them. Is the sum back on the line?For example,aaandbbre on the line (x1= x2), and the sumaa+bb=a + ba + bis also back on this line because the first and the second coordinatesof the sum are equal.Secondly, take a point on the line and multiply it by a scalar. Isthis scalar product on the line? For example, ifaais on the line and c is a real number, thencaa=cacais back on the line because the first and second coordinates are equal.Solutions to True or False Questions 6Since S is closed under addition and scalar multiplication, S is asubspace of R2. Solutions to True or False Questions 7Solution to Question 2: The set S = {(x1,x2,x3):x1+2x2+3x3=6} is a plane in 3-space.x2x3x1(6, 0, 0)(0, 3, 0)(0, 0, 2)Note that the plane does not contain the origin. Therefore, it doesnot contain the zero vector and is not a subspace of R3. Solutions to Exercises 8Solutions to ExercisesExercise 1. Note that column 2 is a multiple of column 1. That is,−2col1+1col2=0.Indeed,−2col1+1col2+0col3+0col4+0col5=0.Thus,−21000is in the nullspace of A. The free column 4 can be written as a linearcombination of the pivot columns that proceed it. Indeed,col4=−1col1+1col3,so1col1+0col2 − 1col3+1col4+0col5=0.Solutions to Exercises 9Thus,10−110is in the nullspace of A.Finally, the free column five can be written as a linear combinationof the pivot columns that proceed it. Indeed,col5=2col1 − 3col3.Thus,−2col1+0col2+3col3+0col4+1col5=0and−20301Solutions to Exercises 10is in the nullspace of matrix A. Thus, a basis for the nullspace ofmatrix A isB =−2100010−110−20301.Exercise 1Solutions to Exercises 11Exercise 2. Starting with matrixA =12−1120 1 352 1 7Subtract 2 times row 1 from row 2. Also, subtract 5 times row 1 fromrow 3.12−110 −4310 −862Subtract 2 times row 2 from row 3.12−110 −43100 00= U.Therefore, there are two pivots columns. The rank of A is 2.Exercise 2Solutions to Exercises 12Exercise 3. In the augmented matrix110−2001 1000 0Note that there are 2 pivot columns and 1 free column. Note thatthe second column can be written as a linear combination of the pivotcolumns proceeding it. Indeed,col2=1col1,so−1col1+1col2+0col3=0.Therefore−110Solutions to Exercises 13is in the nullspace of the coefficient matrix andB =−110as a basis for the nullspace.Secondly, by letting x2=0inx1= −2 − x2x3=1,we have a particular solutionxp=−201.Thus, the ”complete” solution is given byx =−201+ α−110,Solutions to Exercises 14where α is any real number. Exercise
View Full Document
Unlocking...