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18.06 Linear Algebra, Spring 2005Please use the following citation format:Gilbert Strang, 18.06 Linear Algebra, Spring 2005. (MassachusettsInstitute of Technology: MIT OpenCourseWare). (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike.Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: /terms18.06 Linear Algebra, Spring 2005Transcript – Lecture 12OK. This is lecture twelve.We've reached twelve lectures. And this one is more than the others about applications of linear algebra.And I'll confess. When I'm giving you examples of the null space and the row space, I create a little matrix. You probably see that I just invent that matrix as I'm going.And I feel a little guilty about it, because the truth is that real linear algebra uses matrices that come from somewhere. They're not just, like, randomly invented by the instructor.They come from applications. They have a definite structure.And anybody who works with them gets, uses that structure.I'll just report, like, this weekend I was at an event with chemistry professors. OK, those guys are row reducing matrices, and what matrices are they working with? Well, their little matrices tell them how much of each element goes into the --or each molecule, how many molecules of each go into a reaction and what comes out. And by row reduction they get a clearer picture of a complicated reaction. And this weekend I'm going to --to a sort of birthday party at Mathworks. So Mathworks is out Route 9 in Natick.That's where Matlab is created. It's a very, very successful, software, tremendously successful. And the conference will be about how linear algebra is used. And so I feel better today to talk about what I think is the most important model in applied math. And the discrete version is a graph. So can I draw a graph? Write down the matrix that's associated with it, and that's a great source of matrices. You'll see. So a graph is just, so a graph --to repeat --has nodes and edges. OK.And I'm going to write down the graph, a graph, so I'm just creating a small graph here.As I mentioned last time, we would be very interested in the graph of all, websites.Or the graph of all telephones. I mean --or the graph of all people in the world. Here let me take just, maybe nodes one two three --well, I better put in an --I'll put in that edge and maybe an edge to, to a node four, and another edge to node four. How's that? So there's a graph with four nodes.So n will be four in my --n equal four nodes.And the matrix will have m equal the number --there'll be a row for every edge, so I've got one two three four five edges. So that will be the number of rows. And I have to to write down the matrix that I want to, I want to study, I need to give a direction to every edge, so I know a plus and a minus direction. So I'll just do that with an arrow. Say from one to two, one to three, two to three, one to four, three to four.That just tells me, if I have current flowing on these edges then I know whether it's --to count it as positive or negative according as whether it's with the arrow or against the arrow. But I just drew those arrows arbitrarily. OK.Because I --my example is going to come --the example I'll --the words that I will use will be words like potential, potential difference, currents. In other words, I'm thinking of an electrical network.But that's just one possibility.My applied math class builds on this example.It could be a hydraulic network, so we could be doing, flow of water, flow of oil. Other examples, this could be a structure.Like the --a design for a bridge or a design for a Buckminster Fuller dome. Or many other possibilities, so many. So l-but let's take potentials and currents as, as a basic example, and let me create the matrix that tells you exactly what the graph tells you. So now I'll call it the incidence matrix, incidence matrix. OK. So let me write it down, and you'll see, what its properties are.So every row corresponds to an edge.I have five rows from five edges, and let me write down again what this graph looks like.OK, the first edge, edge one, goes from node one to two. So I'm going to put in a minus one and a plus one in th-this corresponds to node one two three and four, the four columns. The five rows correspond --the first row corresponds to edge one. Edge one leaves node one and goes into node two, and that --and it doesn't touch three and four.Edge two, edge two goes --oh, I haven't numbered these edges.I just figured that was probably edge one, but I didn't say so. Let me take that to be edge one. Let me take this to be edge two. Let me take this to be edge three. This is edge four. Ho, I'm discovering --no, wait a minute.Did I number that twice? Here's edge four. And here's edge five. OK? All right. So, so edge one, as I said, goes from node one to two.Edge two goes from two to three, node two to three, so minus one and one in the second and third columns.Edge three goes from one to three.I'm, I'm tempted to stop for a moment with those three edges.Edges one two three, those form what would we, what do you call the, the little, the little, the subgraph formed by edges one, two, and three? That's a loop. And the number of loops and the position of the loops will be crucial.OK. Actually, here's a interesting point about loops. If I look at those rows, corresponding to edges one two three, and these guys made a loop. You want to tell me --if I just looked at that much of the matrix it would be natural for me to ask, are those rows independent? Are the rows independent? And can you tell from looking at that if they are or are not independent? Do you see a, a relation between those three rows? Yes.If I add that row to that row, I get this row.So, so that's like a hint here that loops correspond to dependent, linearly dependent column --linearly dependent rows. OK, let me complete the incidence matrix. Number four, edge four is going from node one to node four.And the fifth edge is going from node three to node four.OK. There's my matrix.It came from the five edges and the four nodes.And if I had a big graph, I'd have a big matrix.And what questions do I ask about matrices? Can I ask --here's the review now. There's a matrix that comes from somewhere.If, if it was a big graph, it would be a large matrix, but a lot of zeros, right? Because every row only has two non-zeros.So the number of --it's a very sparse