Right answers, and common wrong answers, to midterm #1 -------------------------------------------------------------------------- True/False. If A^2=I, then A must be invertible. TRUE. By definition, A is invertible if there is a square matrix B such that AB=I=BA. This is satisfied if we choose B=A, therefore A satisfies the definition of invertibility, so it is invertible. Marco ----- 3 points were given for the right answer, and additional points for a CLEAR explaination. Answers like "A^2=I, A=A^{-1}, so A is invertible" usually were not given more than 3 points, becuase they don't explain WHY A is invertible or WHY the inverse of A exists (and is equal to A). Grading the problems I saw various ways in which the students concluded "A=A^{-1}", and I set up my grading scheme so that those who got to this conclusion without at least hinting at a correct reasoning or by a wrong reasoning received only 3 points. However there were several cases where the statement "A=A^-1" was derived in an incorrect way. Among the most common ones: A^2=I, therefore (multiplying with A^{-1}) we have A=A^{-1}, therefore A^{-1} exists (since A exists), therefore A is invertible. This is wrong because one uses the existence of A^{-1} in this argument, but the existence of A^{-1} is exactly what one has to show. Other common mistakes: AA=I. The identity is invertible, so A*A is invertible and equal to A^{-1}*A^{-1}. Therefore A^{-1} exists and A is invertible. The statement that A*A is invertible is legitimate, but to conclude that its inverse is A^{-1}*A^{-1} one needs first to know that A is invertible. -------------------------------------------------------------------------- True/False. There exists a system of three linear equations in three unknowns that has exactly three solutions. FALSE. A linear system of equations can only have 0, 1 or infinitely many solutions. Paulo ----- In R^3 this can be exemplified by intersection of planes, which are the solutions set of each equation. Three planes may not intersect at all (eg. when they are all parallel and disjoint), have one point of intersection and have infinitely many points of intersection (eg. when they intersect along a line or are all coincident). The main problems with student's solutions fall into three categories. 1. Stating that the result is true and showing that a system can have "at least three" solutions. Observe the problem asked for "exactly three solutions". 2. Stating that the result is false and going about finding a system with infinitely many solutions. Observe that, the existence of one system with infinitely many solutions, is not a counter-example for the statement. 3. Stating that it is false, and exemplifying it with lines in R^3 as opposed to planes. Note that the solution set of one equation in R^3 is a plane, not a line. -------------------------------------------------------------------------- True/False. If A^2=I, then A must be I or -I. FALSE. Two counterexamples are [0 1] and [1 0] [1 0] [0 -1]. J. Ben ------------ The only scores I gave were 10, 5, 3 or 0. I awarded 5 points if the student answered "false" and indicated that A could be a permutation matrix, but didn't give a specific example. Otherwise, a correct answer with no example was awarded 3 points. -------------------------------------------------------------------------- True/False. Let a, b, c be three unit vectors in R^2. If a is orthogonal to b, and b is orthogonal to c, then a = c. FALSE. c can be in the opposite direction from a: if a is unit, orthogonal to b, then so is -a. Alice ----- Main mistakes: failure to read the word "unit" vector, thus giving non-unit counterexamples; counterexamples in R^3 rather than R^2; also, quite a few people gave as counterexamples vectors with 3 entries, the last being zero on all of them. a few people seem confused about who's supposed to be orthogonal to whom, or just mislabeled their picture. -------------------------------------------------------------------------- True/False. Let A,B be square matrices such that A^2 = AB = B^2 = 0. Then (A+B)^2 = 0. FALSE. Since (A+B)^2 = A^2 + AB + BA + B^2 = 0 + 0 + BA + 0, we need to find two matrices A,B satisfying the above but with BA not zero. One example is A = [0 0 0], B = [0 1 0] [0 0 1] [0 0 0] [0 0 0] [0 0 0] Peter ----- To find a counterexample, you needed to use 3 by 3 matrices or larger, and this confused many people. Other common errors were 1: Simple saying (A+B)^2 = A^2+2AB+B^2 2: Saying that A^2=AB implies A=B. This is not true since A need not be invertible (and in fact cannot be). 