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 ShortsP/1 /11 /15 /15 /12 /8 /21 /83Good luck!1Problem 1. LetA =0 1 12 1 01 1 1.(a) [8 points] Determine A−1.(b) [3 points] Using A−1, solve Ax =101.Solution 1. (a) Using the Gauss–Jordan method, we find:0 1 1 1 0 02 1 0 0 1 01 1 1 0 0 1R1↔R3 1 1 1 0 0 12 1 0 0 1 00 1 1 1 0 0R2→R2−2R1 1 1 1 0 0 10 −1 −2 0 1 −20 1 1 1 0 0R3→R3+R2 1 1 1 0 0 10 −1 −2 0 1 −20 0 −1 1 1 −2R2→−R2R3→−R3 1 1 1 0 0 10 1 2 0 −1 20 0 1 −1 −1 2R1→R1−R3R2→R2−2R3 1 1 0 1 1 −10 1 0 2 1 −20 0 1 −1 −1 2R1→R1−R2 1 0 0 −1 0 10 1 0 2 1 −20 0 1 −1 −1 2Hence,A−1=−1 0 12 1 −2−1 −1 2.(b) We obtainx = A−1101=−1 0 12 1 −2−1 −1 2101=001.[Note that this is actually obvious when thinking in terms of the column picture of thelinear system.]2Problem 2. Consider the matrixA =1 1 11 4 41 4 8.(a) [10 points] Calculate the LU decomposition of A.(b) [5 points] Solve1 1 11 4 41 4 8x1x2x3=337without reducing the augmented matrix, but using the LU decomposition.Solution 2. (a) We first find U by Gaussian elimination:1 1 11 4 41 4 8R2→R2−R1R3→R3−R1 1 1 10 3 30 3 7R3→R3−R2 1 1 10 3 30 0 4We determine the matrix L from the row operations performed to get the LU decom-position1 1 11 4 41 4 8=1 0 01 1 01 1 11 1 10 3 30 0 4.(b) We solve Ax = b by first solving Lc = b via forward substitution. We find1 0 01 1 01 1 1c =337=⇒ c =304.Finally, we solve U x = c via backward substitution.1 1 10 3 30 0 4x =304=⇒ x =3−11.3Problem 3. Consider the following system of linear equations:x1−2x2+x3= 1−x1+2x2+x3+2x4= 1−2x1+4x2+4x3+6x4= 4(a) [2 points] Write down the augmented matrix corresponding to this system.(b) [7 points] Determine the row reduced echelon form of the augmented matrix.(c) [6 points] Use your result in (b) to find a parametric description of the set of solutionsto the system of linear equations.Solution 3. (a) The augmented matrix is1 −2 1 0 1−1 2 1 2 1−2 4 4 6 4.(b) Gauss–Jordan elimination produces:1 −2 1 0 1−1 2 1 2 1−2 4 4 6 4R2→R2+R1R3→R3+2R1 1 −2 1 0 10 0 2 2 20 0 6 6 6R3→R3−3R2 1 −2 1 0 10 0 2 2 20 0 0 0 0R2→12R2 1 −2 1 0 10 0 1 1 10 0 0 0 0R1→R1−R2 1 −2 0 −1 00 0 1 1 10 0 0 0 0Hence, the row reduced echelon form is1 −2 0 −1 00 0 1 1 10 0 0 0 0.(c) The variables x2and x4are free, and we find the parametric description of the solutionsas follows:x1= 2x2+ x4x2is freex3= 1 − x4x4is freeOptionally, and equivalently, we can write the set of solutions as2x2+ x4x21 − x4x4: x2, x4in R=0010+ span2100,10−11.[Note that the final span is the null space of the coefficient matrix.]4Problem 4. Letw =1h3h, v1=111, v2=123.(a) [8 points] For which value of h is w a linear combination of v1and v2?(b) [4 points] For the value of h found in (a), write down the linear combination of v1andv2which gives w.Solution 4. (a) w is a linear combination of v1and v2if and only if the system withaugmented matrix1 1 11 2 h1 3 3his consistent. From the echelon form1 1 11 2 h1 3 3hR2→R2−R1R3→R3−R1 1 1 10 1 h − 10 2 3h − 1R3→R3−2R2 1 1 10 1 h − 10 0 h + 1we find that the system is consistent if and only if h = −1.Hence, w is a linear combination of v1and v2if and only if h = −1.(b) The coefficients of such a linear combinations are given by the solutions of the system.In the case h = −1, this system has augmented matrix1 1 10 1 h − 10 0 h + 1=1 1 10 1 −20 0 0.A simple backward substitution gives the solution x2= −2 and x1= 3. The corre-sponding linear combination is3v1− 2v2= w.5Problem 5. [8 points] Determine which of the following sets are a subspace of the vectorspace of all 2 × 2 matrices. In each case, give a short reason.(a) W1=2a bb 3a: a, b in R(b) W2=2a bb 3a: a, b in R and a + b = 1Solution 5. (a) W1is a subspace because we can write it as a span:W1= span2 00 3,0 11 0(b) W2is not a subspace because it does not contain the zero vector. Indeed, if2a bb 3a=0 00 0,then a = 0 and b = 0. But this contradicts a + b = 1.6SHORT 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 =1 00 10 1. Compute ATA.ATA =1 0 00 1 11 00 10 1=1 00 2Short Problem 2. Let A be a matrix such that, for everyxyzin R3, Axyz=2yxx − z.Then, what is A?A =0 2 01 0 01 0 −1Short Problem 3. Let C be a 3 × 4 matrix such that C has two pivot columns, and let dbe a vector in R3. Is it true that, if the equation Cx = d has a solution, then it has infinitelymany solutions?(a) True.(b) False.(c) Unable to determine.This is true, because there is a free variable (actually, it is two). Hence, the system hasinfinitely many solutions unless it is inconsistent.7Short Problem 4. LetA =a a + 1a + 1 a.For which choice(s) of a is the matrix A not invertible?A 2 × 2 matrixa bc dis not invertible if and only if ad − bc = 0. Since a2− (a + 1)2=−2a − 1, A is not invertible if and only if a = −12.Short Problem 5. There is one vector which every subspace of R2has to contain. Whichvector is that?00Short Problem 6. Let W1be the set of all polynomials p(t)
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