matrix.The number of non-zeros is exactly two times five, it's two m. Every row only has two non-zeros. And that's with a lot of structure. And --that was the point I wanted to begin with, that graphs, that real graphs from real --real matrices from genuine problems have structure. OK.We can ask, and because of the structure, we can answer, the, the main questions about matrices.So first question, what about the null space? So what I asking if I ask you for the null space of that matrix? I'm asking you if I'm looking at the columns of the matrix, four columns, and I'm asking you, are those columns independent? If the columns are independent, then what's in the null space? Only the zero vector, right? The null space contains --tells us what combinations of the columns --it tells us how to combine columns to get zero.Can --and is there anything in the null space of this matrix other than just the zero vector? In other words, are those four columns independent or dependent? OK.That's our question.Let me, I don't know if you see the answer.Whether there's --so let's see.I guess we could do it properly. We could solve Ax=0. So let me solve Ax=0 to find the null space. OK.What's Ax? Can I put x in here in, in little letters? x1, x2, x3, x4, that's --it's got four columns.Ax now is that matrix times x. And what do I get for Ax? If the camera can keep that matrix multiplication there, I'll put the answer here. Ax equal --what's the first component of Ax? Can you take that first row, minus one one zero zero, and multiply by the x, and of course you get x2-x1. The second row, I get x3-x2. From the third row, I get x3-x1. From the fourth row, I get x4-x1. And from the fifth row, I get x4-x3. And I want to know when is the thing zero. This is my equation, Ax=0. Notice what that matrix A is doing, what we've created a matrix that computes the differences across every edge, the differences in potential.Let me even begin to give this interpretation.I'm going to think of this vector x, which is x1 x2 x3 x4, as the potentials at the nodes. So I'm introducing a word, potentials at the nodes. And now if I multiply by A, I get these --I get these five components, x2-x1, et cetera. And what are they? They're potential differences. That's what A computes.If I have potentials at the nodes and I multiply by A, it gives me the potential differences, the differences in potential, across the edges. OK.When are those differences all zero? So I'm looking for the null space.Of course, if all the (x)s are zero then I get zero.That, that just tells me, of course, the zero vector is in the null space. But w-there's more in the null space. Those columns are --of A are dependent, right --because I can find solutions to that equation. Tell me --the null space.Tell me one vector in the null space, so tell me an x, it's got four components, and it makes that thing zero.So what's a good x to do that? One one one one, constant potential. If the potentials are constant, then all the potential differences are zero, and that x is in the null space.What else is in the null space? If it --yeah, let me ask you just always, give me a basis for the null space. A basis for the null space will be just that.1 That's --, that's it.That's a basis for the null space.The null space is actually one dimensional, and it's the line of all vectors through that one. So there's a basis for it, and here is the whole null space.Any multiple of one one one one, it's the whole line in four dimensional space. Do you see that that's the null space? So the, so the dimension of the null space of A is one. And there's a basis for it, and there's everything that's in it.Good. And what does that mean physically? I mean, what does that mean in the application? That guy in the null space.It means that the potentials can only be determined up to a constant. Potential differences are what make current flow. That's what makes things happen. It's these potential differences that will make something move in the, in our network, between x2-between node two and node one. Nothing will move if all potentials are the same. If all potentials are c, c, c, and c, then nothing will move. So we're, we have this one parameter, this arbitrary constant that raises or drops all the potentials.It's like ranking football teams, whatever.We have a, there's a, there's a constant --or looking at temperatures, you know, there's a flow of heat from higher temperature to lower temperature. If temperatures are equal there's no flow, and therefore we can measure --we can measure temperatures by, Celsius or we can start at absolute zero.And that arbitrary --it's the same arbitrary constant that, that was there in calculus. In calculus, right, when you took the integral, the indefinite integral, there was a plus c, and you had to set a starting point to know what that c was. So here what often happens is we fix one of the potentials, like the last one.So a typical thing would be to ground that node.To set its potential at zero. And if we do that, if we fix that potential so it's not unknown anymore, then that column disappears and we have three columns, and those three columns are independent.So I'll leave the column in there, but we'll remember that grounding a node is the way to get it out.And grounding a node is the way to --setting a node --setting a potential to zero tells us the, the base for all potentials. Then we can compute the others. OK. But what's the --now I've talked enough to ask what the rank of the matrix is? What's the rank then? The rank of the matrix.So we have a five by four matrix.We've located its null space, one dimensional.How many independent columns do we have? What's the rank? It's three.And the first three columns, or actually any three columns, will be independent. Any three potentials are independent, good variables. The fourth potential is not, we need to set, and typically we ground that node. OK.Rank is three. Rank equals three.OK. Let's see, do I want to ask you about the column space? The column space is all combinations of those columns. I could say more about it and I will. Let me go to the null space of A transpose, because the equation A transpose y equals zero is probably the most fundamental equation of applied mathematics. All right, let's talk about that. That deserves our attention. A transpose y equals zero. Let's --let me put it on here. OK. So A transpose y equals zero.So now I'm