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APPM 2360: Midterm Exam 2 October 24, 2018 1.5 Hours on the Front Of APPM 2360: Midterm exam 2 October 24, 2018 1.5 hours On the front of your Bluebook write: (1) your name, (2) your instructor's name, (3) your section number and (4) a grading table. Text books, class notes, cell phones and calculators are NOT permitted. A one page (letter sized 1 side only) crib sheet is allowed. Problem 1: (20 points) Are the following statements true or false? (Only answer TRUE or FALSE is needed; no justification is required.) If a statement is not always true, reply `FALSE'. Give your answers in the bluebook (and not on this problem sheet). 4 (a) The vectors v~1 = (1; 0; 0; 1), v~2 = (0; 1; 0; 0) and v~3 = (0; 0; 1; 0) span R . (b) The dimension of the space of diagonal n × n real matrices with respect to the usual matrix addition and multiplication by real numbers is n2. (c) If fv1; v2; v3g is a basis of a vector space V , then v1 and v3 are linearly independent. (d) If A is an m × n matrix and RREF(A) has some zero rows, then rank(A) is always less than n. (e) Every nonzero m × n matrix has at least one pivot. Solution: (a)F (b)F (c)T (d)F (e)T Problem 2: (20 points) (a) Given n functions f1(x), f2(x), ::: , fn(x), give the definition for these functions being linearly independent. (b) Consider the specific case of the functions fx; 1 + x; 2 + 3xg. Determine whether these three functions are linearly independent, or not. (c) Define the Wronskian W (x) of a general set of n functions f1(x), f2(x), ::: , fn(x). (d) Apply the Wronskian to the set of functions listed in part (b) of this problem. What can you conclude from the resulting function W (x) regarding linear independence? Solution: 1 (a) The functions are linearly independent if and only if the relation c1f1(x) + c2f2(x) + ··· + cnfn(x) ≡ 0 (identically equal to zero) for constants c1; c2; : : : ; cn can only be satisfied by c1 = c2 = ··· = cn = 0. (b) The relation to test becomes in this case c1x + c2(1 + x) + c3(2 + 3x) ≡ 0, which we rewrite as (c2 + 2c3) · 1 + (c1 + c2 + 3c3) · x ≡ 0. For this to hold, the coefficients for both 1 and x must be zero, i.e. 0 c 1 0 1 2 1 0 c = : 1 1 3 @ 2 A 0 c3 This is a non-square linear system, so it has either no or infinitely many solutions. No solution cannot be the case, since we know the system to have the zero solution. It must therefore also have nontrivial solution(s), confirming linear dependence. (One can easily see that, for example, c1 = 1, c2 = 2, c3 = −1 solves the system, but finding an explicit solution is unnecessary for solving the present problem). (c) f1(x) f2(x) : : : fn(x) 0 0 0 f1(x) f2(x) : : : fn(x) W (x) = :::::::::::: (n−1) (n−1) (n−1) f1 (x) f2 (x) : : : fn (x) (d) The Wronskian becomes x 1 + x 2 + 3x W (x) = 1 1 3 ≡ 0: 0 0 0 Had W (x) not been identically zero, we could have concluded that the functions are inde- pendent. With W (x) ≡ 0, this test using the Wronskian is inconclusive. Problem 3: (20 points) Consider the following matrices: 0 1 0 0 1 0 0 2 A = C = −4 3 0 −1 1 2 @ A −4 2 1 0 1 1 1 1 3 1 B = 0 1 1 D = @ A 1 3 0 0 1 Find the following quantities: (a) C + 2B (b) ABT (c) The eigenvalues of B and their multiplicities. (d) The eigenvalues and eigenvectors of D. Solution: 2 (a) 0 3 2 2 1 C + 2B = @ −4 5 2 A −4 2 3 (b) 0 1 0 0 1 0 0 2 2 2 2 ABT = 1 1 0 = −1 1 2 @ A 2 3 2 1 1 1 (c) Solve equation det(B − λI) = 0 for λ i.e. 1 − λ 1 1 0 1 − λ 1 = 0 0 0 1 − λ 3 which is (1 − λ) = 0. Thus, λ = 1 with multiplicity 3 or λ1 = λ2 = λ3 = 1. Alternatively, the eigenvalues of an upper (or lower) triangular matrix can be read off from the diagonal elements, giving the same result. 