APPM 2360 Exam 2 Solutions Wednesday, March 9, 2016, 7:00pm 8:30pm

Transcription

1 APPM 2360 Exam 2 Solutions Wednesday, March 9, 206, 7:00pm 8:30pm ON THE FRONT OF YOUR BLUEBOOK write: () your name, (2) your student ID number, (3) recitation section (4) your instructor s name, and (5) a grading table. Books, class notes, cell phones, and calculators are NOT permitted. A one-page one-sided hand written crib sheet is allowed.. (30 points, Short Answer Problem). Please box your answer. You do not need to show your work (no partial credit is awarded on this question). (a) (5 points) If A = 0 [ ] and B =, calculate (AB) 0 T. 0 [ ] [ ] [ ] (AB) T = B T A T = = [ ] [ ] 2 2 (b) (5 points) If A = and B =, calculate the determinant 2A B (you should use properties of determinants). We have A = 2 0 = 2 0 = 2 and B = 2 2 = 4 = 3. Using properties of determinants, 2A 2 B = 2 2 A 2 B = 4 A 2 B = 4 A 2 B = ( 3) = -6/3. You could calculate it explicitly, in which case A 2 = ( ) and B = ( 2 ) 3 2 and so 2A 2 B = 2 3( ) so the determinant of this is ( 2 3) 2 2 = 6/3. Award full credit for the right answer regardless of method (though the first method was a lot faster). (c) Suppose that the augmented matrix [A b] corresponding to a linear system A x = b can be reduced, via row operations, to 0 0 k k k 2 where k is a parameter. For which value(s) of k, if any, does the system have... (i) (5 points) no solution? (ii) (5 points) exactly one solution? (iii) (5 points) infinitely many solutions? If k 0,, all columns have a pivot and therefore there is a unique solution. If k =, the third row is a consistent zero row and there is a free variable, so there are infinite solutions. Finally, if k = 0, adding the second row to the third row results in the RREF which is inconsistent. Therefore the answer is (i) k = 0, (ii) k 0,, (iii) k =. (d) (5 points) You are given the matrix C, its inverse C and a vector u as follows: C = C = u = ,, Solve for x if C x = u.

2 x = C u = 0 You do not need to use C since C was given. You could make an augmented matrix [C u] and put it into RREF, but this is a lot more work. 2. (30 points, True-false/short-answer) Please box your answer and clearly indicate True or False (except (a)(ii), which should be a number). You do not need to show your work (no partial credit is awarded on this question). (a) Let PE 4 define the space of all even-symmetric polynomials of degree 4. elements v PE 4 therefore have the form. v = a 0 + a 2 t 2 + a 4 t 4. All vector where a 0, a 2, a 4 R. (i) (5 points) Is PE 4 is a vector subspace of C(, )? [Recall C(, ) is the set of continuous functions that are defined on (, )] True. 0 PE 4. If v, v 2 PE 4 then α v + α 2 v 2 PE 4 (ii) (5 points) What is the dimension of PE 4? Only 3 linearly independent vectors required to span space, thus dim=3. (b) Let M 2 2 define the vector space of all 2 2 matrices with real entries. Consider the following vectors v, v 2, v 3, v 4 in M 2 2 : v = [ ], v 2 = (i) (5 points) Is the vector [ 0 0 in the Span(v, v 2, v 3, v 4 )? True. M 2 2 R 4 v = ] [, v 3 = 0 [ ] ] [, v 4 = 0 The determinant is nonzero implying a unique solution (or observe that if we swap rows 3 and 4, we are in REF, almost RREF, and have 4 pivot columns and are therefore full rank). (ii) (5 points) Do the vectors (v, v 2, v 3, v 4 ) form a basis for M 2 2? True, the dimension of the space is 4. The matrices are linearly independent. (c) (0 points) Is the space of solutions to y +2y +y = 4e t (given by y(t) = c e t +c 2 te t +e t ) a vector space? False. The ODE is nonhomogeneous. In particular, y = 0 is not a solution. 3. (30 points) Consider the system of equations x + 2x 2 3x 3 = 0 2x + x 2 3x 3 = 0 () x + x 2 = 0 (a) (5 points) Write the given system of equations as an augmented matrix and row reduce to RREF to solve the system. (b) (5 points) Using only your results from part (a), what is the determinant of the coefficient matrix of the system? Explain your answer in one or two sentences. 2 ].

3 (c) (5 points) What is the rank of the coefficient matrix of the system? No justification is necessary. (d) (5 points) The following images represent configurations of three planes in R 3 : a b c d e Which of the above images (a,b,...,e) could describe the system of equations in Equation (), where each plane represents the solutions to one equation, and the intersection of the three planes correspond to the solutions of the system of equations? Select all that could be applicable. (a) so that if x = x x 2 x is a solution to A x = 0 (where A is the coefficient matrix of the system), then we must have that x x 3 = 0 = x = x 3 and x 2 x 3 = 0 = x 2 = x 3, x 3 free. That is, x = x x 2 = x 3 x 3 = x 3 x 3 x 3 so that the solution space is Span{ v} where v =. (b) det A must be zero since the homogeneous system has more than just the trivial solution x = 0, which is equivalent to the coefficient matrix A being singular. Alternatively, note that A 0 if and only if the determinant of its RREF has non-zero determinant, and the RREF has zero determinant by inspection; or equivalently, that A 0 if and only if the RREF is the identity. (c) rank A = 2 (there are 2 pivots in the RREF) (d) The solution space is a line through the origin in R 3 with direction v as above. Of the images, (b) [and only (b)] describes the system of equations. 4. (30 points) Consider the ODE system: x (t) = x xy y (t) = xy y (a) (2 points) Determine the horizontal and vertical nullclines of the ODE system. (b) (6 points) Determine all equilibrium points of the ODE system. (c) (2 points) Plot the directions of the slopes along the horizontal and vertical nullclines. You may find it easier to supply two separate figures for the horizontal and vertical nullclines. (a) Solution curves are given by the vector x(t)î + y(t)ĵ, where î and ĵ are the standard unit vectors in R 2. The slope of solution is s(t) = x (t)î + y (t)ĵ. H-nullcline occurs when the vertical component of s(t) is zero (y (t) = 0). This is when xy y = y(x ) = 0 so the H-nullcline consists of the lines y = 0 and x =. V-nullcline occurs when the horizontal component of s(t) is zero (x (t) = 0). This is when 3