finding the null space of A transpose.Oh, and if I ask you its dimension, you could tell me what it is. What's the dimension of the null space of A transpose? We now know enough to answer that question. What's the general formula for the dimension of the null space of A transpose? A transpose, let me even write out A transpose. This A transpose will be n by m, right? n by m.In this case, it'll be four by five.Those columns will become rows. Minus one zero minus one minus one zero is now the first row. The second row of the matrix, one minus one and three zeros. The third column now becomes the third row, zero one one zero minus one. And the fourth column becomes the fourth row.OK, good. There's A transpose.That multiplies y, y1 y2 y3 y4 and y5.OK. Now you've had time to think about this question. What's the dimension of the null space, if I set all those --wow.Usually --sometime during this semester, I'll drop one of these erasers behind there. That's a great moment.There's no recovery. There's --centuries of erasers are back there. OK. OK, what's the dimension of the null space? Give me the general formula first in terms of r and m and n.This is like crucial, you --we struggled to, to decide what dimension meant, and then we figured out what it equaled for an m by n matrix of rank r, and the answer wasm-r, right? There are m=5 components, m=5 columns of A transpose. And r of those columns are pivot columns, because it'll have r pivots.It has rank r. And m-r are the free ones now for A transpose, so that's five minus three, so that's two. And I would like to find this null space. I know its dimension. Now I want to find out a basis for it.And I want to understand what this equation is.So let me say what A transpose y actually represents, why I'm interested in that equation.I'll put it down with those old erasers and continue this.Here's the great picture of applied mathematics.So let me complete that. There's a matrix that I'll call C that connects potential differences to currents.So I'll call these --these are currents on the edges, y1 y2 y3 y4 and y5. Those are currents on the edges. And this relation between current and potential difference is Ohm's Law.This here is Ohm's Law. Ohm's Law says that the current on an edge is some number times the potential drop. That's --and that number is the conductance of the edge, one over the resistance. This is the old current is, is, the relation of current, resistance, and change in potential. So it's a change in potential that makes some current happen, and it's Ohm's Law that says how much current happens. OK.And then the final step of this framework is the equation A transpose y equals zero. And that's --what is that saying? It has a famous name.It's Kirchoff's Current Law, KCL, Kirchoff's Current Law, A transpose y equals zero. So that when I'm solving, and when I go back up with this blackboard and solve A transpose y equals zero, it's this pattern of --that I want you to see. That we had rectangular matrices, but --and real applications, but in those real applications comes A and A transpose.So our four subspaces are exactly the right things to know about. All right.Let's know about that null space of A transpose.Wait a minute, where'd it go? There it is. OK.OK. Null space of A transpose.We know what its dimension should be.Let's find out --tell me a vector in it.Tell me --now, so what I asking you? I'm asking you for five currents that satisfy Kirchoff's Current Law. So we better understand what that law says. That, that law, A transpose y equals zero, what does that say, say in the first row of A transpose? That says --the so the first row of A transpose says minus y1 minus y3 minus y4 is zero. Where did that equation come from? Let me --I'll redraw the graph. Can I redraw the graph here, so that we --maybe here, so that we see again --there was node one, node two, node three, node four was off here. That was, that was our graph.We had currents on those. We had a current y1 going there. We had a current y --what were the other, what are those edge numbers? y4 here and y3 here. And then a y2 and a y5.I'm, I'm just copying what was on the other board so it's ea-convenient to see it. What is this equation telling me, this first equation of Kirchoff's Current Law? What does that mean for that graph? Well, I see y1, y3, and y4 as the currents leaving node one. So sure enough, the first equation refers to node one, and what does it say? It says that the net flow is zero.That, that equation A transpose y, Kirchoff's Current Law, is a balance equation, a conservation law.Physicists, be overjoyed, right, by this stuff.It, it says that in equals out. And in this case, the three arrows are all going out, so it says y1, y3, and y4 add to zero. Let's take the next one.The second row is y1-y2, and that's all that's in that row. And that must have something to do with node two. And sure enough, it says y1=y2, current in equals current out. The third one, y2 plus y3 minus y5 equals zero. That certainly will be what's up at the third node. y2 coming in, y3 coming in, y5 going out has to balance. And finally, y4 plus y5 equals zero says that at this node, y4 plus y5, the total flow, is zero. We don't --you know, charge doesn't accumulate at the nodes.It travels around. OK.Now give me --I come back now to the linear algebra question.What's a vector y that solves these equations? Can I figure out what the null space is for this matrix, A transpose, by looking at the graph? I'm happy if I don't have to do elimination. I can do elimination, we know how to do, we know how to find the null space basis.We can do elimination on this matrix, and we'll get it into a good reduced row echelon form, and the special solutions will pop right out. But I would like to --even to do it without that. Let me just ask you first, if I did elimination on that, on that, matrix, what would the last row become? What would the last row --if I do elimination on that matrix, the last row of R will be all zeros, right? Why? Because the rank is three. We only going to have three pivots. And the fourth row will be all zeros when we eliminate. So elimination will tell us what, what we spotted earlier, what's the null space --all the, all the information, what are the dependencies. We'll find those by elimination, but here in a real example, we can find them by thinking. OK. Again, my question is, what is a solution y? How could