Math 0 Spring 04 Midterm 0/6/4 Lecturer: Jesus Martinez Garcia Time Limit: 50 minutes Name (Print: Teaching Assistant This exam contains 9 pages (including this cover page and 4 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 You may not use your books, notes, or any calculator on this exam You are required to show your work on each problem on this exam The following rules apply: If you use a theorem of lemma you must indicate this and explain why the theorem may be applied Organize your work, in a reasonably neat and coherent way, in the space provided Work scattered all over the page without a clear ordering will receive very little credit Mysterious or unsupported answers will not receive full credit A correct answer, unsupported by calculations, explanation, or algebraic work will receive no credit; an incorrect answer supported by substantially correct calculations and explanations might still receive partial credit Problem Points Score 5 5 3 5 4 5 Total: 00 If you need more space, use the back of the pages; clearly indicate when you have done this Do not write in the table to the right
Math 0 Midterm - Page of 9 0/6/4 (5 points Solving linear systems Consider the following linear system: where k R x y +(k z = k x y +(4k 4z = 4 k 3x +4y +( k + z = + 4k, (a (5 points Write the augmented matrix (A b of the linear system The augmented matrix is (A b = (k k 4k 4 4 k 3 4 k + + 4k (b (5 points Consider k is fixed Find rref(a b We perform Gauss Jordan elimination: (k k 4k 4 4 k 3 4 k + + 4k k k 0 (k + k 0 k + k Now, if k ±, then If k = ±, then 0 3(k 3 0 (k + k 0 0 (k (I:+(II (III:+(II rref(a b = (k k 4(k 4 k 3 4 (k + 4k (I: 3(III (II: (III 0 3(k 3 0 (k + k 0 0 (k 0 0 0 0 0 k 0 0 (k 0 0 0 0 0 k 0 0 0 0 0 0 0 0 k 0 0 (k (II: (I (III:+3(I (III: 3 (c (5 points Analyse whether the linear system has solutions or not according to the values of k For each k such that solutions exist, find all solutions If k = ±, then rk(a = 3 = rk(a b, and by the rank-solutions theorem, the system has no solutions If k ±, then rk(a b = rk(a = 3, and by the rank-solutions theorem
Math 0 Midterm - Page 3 of 9 0/6/4 the system has a unique solution: x 0 y = k z k
Math 0 Midterm - Page 4 of 9 0/6/4 (5 points Inverses and transformations on the plane Let A = (, B = ( (a (0 points Decide if A and B are invertible If they are, find the inverses If they are not, say why not : Recall that a matrix M is invertible, if and only if rk(m =, if and only if rref(m = I Therefore A is not invertible, since its reduced row echelon form is ( 0 0 The matrix B is indeed invertible We find the inverse by means of finding the reduced row echelon form of the following matrix: ( 0 0 ( ( 0 ( 0 +(I ( 0 0 ( 0 0 ( (I 0 0 Observe, in particular that B = B (b (5 points The matrices A and B define certain well known linear transformations T A and T B on the plane Say which transformations they are (ie give their name and say why For A we have: ( ( u A = u u ( ( u u u = ( ( Therefore A is the matrix of a projective transformation ( over the line generated by the vector (, a b or (, For B notice that B =, where a b a + b =, so it gives the matrix of a reflection (c (0 points A and B fix all points in some lines L A and L B respectively, ie A x = x x L A and B x = x x L B Give the equations of L A and L B Since A is a projection over the line generated by (,, the points in this line are fixed by A and the equation of the line is x y = 0, since ( ( ( x x x For B we look for such that B = y y y ( x = ( ( x y y ( x y, ie = t ( = ( x + y x y
Math 0 Midterm - Page 5 of 9 0/6/4 Therefore we have the homogeneous linear system of equations { } ( x y = 0 x +( +, y = 0 which is equivalent to { (I : ( x y = 0 (II : x +( + y = 0 } Since α := = + multiplying (I by α, we obtain (II and (II is therefore redundant The line defined by B is the one with equation y = ( x
Math 0 Midterm - Page 6 of 9 0/6/4 3 (5 points Image, kernel and bases Let where x R 5 3 0 T ( x = 3 6 4 x 3 6 6 0 3 (a (0 points Find a basis for the image of T Justify why the vectors you provide are linearly independent and span Im(T The image is spanned by the columns of the matrix of T Call this matrix A and its columns v,, v 5 R 4 Hence Im(A = v,, v 5 We need to find the redundant vectors of A There are several ways to do this, but we will choose the one that uses the reduced row echelon form We apply Gauss Jordan elimination to A: 3 0 3 6 4 3 6 6 0 3 (I: 7(IV (II:+(IV (II: (I (III: (I (IV : 3(I 3 0 0 0 4 0 0 3 0 0 (I: (II (III: (II (IV : ( 3 0 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Swap (II and (III 0 0 0 0 0 0 0 0 0 0 0 0 0 = rref(a = w w w 3 w 4 w 5 = Observe that the relations among the elements of B give us relations among the elements of A: w 4 = w w v 4 = v v w 5 = w w v 5 = v v Therefore v 4 and v 5 are redundant The vectors w, w, w 3 are linearly independent, therefore v, v, v 3 are linearly independent too, and since they span Im(A, the set { v, v, v 3 } is a basis of Im(A (b (5 points Find a basis for the kernel of T linearly independent and span Ker(T Justify why the vectors you provide are From the reduced row echelon form of A, we deduce that the vector x x = x 3 x 4 Ker(A x 5 x
Math 0 Midterm - Page 7 of 9 0/6/4 if and only if x r + s x x 3 x 4 = r s 0 r = r 0 + s 0 0 = r a + s b x 5 s 0 Therefore a, b = Ker(A The vectors a and b are clearly linearly independent since they are not multiples of each other Moreover, by the rank-nullity theorem we have: Therefore { a, b } is a basis of Ker(A dim(ker(a = 5 dim(im(a = 5 3 =
Math 0 Midterm - Page 8 of 9 0/6/4 4 (5 points Properties of linear transformations (a (5 points Let Mat 3 (R be the space of all 3 matrices We can identify Mat 3 (R with R 6 in the following way: a a b a a b : a 3 a 3 b b, or more concisely, in vector form: b b 3 a b : a Since the sum of matrices and the product of a matrix by a scalar in Mat 3 (R corresponds to that of R 6, we can think of Mat 3 (R as a vector space Show that the following map is a linear transformation: f : Mat 3 (R = R 6 R 3, f a b = a b The map f is linear if it satisfies (i f( u + u = f( u + f( u for all u, u Mat 3 (R, (i f(k u = kf( u for all u Mat 3 (R and k R Indeed, this is the case We let u = a b, u = a b and k R Then (i f( u + u = f a b + a b = f a + a b + b = = ( a + a ( b + b = ( a b + ( a b = f( u + f( u b (ii f(k u = f k a b = f k a k b = Therefore, f is a linear transformation = (k a (k b = k( a b = kf( u (b (0 points Consider the following statements If they are true, provide a proof If they are false, provide a counter-example (i Given square matrices A and B of the same size, if A B = 0, then A = 0 or B = 0 (ii Let A be a square matrix Then Ker(A Ker(A
Math 0 Midterm - Page 9 of 9 0/6/4 (i False Take A = ( 0, B = 0 0 (ii True Let x Ker(A Then A x = 0, so Therefore x Ker(A ( 0 0 Then AB = 0 0 A x = A(A x = A 0 = 0