4 x xy = x( y) = 0 so the V-nullcline consists of the lines x = 0 and y =. (b) The equilibrium points satisfy x xy = 0 = xy y. This reduces to x( y) = 0 = y(x ) which is satisfied for (0, 0) and (, ). To see this, notice that lhs is zero when x = 0 or y =. Then correspondingly to make rhs zero we need y = 0 or x =, respectively. (c) Along the H-nullcline, slopes point left or right and along the V-nullcline they point up or down. To figure out the direction, notice that along the H-nullcline, the horizontal slope s h = (x xy)î = x( y)î. For x > 0, this points right when y < and points left when y >. When y = 0, then this points left for x < 0 and right for x > 0. For the V-nullcline, the vertical component of slope is given by s v = ( y + xy)ĵ = y(x )ĵ. Here, for y > 0, x > corresponds to up and x < to pointing down. For x = 0, slopes point down when y > 0 and up when y < 0. The slopes are plotted in Figure below. Figure. Slopes along the H and V nullclines 5. (30 points) For this problem, you are given a vector space V and a subset S. In each instance, answer the following questions:. (5 points each) Are the members of S linearly independent or linearly dependent? Show your work. 2. (5 points each) Does the set S constitute a basis of the indicated vector space V? Why or why not? Note: a correct answer without justification will not be awarded full credit. (a) (b) S = 2, 3 3 0, 0 0 2, S = { x 3, x +, 4 }, V = P 3 4 V = R3

5 (c) S = { cos 2 (x), sin 2 (x), }, V = C(, ) Solution to a:. These vectors are linearly independent. To demonstrate linear independence, build a matrix from these column vectors, and calculate its determinant. This is easily done by using the 2nd row, yielding: A = Equally easy is the 3rd column: [ ] A = 2 3 = 2 2 0, [ ] A = = 2. (2) 2 As the determinant of A is nonzero, these vectors are linearly independent. 2. These vectors are linearly independent and span R 3, hence they do constitute a basis set of R 3. For full credit, you must verify that they span R 3. The quickest way to do this is observe that R 3 has dimension 3, so any 3 linearly independent vectors forms a basis. A slower but equally valid method is to set a generic matrix b R 3, put [A b] into RREF, observe that the RREF of A is the identity, and conclude that we can solve A x = b for any b, therefore the columns of A span R 3. (This is equivalent to noting that A exists, which we know since A 0.) Solution to b:. These functions are linearly independent. To see this, build the matrix A from these three functions and their first and second derivatives. Then, calculate the Wronskian (i.e., the determinant of this matrix). The determinant calculation is easiest to do using the third column of A, yielding A = x3 x + 4 3x 2 0 6x 0 0, W (x) = A = 4 3x2 6x 0 = 24x. (3) The Wronskian is nonzero for all values of x except x = 0, which indicates that these functions are linearly independent. Or, observe the clear bijection between P 3 and R 4, and write A = which can be put into RREF in one step, showing that we have three pivot columns, hence rank is 3, so this is full rank, and these vectors are linearly independent. 2. By taking linear combinations of these three functions, we can build any polynomial of the form ax 3 + bx + c. Thus, their span is Span(S) = { ax 3 + bx + c a, b, c R } P 3 (4) While these functions are linearly independent, they do not span P 3 since x 2 is not in the span (or, justify it by using that there are only 3 elements in S, while V is 4-dimensional). Thus, these functions do not constitute a basis for P 3. Solution to c: 5

6 . These functions are linearly dependent. We might first check for linear independence by building a matrix A from these three functions and their first and second derivatives. After that, we could calculate the Wronskian using the third column of A to find that: A = cos 2 (x) sin 2 (x) 2cos(x)sin(x) 2cos(x)sin(x) 0 2 [ sin 2 (x) cos 2 (x) ] 2 [ sin 2 (x) cos 2 (x) ] 0, (5) and W = A = 2cos(x)sin(x) 2cos(x)sin(x) 2 [ sin 2 (x) cos 2 (x) ] 2 [ sin 2 (x) cos 2 (x) ] = 0. (6) Since the Wronskian is zero, we can t conclude anything about linear dependence or independence. Instead, we look at the functions and note that is a linear combination of sin 2 x and cos 2 x. Namely, sin 2 x + cos 2 x =. (7) Because the third function is a linear combination of the first two, these functions are not linearly independent. 2. By taking linear combinations of these functions, we can see that Span(S) = { a sin 2 x + b cos 2 x a, b R } C(, ) (8) These functions are linearly dependent and they do not span C(, ) (either fact by itself is sufficient justification; or note V has infinite dimension). This means that they do not constitute a basis for C(, ). 6