current travel around this network without collecting any charge at the nodes? Tell me a y. OK.So a basis for the null space of A transpose.How many vectors I looking for? Two.It's a two dimensional space. My basis should have two vectors in it. Give me one. One set of currents. Suppose, let me start it.Let me start with y1 as one. OK.So one unit of --one amp travels on edge one with the arrow. OK, then what? What is y2? It's one also, right? And of course what you did was solve Kirchoff's Current Law quickly in the second equation.OK. Now we've got one amp leaving node one, coming around to node three.What shall we do now? Well, what shall I take for y3 in other words? Oh, I've got a choice, but why not make it what you said, negative one.So I have just sent current, one amp, around that loop.What shall y4 and y5 be in this case? We could take them to be zero. This satisfies Kirchoff's Current Law. We could check it patiently, that minus y1 minus y3 gives zero.We know y1 is y2. The others, y4 plus y5 is certainly zero. Any current around a loop satisfies --satisfies the Current Law. OK. Now you know how to get another one. Take current around this loop. So now let y3 be one, y5 be one, and y4 be minus one.And so, so we have the first basis vector sent current around that loop, the second basis vector sends current around that loop. And I've --and those are independent, and I've got two solutions --two vectors in the null space of A transpose, two solutions to Kirchoff's Current Law.Of course you would say what about sending current around the big loop. What about that vector? One for y1, one for y2, nothing f-on y3, one for y5, and minus one for y4. What about that? Is that, is that in the null space of A transpose? Sure. So why don't we now have a third vector in the basis? Because it's not independent, right? It's not independent. This vector is the sum of those two. If I send current around that and around that --then on this edge y3 it's going to cancel out and I'll have altogether current around the whole, the outside loop. That's what this one is, but it's a combination of those two.Do you see that I've now, I've identified the null space of A transpose --but more than that, we've solved Kirchoff's Current Law.And understood it in terms of the network.OK. So that's the null space of A transpose. I guess I --there's always one more space to ask you about. Let's see, I guess I need the row space of A, the column space of A transpose. So what's N, what's its dimension? Yup? What's the dimension of the row space of A? If I look at the original A, it had five rows.How many were independent? Oh, I guess I'm asking you the rank again, right? And the answer is three, right? Three independent rows. When I transpose it, there's three independent columns. Are those columns independent, those three? The first three columns, are they the pivot columns of the matrix? No. Those three columns are not independent. There's a in fact, this tells me a relation between them. There's a vector in the null space that says the first column plus the second column equals the third column.They're not independent because they come from a loop. So the pivot columns, the pivot columns of this matrix will be the first, the second, not the third, but the fourth. One, columns one, two, and four are OK. Where are they --those are the columns of A transpose, those correspond to edges.So there's edge one, there's edge two, and there's edge four. So there's a --that's like --is a, smaller graph.If I just look at the part of the graph that I've, that I've, thick --used with thick edges, it has the same four nodes. It only has three edges.And the, those edges correspond to the independent guys. And in the graph there --those three edges have no loop, right? The independent ones are the ones that don't have a loop. All the --dependencies came from loops. They were the things in the null space of A transpose. If I take three pivot columns, there are no dependencies among them, and they form a graph without a loop, and I just want to ask youwhat's the name for a graph without a loop? So a graph without a loop is --has got not very many edges, right? I've got four nodes and it only has three edges, and if I put another edge in, I would have a loop. So it's this graph with no loops, and it's the one where the rows of A are independent.And what's a graph called that has no loops? It's called a tree. So a tree is the name for a graph with no loops. And just to take one last step here. Using our formula for dimension. Using our formula for dimension, let's look --once at this formula.The dimension of the null space of A transpose is m-r.OK. This is the number of loops, number of independent loops. m is the number of edges.And what is r? What is r for our --we'll have to remember way back. The rank came --from looking at the columns of our matrix. So what's the rank? Let's just remember. Rank was --you remember there was one --we had a one dimensional --rank was n minus one, that's what I'm struggling to say. Because there were n columns coming from the n nodes, so it's minus, the number of nodes minus one, because of that C, that one one one one vector in the null space.The columns were not independent.There was one dependency, so we needed n minus one.This is a great formula. This is like the first shall I, --write it slightly differently? The number of edges --let me put things --have I got it right? Number of edges is m, the number --r-is m-r, OK. So, so I'm getting --let me put the number of nodes on the other side. So I --the number of nodes --I'll move that to the other side --minus the number of edges plus the number of loops is --I have minus, minus one is one.The number of nodes minus the number of edges plus the number of loops is one. These are like zero dimensional guys. They're the points on the graph. The edges are like one dimensional things, they're, they connect nodes. The loops are like two dimensional things. They have, like, an area. And this count works for every graph.And it's known as Euler's Formula.We see Euler again, that guy never stopped.OK. And can we just check --so what I saying? I'm saying that linear algebra proves Euler's Formula. Euler's Formula is this great topology fact about any graph. I'll draw, let me draw another graph, let me draw a graph with more edges and loops. Let me put in lots of --OK. I just drew a graph there.So