3: Saying A^2=0 implies A=0 (I only saw a couple of these, but I did see them!) -------------------------------------------------------------------------- True/False. A system of four equations in three unknowns is always inconsistent (has no solutions). FALSE. See below for counterexamples. Ari --- One good solution to my problem was the system: x = 1 y = 1 z = 1 x + y + z = 3 Another of my favorites was: x + y + z = 1 2x + 2y + 2z = 2 3x + 3y + 3z = 3 4x + 4y + 4z = 4 The most frequent mistakes on my problem were: - not giving an example, or giving an example that was just arbitrary letters - giving an example that was an inconsistent system (how many points they got for this depended on whether they'd made an arithmetic error adding up rows or just come up with a bunch of random numbers) - incorrect reasoning (e.g., "The fourth column must be a linear combination of the other three, so it's consistent," or "Four lines in R^3 can intersect at a point.") In all cases, partial credit depended on the quality of their explanation and how clear it was that they actually understood what was going on. An inconsistent system could yield only four points if that was all they wrote. If they gave a convincing explanation and an obvious arithmetic mistake made their system inconsistent, they could get as many as eight. -------------------------------------------------------------------------- The "solve for choke" problem. Answer. We know that there is some nonzero vector a[1 4 0 -2] + b[0 2 2 6] (forgive the use of row vectors, that can also be written for typesetting purposes) c[1 4 3 0] + d[2 6 1 x]. If d isn't zero, then we're in business, [2 6 1 x] = 1/d (a[1 4 0 -2] + b[0 2 2 6] - c[1 4 3 0]) because now we know that the mystery vector is in the span of the other three. So if we put all four in a matrix and do Gaussian elimination, we should be able to totally kill the [2 6 1 x] vector. Gaussian eliminate, see what's left over, and require it to be zero, giving an equation for x. What if d is zero? Then c can't be zero (this is where we use that it was a _nonzero_ vector in both spans). So c[1 4 3 0] = a[1 4 0 -2] + b[0 2 2 6] and dividing by c, we find out that [1 4 3 0] is in the span of [1 4 0 -2], [0 2 2 6]. But it's not (as we can discover during the same Gaussian elimination process). Yuan ---- Some notice the following relationship by inspection, (1, 4, 0, -2) - (0, 2, 2, 6) = (1 ,2, -2, -8) and -(1,4,3,0) + (2, 6, 1, x) = (1,2,-2,x), then set x = -8 to give the answer. This approach needs extensive explanation on why no other choice for x, which were missed in their answers. Some try to transform the four vectors into an orthogonal set by Gram-Schmidt process, then get lost. A few notice correctly that the last vector containing x must be in the span of the first three (one or two obtain this by dividing d on both sides of the equations), but they did not explain the reason (e.g. why d is not 0 here). -------------------------------------------------------------------------- Compute the 48th power of [a a] [0 b]. There were maybe seven different correct answers given to this: the most common was [a^48 a^48 + a^47 b + a^46 b^2 + ... + a b^47] [ 0 b^48] Darren, J. Ben, and Allen ------------------------- We split the 20 points into 8 points for getting the NW, SW, and SE entries right, and 12 for the NE entry alone. Essentially everyone got the 0 entry correct. Almost everyone who got it right did so by looking at the first few powers and guessing the generalization to Nth powers. If you either went up to the 5th power, or not as far but gave a sentence explaining why your generalization looked right, you got full credit. Some people went to the 4th power, then guessed wrong, but everyone who went as far as 5th power guessed correctly at that point. An incredibly common error -- and a really bad one! Take note, and don't do this again! -- was to screw up exponentiation. Essentially, people replaced 48 = 2^5 + 2^4 by 2*5 + 2*4, and computed the 18th power, believing M^32 = M^(2^5) = (M^2)^5 = M^10. This should have set off *big* warning bells. One very sad mistake was to correctly guess the formula for the Nth power, then plug in N=18. Other powers seen were the 24th, 96th, 47th, 9th (beyond 2*5 + 2*4 to 5+4), 33rd (the 32=16+16 became 16+1), 23rd (from 16+17=23), and 42nd (a simple miscopying of 48). In general, if your answer was wrong, and your algebra was wrong, and you included no English explaining what you were doing, you'd get very few points. If your answer was correct -- and we got some answers that looked very weird, but were correct -- you'd only lose points if you had no explanation, and then only a couple. (By the way, there is a good formula for the NE corner of the Nth power, which nobody got: a (a^N - b^N) / (a-b). See what happens if you multiply by the original matrix again.)