No books, notes, any calculator, or electronic devices are allowed on this exam. Show all of your steps in each answer to receive a full credit.

Transcription

1 MTH Fall 2016 Exam 1 10/05/16 Name (Print): PID: READ CAREFULLY THE FOLLOWING INSTRUCTION Do not open your exam until told to do so. This exam contains 7 pages (including this cover page) and 7 problems. Check to see if any pages are missing. Enter all requested information on the top of this page, and put your initials on the top of every page, in case the pages become separated. No books, notes, any calculator, or electronic devices are allowed on this exam. Show all of your steps in each answer to receive a full credit. If you need more space, use the back of the pages; clearly indicate when you have done this. Do first all of the problems you know how to do immediately. Do not spend too much time on any particular problem. Return to difficult problems later. You will be given exactly 50 minutes for this exam.

2 MTH Exam 1 - Page 2 of 7 10/05/16 Problem Points Score Total: 100

3 MTH Exam 1 - Page 3 of 7 10/05/16 1. (5 points) True or False (Circle T or F)? No explaination is needed. T F (A B) 3 = A 3 3A 2 B + 3AB 2 B 3 holds for any n n matrices A and B. (The above would be true if A and B commute. For instance there is a term of the form BAA in (A B) 3, in which B appears before A and cannot be moved to the right of A.) [ ] T F is in reduced row echelon form T F The sum of two elementary matrices is an elementary matrix. T F det(ca) = c det(a) holds for any n n matrix A. T F A vector space is always a subspace of itself. 2. (20 points) Consider the system of linear equations x y z = a x + y 2z = b 3x y = c (a) (10 points) Find the condition on a, b, and c such that the system will have infinitely many solutions. Solution: We have the augmented matrix a a b R 2 R b a R c 3 3R c 3a R 3 R a b a a b + c The above coefficient matrix will have two pivots when we put it in reduced row echelon form. So we need -2a -b + c = 0 in order for the equation given by the zero row in the coefficient matrix to be consistent (the other two rows will always be consistent). (b) (10 points) Find a solution when a = 1, b = 1, and c = 1. Solution: We have R R 1 +R So we get the system x 1 2 z = 0 y 3 2 z = 1

4 MTH Exam 1 - Page 4 of 7 10/05/16 or in other words x = 1 2 z y = z. with z an arbitrary free variable. So the set of all solutions is { (1 2 α, ) T 2 α, α : α R}. (Here, we set z = α.) So pick any α, say α = 0, to get the solution (0, 1, 0) T.

5 MTH Exam 1 - Page 5 of 7 10/05/16 3. (15 points) Find the inverse of the following matrix Solution: We form an augmented matrix and row reduce: So the inverse matrix is 4. (15 points) Compute det(b 5 ), where B = Solution: Since det(b 5 ) = (det B) 5, it suffices to compute det B. We have R 2 +R 1 R 3 R R R 3 R 2 R Call the last matrix B. Then det B = det B, since the first two set of operations were Type III (preserves determinant) and the last operation was Type I (changes the sign). Since B is upper triangular, then det B = = 6. So det B 5 = ( det B ) 5 = ( 6) 5.

6 MTH Exam 1 - Page 6 of 7 10/05/16 5. (15 points) Let U = {f(x) C[0, 1] : f(0) = f(1)}. Show that U is a subspace of C[0, 1]. Solution: We need to show that if f, g U, then αf + βg U for arbitrary scalars α, β. (For β = 0, this shows U is closed under scalar multiplication and for α = β = 1, this shows U is closed under addition.) So since f, g U, we have f(0) = f(1) g(0) = g(1). If we take α times the first equation and it to β times the second equation we get αf(0) + βg(0) = αf(1) + βg(1). Since αf + βg is continuous and satisfies (αf + βg)(0) = (αf + βg)(1), it belongs to U. 6. (15 points) Show that the functions x + 1, x 2 + x, x 3 + x 2, x 3 are linearly independent in P 4. Solution: Suppose 0 = α 1 (x + 1) + α 2 (x 2 + x) + α 3 (x 3 + x 2 ) + α 4 x 3 = α 1 + (α 1 + α 2 )x + (α 2 + α 3 )x 2 + α 4 x 3 The only way for a polynomial to be zero is if all its coefficients are zero. This implies α 1 = 0 α 1 + α 2 = 0 α 2 + α 3 = 0 α 4 = 0. In other words, α α α 3 = 0 0. (1) α 4 0 The above coefficient matrix is invertible since it has determinant 1 (it is lower triangular and so this its determinant is the product of its diagonal entries). So the only solution to (1) is the zero vector, i.e. α i = 0 for all i. So the only linear combination of our polynomials equal to zero is the trivial linear combination, hence they are linearly independent.

7 MTH Exam 1 - Page 7 of 7 10/05/16 7. (15 points) Suppose that S = {v 1, v 2,..., v k, v k+1 } is a linearly independent set of vectors. Show that {v 1 + v k+1, v 2 + v k+1,..., v k + v k+1 } is also a linearly independent set of vectors. Solution: Suppose This is equivalent to α 1 (v 1 + v k+1 ) + α 2 (v 2 + v k+1 ) + α k (v k + v k+1 ) = 0 (2) α 1 v 1 + α 2 v 2 + α k v k + (α α k )v k+1 = 0. (3) Since the v 1,..., v k+1 are linearly independent by hypothesis, then all coefficients of (3) are zero. In particular, α 1 =... = α k = 0. But this implies that (2) is the trivial linear combination. So {v 1 + v k+1, v 2 + v k+1,..., v k + v k+1 } is linearly independent.