what are the, what are the quantities in that formula? How many nodes have I got? Looks like five. How many edges have I got? One two three four five six seven. How many loops have I got? One two three.And Euler's right, I always get one.That, this formula, is extremely useful in understanding the relation of these quantities --the number of nodes, the number of edges, and the number of loops. OK.Just complete this lecture by completing this picture, this cycle. So let me come to the --so this expresses the equations of applied math. This, let me call these potential differences, say, E. So E is A x.That's the equation for this step.The currents come from the potential differences. y is C E. The potential --the currents satisfy Kirchoff's Current Law. Those are the equations of --with no source terms. Those are the equations of electrical circuits of many --those are like the, the most basic three equations. Applied math comes in this structure. The only thing I haven't got yet in the picture is an outside source to make something happen.I could add a current source here, I could, I could add external currents going in and out of nodes.I could add batteries in the edges.Those are two ways. If I add batteries in the edges, they, they come into here.Let me add current sources. If I add current sources, those come in here. So there's a, there's where current sources go, because the F is a like a current coming from outside. So we have our edges, we have our graph, and then I send one amp into this node and out of this node --and that gives me, a right-hand side in Kirchoff's Current Law. And can I --to complete the lecture, I'm just going to put these three equations together.So I start with x, my unknown.I multiply by A. That gives me the potential differences. That was our matrix A that the whole thing started with. I multiply by C.Those are the physical constants in Ohm's Law.。
MIT线性代数2014期末考题
1. (15 points) (a) If A is a 3 by 4 matrix, what does this tell us about its nullspace? (b) If we also know that 1 Ax = 1 1 has no solution, what do we know about the rank of A? (c) If Ax = b and AT y = 0, find y T b by using those equations. This says that the space of A and the are .
(a) Give a basis for the nullspace of A (that matrix is not shown) and a basis for the row space of A. (b) When does Ax = b have a solution? Give a basis for the column space of A. (c) Give a basis for the nullspace of AT .
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8. (15 points) A 4-node graph has all six possible edges. Its incidence matrix A and its Laplacian matrix AT A are A= −1 1 0 −1 0 1 −1 0 0 0 −1 1 0 −1 0 0 0 −1 0 0 1 0 1 1 3 −1 −1 −1 −1 3 −1 −1 AT A = −1 −1 3 −1 −1 −1 −1 3
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6. (10 points) (a) Find the eigenvalues of this matrix A (the numbers in each column add to zero). 1 0 −1 2 A = 1 −1 1 . 1 0 2 −1 (b) If you solve
线性代数及其应用_英文版_答案__第三章
1593.1 SOLUTIONSNotes : f time is needed for other topics, this chapter may be omitted. Section 5.2 contains enoughinformation about determinants to support the discussion there of the characteristic polynomial of a matrix. In section 5.1, some exercises in this section provide practice in computing determinants, while others allow the student to discover the properties of determinants which will be studied in the next section. Determinants are developed through the cofactor expansion, which is given in Theorem 1. Exercises 33–36 in this section provide the first step in the inductive proof of Theorem 3 in the next section.A “Checkpoint ” in the Study Guide leads students to discover that if the k th column of the identity matrix is replaced by a vector x , then the determinant of the resulting matrix is the k th entry of x . This idea is used in the proof of Cramer’s Rule, in Section 3.3. 1. Expand across along the first row: 3043222232323043(13)4(10)1510105051=−+=−+=−−− Expand down the second column:122232304223434232(1)0(1)3(1)53(3)5(2)101012251+++=−⋅+−⋅+−⋅=−−−=−−− 2. Expand across the first row: 0513040434300515(4)1(22)2412124241−−−=−+=−+= Expand down the second column:122232051400101430(1)5(1)(3)(1)45(4)3(2)4(4)221214241+++−=−⋅+−⋅−+−⋅=−−−−−= 3. Expand across the first row: 2431232313122(4)32(9)4(5)(3)(11)5411114141−=−−+=−+−+=−−−−160 CHAPTER 3 • DeterminantsExpand down the second column: 122232243322323312(1)(4)(1)1(1)44(5)1(5)4(5)5111132141+++−=−⋅−+−⋅+−⋅=−+−−−=−−−− 4. Expand across the first row: 1351121212111351(2)3(1)5(5)20423234342=−+=−−+= Expand down the second column:122232135211515211(1)3(1)1(1)43(1)1(13)4(9)20323221342+++=−⋅+−⋅+−⋅=−+−−−= 5. Expand across the first row: 2340545440523(4)2(5)3(1)4(4)23165651516−=−+−=−−−−=− 6. Expand across the first row: 524350503355(2)45(1)2(10)4(6)1472724247−−−−=−−+=++−=−−− 7. Expand across the first row: 4305262656524304(1)3(0)4739397973=−+=−= 8. Expand across the first row: 81634344038168(6)1(11)6(8)11253532325=−+=−+−=−−−− 9. First expand across the third row, then expand across the first row of the remaining matrix:3113600505172572(1)27252(1)510(1)102000313188318++−=−⋅−=⋅−⋅==Chapter 3 • Supplementary Exercises 16110. First expand across the second row, then expand either across the third row or down the second columnof the remaining matrix.2312521220030(1)3265267554544+−−=−⋅−−− 31332212(3)(1)5(1)4(3)(5(2)4(2))66526++⎛⎞−−=−−⋅+−⋅=−+−=−⎜⎟−−⎝⎠or2312521220030(1)3265267554544+−−=−⋅−−− 12222512(3)(1)(2)(1)(6)5454++⎛⎞=−−⋅−+−⋅−⎜⎟⎝⎠()(3)2(17)6(6)6=−−−−=− 11. Following the text’s instruction, a good strategy is to expand down the first column of the matrix, andrepeat the process until the determinant is expressed as the product of the diagonal entries of the original matrix:11113584237023715(1)30153(1)(2)0015222++−−−−−=−⋅=⋅−⋅− = 3(–2)(2) = –12 Of course, with Theorem 2 available, the best strategy is to use it and simply compute the product of thediagonal entries in the matrix. 12. Following the text’s instruction, a good strategy is to expand along the first row of the matrix, and repeatthe process until the determinant is expressed as the product of the diagonal entries of the original matrix:1111400010071003(1)46304(1)(1)2630438435843++−−=−⋅=⋅−⋅−−−−−− = 4(–1)( –9) = 36 Of course, with Theorem 2 available, the best strategy is to use it and simply compute the product of thediagonal entries in the matrix. 