Math 415 - Midterm 1Thursday, September 25, 2014Circle your section:Philipp Hieronymi 2pm 3pmArmin Straub 9am 11amName:NetID:UIN:Problem 0. [1 point] Write down the number of your discussion section (for instance, AD2or ADH) and the first name of your TA (Allen, Anton, Babak, Mahmood, Michael, Nathan,Tigran, Travis).Section: TA:To be completed by the grader:0 1 2 3 4 5 6 ShortsP/1 /15 /10 /15 /10 /10 /12 /21 /94Good luck!1Instructions• No notes, personal aids or calculators are permitted.• This exam consists of 9 pages. Take a moment to make sure you have all pages.• You have 75 minutes.• Answer all questions in the space provided. If you require more space to write youranswer, you may continue on the back of the page (make it clear if you do).• Explain your work! Little or no points will be given for a correct answer with noexplanation of how you got it.• In particular, you have to write down all row operations for full credit.Problem 1. LetA =1 0 21 1 01 0 1.(a) [8 points] Determine A−1.(b) [2 points] Check whether AA−1= I3.Solution: We have:1 0 2 1 0 01 1 0 0 1 01 0 1 0 0 1R2→R2−R1,R3→R3−R1−−−−−−−−−−−−−−→1 0 2 1 0 00 1 −2 −1 1 00 0 −1 −1 0 1R1→R1+2R3,R2→R2−2R3−−−−−−−−−−−−−−−→1 0 0 −1 0 20 1 0 1 1 −20 0 −1 −1 0 1R3→−R3−−−−−→1 0 0 −1 0 20 1 0 1 1 −20 0 1 1 0 −1Thus, A−1=−1 0 21 1 −21 0 −1.2Problem 2. Consider the matrixA =3 2 03 1 26 7 5.(a) [10 points] Calculate the LU decomposition of A.(b) [5 points] Solve3 2 03 1 26 7 5x1x2x3=4710without reducing the augmented matrix, but using the LU decomposition.Solution:(a) First, we transform A to echelon form (an upper triangular matrix) using upward rowoperations:3 2 03 1 26 7 5R2→R2−R1−−−−−−−→3 2 00 −1 26 7 5R3→R3−2R1−−−−−−−→3 2 00 −1 20 3 5R3→R3+3R2−−−−−−−→3 2 00 −1 20 0 11= UTo get L, we have to apply the inverse of the row operations in the reverse order to I:1 0 00 1 00 0 1R3→R3−3R2−−−−−−−→1 0 00 1 00 −3 1R3→R3+2R1−−−−−−−→1 0 00 1 02 −3 1R2→R2+R1−−−−−−−→1 0 01 1 02 −3 1= L[Note: for the exam, you may also immediately write down L. The above details giveyou another chance to see why that is.](b) Let c = Ux. First we solve Lc = b:c1= 4c1+ c2= 7 ⇒c2= 32c1− 3c2+ c3= 10 ⇒c3= 11Finally, we solve Ux = c to find x:11x3= 11 ⇒x3= 1−x2+ 2x3= 3 ⇒x2= −13x1+ 2x2= 4 ⇒x1= 2Thus:x =2−11[Note: make sure to quickly check your answer by computing Ax.]3Problem 3. LetB =1 3 −5 41 4 −8 7−3 −7 9 −6.(a) [8 points] Determine the reduced echelon form of B.(b) [7 points] Use your result in (a) to find a parametic description of the set of solutionsof the following system of linear equations:x1+ 3x2− 5x3= 4x1+ 4x2− 8x3= 7−3x1− 7x2+ 9x3= −6Solution:(a) We have:1 3 −5 41 4 −8 7−3 −7 9 −6R2→R2−R1,R3→R3+3R1−−−−−−−−−−−−−−−→1 3 −5 40 1 −3 30 2 −6 6R3→R3−2R2−−−−−−−→1 3 −5 40 1 −3 30 0 0 0R1→R1−3R2−−−−−−−→1 0 4 −50 1 −3 30 0 0 0(b) Note that B is the augmented matrix corresponding to this system of linear equations.According to the reduced echelon form of B, x1and x2are dependent variables (sincethe first and the second column are pivot columns) and x3is free. We have:x1+ 4x3= −5 ⇒x1= −4x3− 5x2− 3x3= 3 ⇒x2= 3x3+ 3So the set of solutions is:{−4x3− 53x3+ 3x3: x3∈ R} = {−431x3+−530: x3∈ R}Which is a line that does not go through the origin.4Problem 4. [10 points] Consider the vectorsw =2−4−11, v1=1010, v2=0210, v3=1001.Is w in span{v1, v2, v3}? Show your calculations!Solutions: We have to determine if there are real numbers x, y, and z such that xv1+ yv2+ zv3= w.This is equivalent to determining if the following system of equations is consistent:1 0 1 20 2 0 −41 1 0 −10 0 1 1R3→R3−R1−−−−−−−→1 0 1 20 2 0 −40 1 −1 −30 0 1 1R3→R3−1/2R2−−−−−−−−−→1 0 1 20 2 0 −40 0 −1 −10 0 1 1R4→R4+R3−−−−−−−→1 0 1 20 2 0 −40 0 −1 −10 0 0 0In this echelon form, there is no row of the form [0 0 0 | r], where r is non-zero (or in other words,there is no pivot position in the right hand side). Thus, the system is consistent. Therefore, itis possible to write w as a linear combination of v1, v2, v3, i.e., w is in span{v1, v2, v3}.5Problem 5. [10 points] Letv1=10−1, v2=0−13, v3=1h1h2.For which values of h1and h2is v3a linear combination of v1and v2?Solution: Similar to the previous problem, we have to determine when the following systemof equations is consistent:1 0 10 −1 h1−1 3 h2R3→R3+R1−−−−−−−→1 0 10 −1 h10 3 h2+ 1R3→R3+3R2−−−−−−−→1 0 10 −1 h10 0 3h1+ h2+ 1The system is consistent if and only if 3h1+ h2+ 1 = 0, i.e., h2= −3h1− 1.Hence, v3is a linear combination of v1and v2if and only if h2= −3h1− 1.6Problem 6. [12 points] Determine which of the following sets are subspaces and give reasons:(a) W1= {ab: ab = 0},(b) W2= {a + 1a: a in R},(c) W3= {ab: a2+ b2≤ 1}.Solution:(a) W1is not a subspace, since10and01are in W1but10+01=11is notin W1.(b) W2is not a subspace, since00is not in W2.(c) W3is not a subspace, since10is in W3but 210=20is not in W3.7SHORT ANSWERS[21 points overall, 3 points each]Instructions: The following problems have a short answer. No reason needs to be given.If the problem is multiple choice, circle the correct answer (there is always exactly one correctanswer).Short Problem 1. Let A be a matrix such that, for everyxyzin R3, Axyz=y + z02x − z.Then, what is A?A =0 1 10 0 02 0 −1Short Problem 2. Let A =3 0 02 3
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