2 (d) The equation det(D − λI) = 0 is (3 − λ) − 1 = 0, so λ − 3 = ±1 and eigenvalues are λ1 = 4 T and λ2 = 2. Solving (D − λ1I)v~1 = ~0 for v~1 = (x; y) , we find −1 1 x = 0; 1 −1 y T i.e. x = y and v~1 = x(1; 1) is an eigenvector for any x 6= 0. T Similarly, (D − λ2I)v~2 = ~0 with v~2 = (x; y) , is 1 1 x = 0; 1 1 y T i.e. x = −y and v~2 = x(1; −1) is an eigenvector for any x 6= 0. Problem 4: (20 points) 2 1 1 1 3 (a) Find the inverse of the matrix A = 4 −1 −2 6 5 0 0 1 Solution: 2 1 1 1 1 0 0 3 2 1 1 1 1 0 0 3 R2 R2+R1 4 -1 -2 6 0 1 0 5 −−−−−−−! 4 0 -1 7 1 1 0 5 0 0 1 0 0 1 0 0 1 0 0 1 2 1 1 1 1 0 0 3 R2 −1·R2 −−−−−−−! 4 0 1 -7 -1 -1 0 5 0 0 1 0 0 1 2 1 1 0 1 0 -1 3 R2 R2+7·R3 −−−−−−−−−! 4 0 1 0 -1 -1 7 5 R1 R1−R3 0 0 1 0 0 1 2 1 0 0 2 1 -8 3 R1 R1−R2 −−−−−−−! 4 0 1 0 -1 -1 7 5 0 0 1 0 0 1 3 2 2 1 −8 3 so the answer is A−1 = 4 −1 −1 7 5 : 0 0 1 −1 −1 Alternatively, A can be found as (A )ij = Cji= det(A) in terms of cofactors Cij of each element Aij of A in det(A). (b) Solve the following system of equations for x and y: x + y = 2 5x + 4y = 3 1 1 −4 1 1 0 Hint: observe · = . 5 4 5 −1 0 1 1 1 x 2 Solution: Observe that we can write the system of equations as · = : The 5 4 y 3 −4 1 hint tells us that the inverse of the coefficient matrix is , so we just compute 5 −1 −4 1 2 −5 · = : 5 −1 3 7 Problem 5: (20 points) This problem consists of three independent parts. (a) Consider the set S of all functions f(x) continuous on [0; 1] and such that f(0) = f(1). Is this a vector space? Justify your answer. 2 (b) Express the given vector ~v in terms of the given basis fv~1; v~2g of the vector space R , where a 3 −1 ~v = ; v~ = ; v~ = : b 1 2 2 2 3 (c) Consider P4, the vector space of all polynomials p(x) of degree ≤ 4. Are p1(x) = x − 3x + 1, 4 4 3 p2(x) = x − 6x + 3 and p3(x) = x − 2x + 1 linearly independent elements of P4? What is the dimension of the subspace of P4 they span? Give an example of a basis for this subspace. Justify your answers. Solution: (a) This set is a vector space if any linear combination of two such functions is also in S. Let f1(x) 2 S and f2(x) 2 S be two such functions and c1; c2 be two numbers. Then function f(x) = c1f1(x) + c2f2(x) is continuous on [0; 1] and f(0) = c1f1(0) + c2f2(0) = c1f1(1) + c2f2(1) = f(1); so f(x) 2 S. Therefore S is a vector space. (b) We need to find numbers x1 and x2 such that ~v = x1v~1 + x2v~2 i.e. to solve the linear system A~x = ~v where x1 3 −1 ~x = ;A = (v~1jv~2) = x2 2 2 4 Matrix A is invertible since det A = 8, therefore 1 2 1 a 1 2a + b ~x = A−1~v = = 8 −2 3 b 8 −2a + 3b Thus, a (2a + b) 3 (−2a + 3b) −1 = + b 8 2 8 2 3 4 4 3 (c) Consider numbers c1; c2; c3 such that c1(x − 3x + 1) + c2(x − 6x + 3) + c3(x − 2x + 1) = 0 for all x, then we get the system 8 c2 + c3 = 0 > < c1 − 2c3 = 0 −3c1 − 6c2 = 0 > : c1 + 3c2 + c3 = 0 We verify that, if c2 = −c3 and c1 = 2c3, then the last two equations are also satisfied. Thus, one of the three numbers is arbitrary and the polynomials are linearly dependent. However, e.g. p1 and p2 are linearly independent (they have different powers of x). Therefore spanfp1; p2; p3g = spanfp1; p2g and the dimension dim(spanfp1; p2g) = 2. fp1; p2g is a possible basis for this subspace. Alternatively, we can use the (general method of) Gaussian elimination to solve the linear T system A~c = ~0 for ~c = (c1; c2; c3) . The matrix of the coefficients is 0 0 1 1 1 B 1 0 −2 C A = B C @ −3 −6 0 A 1 3 1 and we find RREF(A) with the same result. 5.
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