13. First expand either across the second row or down the second column. Using the second row,23407354035002007348(1)27364850235052312912+−−−−=−⋅−−−−−−162 CHAPTER 3 • DeterminantsNow expand down the second column to find: 232240354357348(1)22(1)3523502301212++−⎛⎞−−⎜⎟−⋅=−−⋅−⎜⎟−⎜⎟−⎝⎠− Now expand either down the first column or across third row. Using the first column, 224352(1)3523012+⎛⎞−⎜⎟−−⋅−⎜⎟⎜⎟−⎝⎠112123356(1)4(1)5(6)(4(1)5(1))61212++⎛⎞−−=−−⋅+−⋅=−−=⎜⎟−−⎝⎠14. First expand either across the fourth row or down the fifth column. Using the fifth column,35632406324904109041(1)18567130003000042324232+−−=−⋅− Now expand across the third row to find: 353163243249041(1)11(1)304130002324232++⎛⎞−⎜⎟−⋅=−⋅−⎜⎟⎜⎟⎝⎠Finally, expand either down the first column or along second row. Using the first column,313241(1)3041232+⎛⎞⎜⎟−⋅−⎜⎟⎜⎟⎝⎠113141243(1)3(1)2(3)(3(11)2(18))93241++⎛⎞−=−⋅+−⋅=−+=⎜⎟−⎝⎠15. 304232051=− (3)(3)(–1) + (0)(2)(0) + (4)(2)(5) – (0)(3)(4) – (5)(2)(3) – (–1)(2)(0) =–9 + 0 + 40 – 0 – 30 –0 = 116. 051430241−= (0)(–3)(1) + (5)(0)(2) + (1)(4)(4) – (2)(–3)(1) – (4)(0)(0) – (1)(4)(5) =0 + 0 + 16 – (–6) – 0 – 20 = 217. 243312141−=− (2)(1)(–1) + (–4)(2)(1) + (3)(3)(4) – (1)(1)(3) – (4)(2)(2) – (–1)(3)(–4) = –2 + (–8) + 36 – 3 – 16 – 12 = –5Chapter 3 • Supplementary Exercises 16318. 135211342= (1)(1)(2) + (3)(1)(3) + (5)(2)(4) – (3)(1)(5) – (4)(1)(1) – (2)(2)(3) =2 + 9 + 40 – 15 – 4 – 12 = 2019.,a b ad bc cd =− ()c dcb da ad bc ab=−=−− The row operation swaps rows 1 and 2 of the matrix, and the sign of the determinant is reversed. 20.,a b ad bc cd =− ()()()a ba kd kcb kad kbc k ad bc kckd=−=−=− The row operation scales row 2 by k , and the determinant is multiplied by k . 21.3418202,56=−=− 343(64)(53)425364k k kk=+−+=−++The row operation replaces row 2 with k times row 1 plus row 2, and the determinant is unchanged. 22.,a b ad bc cd =− ()()a kc b kda kc d cb kd ad kcd bc kcd ad bc cd++=+−+=+−−=− The row operation replaces row 1 with k times row 2 plus row 1, and the determinant is unchanged. 23. 1113841(4)1(2)1(7)5,232−−=−+−=−− 384(4)(2)(7)5232k k kk k k k −−=−+−=−− The row operation scales row 1 by k , and the determinant is multiplied by k . 24. 322(2)(6)(3)263,656a b ca b c a b c =−+=−+ 3223(65)2(66)2(56)263656ab c b c a c a b a b c =−−−+−=−+− The row operation swaps rows 1 and 2 of the matrix, and the sign of the determinant is reversed. 25. By Theorem 2, the determinant of a triangular matrix is the product of the diagonal entries: 10010(1)(1)(1)101k== 26. By Theorem 2, the determinant of a triangular matrix is the product of the diagonal entries: 10010(1)(1)(1)101k==164 CHAPTER 3 • Determinants27. By Theorem 2, the determinant of a triangular matrix is the product of the diagonal entries: 0010()(1)(1)01k k k == 28. By Theorem 2 the determinant of a triangular matrix is the product of the diagonal entries: 1000(1)()(1)01k k k == 29. A cofactor expansion across row 1 gives 010110011011=−=− 30. A cofactor expansion across row 1 gives 001011011101==− 31. A 3 × 3 elementary row replacement matrix looks like one of the six matrices 100100100100101010,010,010,01k ,010,010010101010101k kkkk⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦In each of these cases, the matrix is triangular and its determinant is the product of its diagonal entries,which is 1. Thus the determinant of a 3 × 3 elementary row replacement matrix is 1. 32. A 3 × 3 elementary scaling matrix with k on the diagonal looks like one of the three matrices 00100100010,00,01001010k k k ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦In each of these cases, the matrix is triangular and its determinant is the product of its diagonal entries,which is k. Thus the determinant of a 3 × 3 elementary scaling matrix with k on the diagonal is k . 33. 01,10E ⎡⎤=⎢⎥⎣⎦,a b A c d ⎡⎤=⎢⎥⎣⎦c d EA a b ⎡⎤=⎢⎥⎣⎦ det E = –1, det A = ad – bc ,det EA = cb – da = –1(ad – bc ) = (det E )(det A )34. 10,0E k ⎡⎤=⎢⎥⎣⎦,a b A c d ⎡⎤=⎢⎥⎣⎦ab EA kc kd ⎡⎤=⎢⎥⎣⎦ det E = k , det A = ad – bc ,det EA = a (kd ) – (kc )b = k (ad – bc ) = (det E )(det A )Chapter 3 • Supplementary Exercises 16535. 1,01k E ⎡⎤=⎢⎥⎣⎦,a b A c d ⎡⎤=⎢⎥⎣⎦a kc b kd EA c d ++⎡⎤=⎢⎥⎣⎦ det E = 1, det A = ad – bc ,det EA = (a + kc )d – c (b + kd ) = ad + kcd – bc – kcd = 1(ad – bc ) = (det E )(det A )36. 10,1E k ⎡⎤=⎢⎥⎣⎦,a b A c d ⎡⎤=⎢⎥⎣⎦ab EA kac kbd ⎡⎤=⎢⎥++⎣⎦det E = 1, det A = ad – bc ,det EA = a (kb + d ) – (ka + c )b = kab + ad – kab – bc = 1(ad – bc ) = (det E )(det A ) 37. 31,42A ⎡⎤=⎢⎥⎣⎦ 1555,2010A ⎡⎤=⎢⎥⎣⎦det A = 2, det 5A = 50 ≠ 5det A 38. ,a b A c d ⎡⎤=⎢⎥⎣⎦ ,ka kb kA kc kd ⎡⎤=⎢⎥⎣⎦det A = ad – bc , 22det ()()()()()det kA ka kd kb kc k ad bc k A =−=−=39. a . True. See the paragraph preceding the definition of the determinant.b . False. See the definition of cofactor, which precedes Theorem 1. 40. a . False. See Theorem 1. b . False. See Theorem 2.41. The area of the parallelogram determined by 3,0⎡⎤=⎢⎥⎣⎦u 1,2⎡⎤=⎢⎥⎣⎦v u + v , and 0 is 6, since the base of theparallelogram has length 3 and the height of the parallelogram is 2. By the same reasoning, the area ofthe parallelogram determined by 3,0⎡⎤=⎢⎥⎣⎦u ,2x ⎡⎤=⎢⎥⎣⎦x u + x , and 0 is also 6.X 1Also note that []31det det 6,02⎡⎤==⎢⎥⎣⎦uv and []3det det 6.02x ⎡⎤==⎢⎥⎣⎦u x The determinant of the matrix whose columns are those vectors which define the sides of the parallelogram adjacent to 0 is equal to the area of the parallelogram.166 CHAPTER 3 • Determinants42. The area of the parallelogram determined by a b ⎡⎤=⎢⎥⎣⎦u , 0c ⎡⎤=⎢⎥⎣⎦v , u + v , and 0 is cb , since the base of theparallelogram has length c and the height of the parallelogram is b.X 1Also note that []det det 0a c cb b ⎡⎤==−⎢⎥⎣⎦uv , and []det det .0c a cb b ⎡⎤==⎢⎥⎣⎦v u The determinant of the matrix whose columns are those vectors which define the sides of the parallelogram adjacent to 0 either is equal to the area of the parallelogram or is equal to the negative of the area of the parallelogram.43. [M] Answers will vary. The conclusion should be that det (A + B ) ≠ det A + det B . 