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CR MATH 45 - Math 45 Exam

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College of the Redwoods Mathematics Department Math 45 Linear Algebra Exam 2 Linear Algebra David Arnold c 2000 David Arnold Eureka redwoods cc ca us Copyright Last Revision Date October 23 2000 Version 1 00 2 Essay Questions Directions Place the solution to each of the following exercises on your own paper You must follow directions explicitly and show all work to receive full credit I suggest that you use hand calculations to perform row reduction on the rst problem but after that use your calculator to nd the reduced row echelon form of any matrix Don t waste time performing elementary row operations by hand If you do and you run out of time then you have no one to blame but yourself Exercise 1 Use elementary row operations to place the matrix 2 1 0 A 1 1 2 1 1 1 in upper triangular form Use the appropriate properties from the Strang activity to evaluate the determinant Exercise 2 Use cofactor expansion to nd the determinant of matrix A in problem 1 Exercise 3 Suppose that 1 0 U 0 0 0 2 1 2 0 2 0 0 3 0 1 3 a What is the determinant of U b What is the determinant of U 2 c What is the determinant of U 1 Exercise 4 Let V be a vector space over R a Let U be a subset of V Complete the following de nition U is a subspace of V if and only if b As an example consider the following subset of R2 x1 U x1 x2 1 x2 Prove or disprove U is a subspace of R2 Exercise 5 Let V be a vector space over R a Let v1 vn V Complete the following de nition The vectors v1 vn are linearly independent if and only if b Consider the vectors from R3 1 v1 1 1 2 v2 0 1 and 7 v3 3 5 Prove or disprove The vectors v1 v2 and v3 are linearly independent 3 Exercise 6 Let V be a vector space over R Let v1 vn V a Complete the following de nition The span of vectors v1 vn is b Consider the vectors from R4 1 1 v1 1 0 0 0 v2 1 1 Is the vector and 1 1 v3 0 1 1 1 b 1 1 in the span of v1 v2 and v3 Exercise 7 Find a basis for the nullspace of the 1 A 1 1 matrix 2 2 2 0 1 1 0 1 2 3 2 4 What is the dimension or nullity of the nullspace of A Exercise 8 Find a basis for the column space of matrix A in problem 7 State the rank of matrix A Solutions to Exercises 4 Solutions to Exercises Exercise 1 Swap rows 1 and 2 this negates the determinant 1 2 1 0 A 1 1 2 2 1 1 1 1 1 2 1 0 1 1 Multiply row 1 by 2 and add the result to row 2 This does not change the value of the determinant 1 1 2 3 4 A 0 1 1 1 Multiply row 1 by 1 and add the result to row 3 This does not change the value of the determinant 1 A 0 0 1 2 3 4 0 3 Because the resulting matrix is upper triangular the determinant equals the product of the diagonal elements Thus A 1 3 3 9 Exercise 1 Exercise 2 Expand across the rst row 2 A 1 1 2 Evaluate 1 1 1 0 1 2 1 1 1 2 1 1 1 2 1 A 2 1 2 1 1 2 2 3 1 3 6 3 9 Exercise 2 Exercise 3 a Since U is upper triangular the determinant is the product of its diagonal elements 1 0 2 3 0 1 2 0 U 0 0 2 1 1 1 2 3 6 0 0 0 3 Exercise 3 b The determinant of a product is the product of the determinants so U 2 U U U U 6 6 36 Exercise 3 c Because U 0 and U is square U is nonsingular and invertible Moreover because the determinant of the identity matrix is 1 1 I U U 1 U U 1 Solutions to Exercises 5 Thus U 1 1 1 1 U 6 6 Exercise 4 a U is a subspace of V if and only if two conditions are satis ed 1 U is closed under addition That is for every u v U it must be the case that u v U 2 U is closed under scalar multiplication That is for every u U and R it must be the case that u U Exercise 4 b U is not a subspace of R2 To show this I will demonstrate that it is not closed under addition Let 1 0 U 0 1 Note that 1 0 1 so 1 0 T U Similarly 0 1 1 so 0 1 T U However 1 0 1 0 1 1 but 1 1 1 so 1 1 T U Therefore U is not closed under addition Exercise 5 a The vectors v1 vn are linearly independent if and only if the equation c1 v1 cn vn 0 has only the trivial solution i e c1 cn 0 Exercise 5 b Set up the equation 0 1 2 7 c1 1 c2 0 c3 3 0 0 1 1 5 To solve set up the augmented matrix 1 2 7 1 0 3 1 1 5 Reduce Therefore 0 0 0 1 0 3 0 0 1 2 0 0 0 0 0 c1 3c3 c2 2c3 c3 free Consequently there are nontrivial solutions and the vectors are linearly independent For example 7 0 1 2 1 0 3 3 2 1 0 1 1 5 0 Solutions to Exercises 6 Exercise 6 a The Span v1 vn is the set of all linear combinations of the vectors v1 vn In symbols Span v1 vn c1 v1 cn vn ci R Exercise 6 b Suppose that b is in the span Then we can write 1 0 1 1 1 0 1 1 c1 1 c2 1 c3 0 1 0 1 1 1 Setup the augmented matrix 1 1 1 0 0 1 0 1 1 0 1 1 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 Thus the system is inconsistent so it is not possible to write b as a linear combination of the vectors v1 v2 and v3 Therefore b is not in the span of v1 v2 and v3 Reduce Exercise 7 Set up the equation Ax 0 1 1 1 Set up the augmented matrix 2 0 1 2 1 0 2 1 2 1 2 1 2 1 2 Reduce Thus the solution of Ax 0 is x1 0 3 x2 0 2 x 3 0 4 x4 x5 0 1 3 1 0 2 1 2 4 0 0 0 1 2 0 …


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CR MATH 45 - Math 45 Exam

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