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 or 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 of R3ft.(a) True (b) FalseEssay QuestionsDirections: Answer each of the following. You must follow directions explicitly and show all work to receivefull credit. No calculators allowed!Exercise 1. State the nullspace of matrixA =120−12001 1 −3000 0 0Exercise 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 of a linear system of equations.110−2001 1000 0Write down the “complete” solution of the system without doing any work.Solutions to True or False Questions 3Solutions to True or False QuestionsSolution to Question 1: Draw the graph of s{(x1,x2):x1= x2} in the plane.x1x2x1= x2Take 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 coordinates of the sum are equal.Secondly, take a point on the line and multiply it by a scalar. Is this scalar product on the line? Forexample, ifaais on the line and c is a real number, thencaa=cacais back on the line because the first and second coordinates are equal.Since S is closed under addition and scalar multiplication, S is a subspace of R2. Solution 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 does not contain the zero vector and is not asubspace of R3. Solutions to E xercises 4Solutions 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 linear combination of the pivot columns thatproceed it. Indeed,col4=−1col1+1col3,so1col1+0col2 − 1col3+1col4+0col5=0.Thus,10−110is in the nullspace of A.Finally, the free column five can be written as a linear combination of the pivot columns that proceed it.Indeed,col5=2col1 − 3col3.Thus,−2col1+0col2+3col3+0col4+1col5=0and−20301is in the nullspace of matrix A. Thus, a basis for the nullspace of matrix A isB =−2100010−110−20301.Exercise 1Exercise 2. Starting with matrixA =12−1120 1 352 1 7Solutions to E xercises 5Subtract 2 times row 1 from row 2. Also, subtract 5 times row 1 from row 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 2Exercise 3. In the augmented matrix110−2001 1000 0Note that there are 2 pivot columns and 1 free column. Note that the second column can be written as alinear combination of the pivot columns proceeding it. Indeed,col2=1col1,so−1col1+1col2+0col3=0.Therefore−110is 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,where α is any real number. Exercise
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