44. [M] Answers will vary. The conclusion should be that det (AB ) = (det A )(det B ).45. [M] Answers will vary. For 4 × 4 matrices, the conclusions should be that det det ,T A A = det(–A ) =det A , det(2A ) = 16det A , and 4det (10)10det A A =. For 5 × 5 matrices, the conclusions should be that det det ,T A A = det(–A ) = –det A , det(2A ) = 32det A , and 5det (10)10det .A A = For 6 × 6 matrices, theconclusions should be that det det T A A =, det(–A ) = det A , det(2A ) = 64det A , and 6det (10)10det .A A = 46. [M] Answers will vary. The conclusion should be that 1det 1/det .A A −=3.2 SOLUTIONSNotes : This section presents the main properties of the determinant, including the effects of row operationson the determinant of a matrix. These properties are first studied by examples in Exercises 1–20. The properties are treated in a more theoretical manner in later exercises. An efficient method for computing the determinant using row reduction and selective cofactor expansion is presented in this section and used in Exercises 11–14. Theorems 4 and 6 are used extensively in Chapter 5. The linearity property of the determinant studied in the text is optional, but is used in more advanced courses. Exercises 15-26, 39, and 40 make good test questions because they involve few computations.1. Rows 1 and 2 are interchanged, so the determinant changes sign (Theorem 3b.).2. The constant 2 may be factored out of the Row 1 (Theorem 3c.).3. The row replacement operation does not change the determinant (Theorem 3a.).4. The row replacement operation does not change the determinant (Theorem 3a.).Chapter 3 • Supplementary Exercises 1675.156156156 1440120123 279033003−−−−−=−=−=−−−6. 153153153153 3330181260326032(6)(3)18 2137031031001−−−−−=−=−=−=−=−−−−7.1302130213021302 25740178017801780 35210425003027003027 112304250030270000−−====−−−−−−8.133413341334 0125012501250 254301250000 3752024100000−−−−−−===−−−−−−−9.1130113011301130 0154015401540154(3)3 1285015500010035 3123027300350001−−−−−−−−===−=−−=−−−−−−−10.1310213102 024******* 2623900035 3738702081 35527048213−−−−−−−−−−=−−−−−−−1310213102024*******(24)24000350047700477000350000100001−−−−−−−−−−==−=−−=−−−−168 CHAPTER 3 • Determinants11. First use a row replacement to create zeros in the second column, and then expand down the secondcolumn:253125313133013301356496049604921410410021−−−−−−−==−−−−−−−−− Now use a row replacement to create zeros in the first column, and then expand down the first column:3133132356495023(5)(3)(5)(3)(8)120212121−−−−−−=−−=−=−−= 12. First use a row replacement to create zeros in the fourth column, and then expand down the fourthcolumn:123012301233430343033435466302030242434243−−−==−−−− Now use a row replacement to create zeros in the first column, and then expand down the first column:123123101233433010123(1)3(1)(38)1146113020611−−==−=−−=−−−−−−13. First use a row replacement to create zeros in the fourth column, and then expand down the fourth column:2541254132476203201624624062406776770677−−−−==−−−−−−−−−− Now use a row replacement to create zeros in the first column, and then expand down the first column:0320323216241624(1)(6)(1)(6)(1)653677053−−−−−−−−−=−−−=−−=−−=−14. First use a row replacement to create zeros in the third column, and then expand down the third column:32143214133130313031900342890003443404344−−−−−−−−−==−−−−−−− Now expand along the second row:1333319001((9))(1)(9)(0)044344−−−=−−==−−15. 55(7)35555ab c a b cde f d e f ghigh i=== 16. 33333(7)21ab c a b cde f d e f g h i gh i === 17. 7a b c a b cgh i d e f defg hi=−=− 18. (7)7g h i a b ca b c ab c g h i de f d e f d e f g h i⎛⎞⎜⎟=−=−−=−−=⎜⎟⎜⎟⎝⎠19. 22222222(7)14ab c a b c a b cd ae bf c d e f d e f gh i g highi+++==== 20. 7a d b e c f a b cde f d e f gh ighi+++== 21. Since 23013410121=−≠, the matrix is invertible. 22. Since 5011320053−−−=, the matrix is not invertible. 23. Since20081750038600754−−=, the matrix is not invertible. 24. Since 473605110726−−−=≠−, the columns of the matrix form a linearly independent set. 25. Since 78745010675−−=−≠−−, the columns of the matrix form a linearly independent set.26. Since32205610060304703−−−=−−, the columns of the matrix form a linearly dependent set. 27. a . True. See Theorem 3.b . True. See the paragraph following Example 2.c . True. See the paragraph following Theorem 4.d . False. See the warning following Example 5. 28. a . True. See Theorem 3. b . False. See the paragraphs following Example 2. c . False. See Example 3. d . False. See Theorem 5. 29. By Theorem 6, 555det (det )(2)32B B ==−=−.30. Suppose the two rows of a square matrix A are equal. By swapping these two rows, the matrix A is not changed so its determinant should not change. But since swapping rows changes the sign of thedeterminant, det A = – det A . This is only possible if det A = 0. The same may be proven true for columns by applying the above result to T A and using Theorem 5. 31. By Theorem 6, 1(det )(det )det 1A A I −==, so 1det 1/det .A A −= 32. By factoring an r out of each of the n rows, det ()det .n rA r A = 33. By Theorem 6, det AB = (det A )(det B ) = (det B )(det A ) = det BA . 34. By Theorem 6 and Exercise 31, 111det ()(det )(det )(det )(det )(det )(det )PAP P A P P P A −−−==1(det )(det )1det det P A A P ⎛⎞==⎜⎟⎝⎠det A =35. By Theorem 6 and Theorem 5, 2det (det )(det )(det ).T T U U U U U == Since ,T U U I =det det 1T U U I ==, so 2(det ) 1.U = Thus det U = ±1.36. By Theorem 6 44det (det )A A =. Since 4det 0A =, then 4(det )0A =. Thus det A = 0, and A is notinvertible by Theorem 4. 37. By Theorem 2, det A = 3 and det B = 8, while 60174AB ⎡⎤=⎢⎥⎣⎦. Thus det AB = 24 = 3 × 8 = (det A )(det B ). 38. Compute det A = 0 and det B = –2. Also, 6020AB ⎡⎤=⎢⎥−⎣⎦. Thus det AB = 0 = 0 × –2 = (det A )(det B ).39. a . By Theorem 6, det AB = (det A )(det B ) = 4 × –3 = –12.b . By Exercise 32, 3det 55det 1254500A A ==×=.c . By Theorem 5, det det 3T B B ==−.d . By Exercise 31, 1det 1/det 1/4A A −==. e . By Theorem 6, 333det (det )464A A ===.40. a . By Theorem 6, det AB = (det A )(det B ) = –1 × 2 = –2.b . By Theorem 6, 555det (det )232B B ===.c . By Exercise 32, 4det 22det 16116A A ==×−=−.d . By Theorems 5 and 6, det (det )(det )(det )(det )111T T A A A A A A ===−×−=.e . By Theorem 6 and Exercise 31,11det (det )(det )(det )(1/det )(det )(det )det 1B AB B A B B A B A −−====−. 41. det A = (a + e )d – c (b + f ) = ad + ed – bc – cf = (ad – bc ) + (ed – cf ) = det B + det C . 42. 