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Last name: name: 1 Mid term exam 1 (Notes, books, and calculators are not authorized) Show all your work in the blank space you are given on the exam sheet. Always justify your answer. Answers with no justification will not be graded. Question 1: Find a parametric representation of the line in R4 that passes through P = (1, 2, 3, 4) and is orthogonal to the hyperplane x1 − 2x2 + 3x3 − 4x4 = 5. A normal vector to the hyperplane x1 − 2x2 + 3x3 − 4x4 = 5 is u := (1, −2, 3, −4). The line in question is the collection of points X = (x1 , x2 , x3 , x4 ) so that X − P is parallel to u. This means that there is a real number t so that X − P = tu = (t, −2t, 3t, −4t). The parametric representation of the line is x1 = 1 + t, x2 = 2 − 2t, x3 = 3 + 3t, x4 = 4 − 4t. Question 2: Find a unit vector orthogonal to u = (1, 2, 3) and v = (3, 2, 1). It is a known property of the cross product that the following vector w := u × v is orthogonal to u and v. By definition 2 2 3 1 1 3 ) (1) , , w = ( 3 1 1 3 2 2 = (−4, 8, −4). By normalizing w we obtain the desired vector: w 1 = √ (−4, 8, −4). kwk 96 (2) 2 Mid term exam 1, September 28, 2010 Question 3: Find an equation of the hyperplane in R4 that passes through P = (0, 1, 2, 3) and is normal to u = (−1, −1, 1, 1). By definition H is the collection of points X = (x1 , x2 , x3 , x4 ) so that (X − P )·u. This means 0 =(X − P )·(−1, −1, 1, 1) = (x1 , x2 − 1, x3 − 2, x4 − 3)·(−1, −1, 1, 1) = −x1 − (x2 − 1) + (x3 − 2) + (x4 − 3) = −x1 − x2 + x3 + x4 − 4. The equation of the hyperplane in question 0 = −x1 − x2 + x3 + x4 − 4. Question 4: Prove the triangle inequality: ku + vk ≤ kuk + kvk for all u, v ∈ Rn . By definition of the dot product, we have the following equality: ku + vk2 := (u + v)2 = (u + v)·(u + v) = u·u + u·v + v·u + v·u. Using the Cauchy-Schwarz inequality, we obtain ku + vk2 ≤ kuk2 + kukkvk + kvkkuk + kvk2 = (kuk + kvk)2 . Taking the square root gives the triangle inequality: ku + vk ≤ kuk + kvk. Last name: name: Question 5: Find the inverse of A = 1 3 3 2 4 The determinant of A is |A| = 4 − 16 = −2 6= 0. The matrix is invertible. We use the formula from class to compute the inverse: 1 4 −2 −1 A = −2 −3 1 1 −1 1 1 2 3 B= . Find (a) 3A − 4B (b) AT (c) AT B. −1 1 −1 3 2 1 3 −3 3 −4 −8 −12 −1 −11 −9 3A − 4B = + = −3 3 −3 −12 −8 −4 −15 −5 −7 1 −1 T −1 1 A = 1 −1 −2 0 2 AT B = 2 0 −2 −2 0 2 Question 6: Let A = 4 Mid term exam 1, September 28, 2010 Question 7: Let A be a square matrix. Assume that A has the following property: X T AX ≥ 2kXk2 for all column vector X. Prove that A is invertible. Consider X so that AX = 0. Then 0 = X T AX ≥ 2kXk2 . This proves that kXk = 0, which in turn proves that X = 0. As a result the solution set of AX = 0 is {0}. This proves that A is invertible since A is square. Question 8: Find 2 × 2 nonzero matrices A and B such that AB = 0. Consider the following matrix: A= 0 0 1 , 0 Then AB = 0 B=A Last name: name: 5 −3 13 . Find the LU factorization of A. −5 1 Question 9: Let A = −3 2 2 −4 1 We compute the echelon form 1 A −3 2 of A to get U and L 2 −3 1 2 −4 13 ∼ 0 2 1 −5 0 −3 1 l1 = −3 , 2 −3 1 4 ∼ 0 1 0 0 l2 = 1 , − 23 2 2 0 0 l3 = 0 1 Finally we have 1 L = −3 2 0 1 − 32 0 0 , 1 1 U = 0 0 2 2 0 −3 4 7 −3 4 7 6 Mid term exam 1, September 28, 2010 Question 10: Let u1 = (1, 2, 4), u2 = (2, −3, 1), u3 = (2, 1, −1) in R3 . (a) Show that u1 , u2 , u3 are orthogonal. We compute ui ·uj for i 6= j, i, j ∈ {1, 2, 3}. u1 ·u2 = 2 − 6 + 4 = 0, u1 ·u3 = 2 + 2 − 4 = 0, u2 ·u3 = 4 − 3 − 1 = 0. This proves the statement. (b) consider the 3 × 3 matrix U = [u1 u2 u3 ] where uj , j ∈ {1, 2, 3} are considered as column vectors. Show that U is invertible. x1 Consider X = x2 so that U X = 0. Then x3 x1 u1 + x2 u2 + x3 u3 = 0. Taking the dot product with u1 gives 0 = x1 u1 ·u1 + x2 u2 ·u1 + x3 u3 ·u1 = x1 ku1 k2 = 21x1 . This proves that x1 = 0. By repeating this process with u2 and u3 we obtain that x2 = 0 and x3 = 0. As a result the solution set of U X = 0 is {0}. This means that U is invertible. (c) Let v = (3, 5, 2). Solve U X = v, where v is considered as column vector. (Do not compute the reduced echelon form of the system.) By definition x1 u1 + x2 u2 + x3 u3 = v Taking the dot product with u1 and using the orthogonality of u1 , u2 , u3 gives v·u1 = 21 = x1 u1 ·u1 + x2 u2 ·u1 + x3 u3 ·u1 = x1 ku1 k2 = 21x1 , i.e., x1 = 1. Similarly v·u2 = −7 = x1 u1 ·u2 + x2 u2 ·u2 + x3 u3 ·u2 = x2 ku2 k2 = 14x2 , i.e., x1 = − 21 . v·u3 = 9 = x1 u1 ·u3 + x2 u2 ·u3 + x3 u3 ·u3 = x3 ku3 k2 = 6x3 , i.e., x1 = 32 . Hence the solution is 1 X = − 12 . 3 2