1det ()(1)(1)1det det 1a bA B a d cb a d ad cb A a d B c d++==++−=+++−=++++, sodet (A + B ) = det A + det B if and only if a + d = 0.43. Compute det A by using a cofactor expansion down the third column: 111322233333det ()det ()det ()det A u v A u v A u v A =+−+++113223333113223333det det det det det det u A u A u A v A v A v A =−++−+det det B C =+44. By Theorem 5, det det ().T AE AE = Since ()T T T AE E A =, det det().T T AE E A = Now T E is itself anelementary matrix, so by the proof of Theorem 3, det ()(det )(det ).T T T T E A E A = Thus it is true that det (det )(det ),T T AE E A = and by applying Theorem 5, det AE = (det E )(det A ).45. [M] Answers will vary, but will show that det T A A always equals 0 while det T AA should seldom bezero. To see why T A A should not be invertible (and thus det 0T A A =), let A be a matrix with morecolumns than rows. Then the columns of A must be linearly dependent, so the equation A x = 0 must have a non-trivial solution x . Thus ()(),T T T A A A A A ===x x 00 and the equation ()T A A =x 0 has a non-trivial solution. Since T A A is a square matrix, the Invertible Matrix Theorem now says that T A A is not invertible. Notice that the same argument will not work in general for ,T AA since T A has more rows than columns, so its columns are not automatically linearly dependent.46. [M] Compute det A = 1 and cond A ≈ 23683. Note that this is the 2A condition number, which is used inSection 2.3. Since det A ≠ 0, it is invertible and11914075494012196267195195278203199A −−−⎡⎤⎢⎥−−−⎢⎥=⎢⎥−⎢⎥−−−⎢⎥⎣⎦The determinant is very sensitive to scaling, as 4det1010det 10,000A A == and det 0.1A =4(0.1)det 0.0001.A = The condition number is not changed at all by scaling: cond(10A ) =cond(0.1A ) = cond A ≈ 23683.When4A I =, det A =1 and cond A = 1. As before the determinant is sensitive to scaling: 4det1010det 10,000A A == and 4det 0.1(0.1)det 0.0001.A A == Yet the condition number is not changed by scaling: cond(10A ) = cond(0.1A ) = cond A = 1.3.3 SOLUTIONSNotes : This section features several independent topics from which to choose. The geometric interpretationof the determinant (Theorem 10) provides the key to changes of variables in multiple integrals. Students of economics and engineering are likely to need Cramer’s Rule in later courses. Exercises 1–10 concern Cramer’s Rule, exercises 11–18 deal with the adjugate, and exercises 19–32 cover the geometric interpretation of the determinant. I n particular, Exercise 25 examines students’ understanding of linear independence and requires a careful explanation, which is discussed in the Study Guide . The Study Guide also contains a heuristic proof of Theorem 9 for 2 × 2 matrices.1. The system is equivalent to A x = b , where 5724A ⎡⎤=⎢⎥⎣⎦ and 31⎡⎤=⎢⎥⎣⎦b . Compute 12123753(),(),det 6,det ()5,det ()1,1421A A A A A ⎡⎤⎡⎤=====−⎢⎥⎢⎥⎣⎦⎣⎦b b b b 1212det ()det ()51,.det 6det 6A A x x A A ====−b b2. The system is equivalent to A x = b , where 4152A ⎡⎤=⎢⎥⎣⎦ and 67⎡⎤=⎢⎥⎣⎦b . Compute 12126146(),(),det 3,det ()5,det ()2,7257A A A A A ⎡⎤⎡⎤=====−⎢⎥⎢⎥⎣⎦⎣⎦b b b b 1212det ()det ()52,.det 3det 3A A x x A A ====−b b3. The system is equivalent to A x = b , where 3256A −⎡⎤=⎢⎥−⎣⎦ and 75⎡⎤=⎢⎥−⎣⎦b . Compute 12127237(),(),det 8,det ()32,det ()20,5655A A A A A −⎡⎤⎡⎤=====⎢⎥⎢⎥−−−⎣⎦⎣⎦b b b b 1212det ()det ()322054,.det 8det 82A A x x A A ======b b4. The system is equivalent to A x = b , where 5331A −⎡⎤=⎢⎥−⎣⎦ and 95⎡⎤=⎢⎥−⎣⎦b . Compute 12129359(),(),det 4,det ()6,det ()2,5135A A A A A −⎡⎤⎡⎤===−==−⎢⎥⎢⎥−−−⎣⎦⎣⎦b b b b 1212det ()det ()6321,.det 42det 42A A x x A A −===−===−−b b5. The system is equivalent to A x = b , where 210301012A ⎡⎤⎢⎥=−⎢⎥⎢⎥⎣⎦ and 783⎡⎤⎢⎥=−⎢⎥⎢⎥−⎣⎦b . Compute 123710270217()801,()381,()308,312032013A A A ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥=−=−−=−−⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥−−−⎣⎦⎣⎦⎣⎦b b b 123det 4,det ()6,det ()16,det ()14,A A A A ====−b b b 312123det ()det ()det ()6316147,4,.det 42det 4det 42A A A x x x A A A −=========−b b b6. The system is equivalent to A x = b , where 211102313A ⎡⎤⎢⎥=−⎢⎥⎢⎥⎣⎦ and 422⎡⎤⎢⎥=⎢⎥⎢⎥−⎣⎦b . Compute 123411241214()202,()122,()102,213323312A A A ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥==−=−⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥−−−⎣⎦⎣⎦⎣⎦b b b 123det 4,det ()16,det ()52,det ()4,A A A A ==−==−b b b 312123det ()det ()det ()165244,13, 1.det 4det 4det 4A A A x x x A A A −−===−======−b b b7. The system is equivalent to A x = b , where 6492s A s ⎡⎤=⎢⎥⎣⎦ and 52⎡⎤=⎢⎥−⎣⎦b . Compute 12125465(),(),det ()108,det ()1245.2292sA A A s A s s ⎡⎤⎡⎤===+=−−⎢⎥⎢⎥−−⎣⎦⎣⎦b b b b Since 22det 123612(3)0A s s =−=−≠for s ≠s ≠12122222det ()det ()108541245415,.det det 12(3)6(3)12(3)4(3)A A s s s s x x A A s s s s ++−−−−======−−−−b b 8. The system is equivalent to A x = b , where 3595s A s −⎡⎤=⎢⎥⎣⎦ and 32⎡⎤=⎢⎥⎣⎦b . Compute 12123533(),(),det ()1510,det ()627.2592s A A A s A s s −⎡⎤⎡⎤===+=−⎢⎥⎢⎥⎣⎦⎣⎦b b b bSince 22det 154515(3)0A s s =+=+≠ for all values of s , the system will have a unique solution for allvalues of s . For such a system, the solution will be12122222det ()det ()151********,.det det 15(3)3(3)15(3)5(3)A A s s s s x x A A s s s s ++−−======++++b b 9. The system is equivalent to A x = b , where 236s s A s −⎡⎤=⎢⎥⎣⎦ and 14−⎡⎤=⎢⎥⎣⎦b . Compute 1212121(),(),det ()2,det ()4 3.4634s s A A A s A s s −−−⎡⎤⎡⎤====+⎢⎥⎢⎥⎣⎦⎣⎦b b b b Since 2det 666(1)0A s s s s =+=+= for s = 0, –1, the system will have a unique solution when s ≠ 0, –1.For such a system, the solution will be1212det ()det ()2143,.det 6(1)3(1)det 6(1)A A s s x x A s s s A s s +=====+++b b 10. The system is equivalent to A x = b , where 2136s A s s ⎡⎤=⎢⎥⎣⎦ and 12⎡⎤=⎢⎥⎣⎦b . Compute 12121121(),(),det ()62,det ().2632s A A A s A s s s ⎡⎤⎡⎤===−=⎢⎥⎢⎥⎣⎦⎣⎦b b b b Since 2det 1233(41)0A s s s s =−=−= for s = 0,1/4, the system will have a unique solution whens ≠ 0,1/4. For such a system, the solution will be1212det ()det ()621,.det 3(41)det 3(41)3(41)A A s sx x A s s A s s s −=====−−−b b 11. Since det A = 3 and the cofactors of the given matrix are 11000,11C == 12303,11C =−=−− 13303,11C ==− 21211,11C −−=−= 22011,11C −==−− 23022,11C −=−=− 31210,00C −−== 32013,30C −=−=− 33026,3C −== 010adj 313326A ⎡⎤⎢⎥=−−−⎢⎥⎢⎥⎣⎦ and 11/301adj 11/31.det 12/32A A A−⎡⎤⎢⎥==−−−⎢⎥⎢⎥⎣⎦12. Since det A = 5 and the cofactors of the given matrix are 11211,10C −==− 12210,00C =−= 13222,01C −== 21133,10C =−= 22130,0C == 23111,01C =−=− 31137,21C ==− 32135,21C =−= 33114,22C ==−−137adj 005214A −⎡⎤⎢⎥=⎢⎥⎢⎥−−⎣⎦ and 11/53/57/51adj 001.det 2/51/54/5A A A−−⎡⎤⎢⎥==⎢⎥⎢⎥−−⎣⎦13. Since det A = 6 and the cofactors of the given matrix are 11011,11C ==− 12111,21C =−= 13101,21C == 21541,11C =−=− 22345,21C ==− 23357,21C =−= 31545,01C == 32341,11C =−= 33355,10C ==− 115adj 151175A −−⎡⎤⎢⎥=−⎢⎥⎢⎥−⎣⎦ and 11/61/65/61adj 1/65/61/6.det 1/67/65/6A A A−−−⎡⎤⎢⎥==−⎢⎥⎢⎥−⎣⎦14. Since det A = –1 and the cofactors of the given matrix are 11215,34C == 12012,24C =−= 13024,23C ==− 21673,33C =−=− 22372,24C ==− 23363,23C =−= 31678,21C ==− 32373,01C =−=− 33366,02C == 538adj 223436A −−⎡⎤⎢⎥=−−⎢⎥⎢⎥−⎣⎦ and 15381adj 223.det 436A A A −−⎡⎤⎢⎥==−⎢⎥⎢⎥−−⎣⎦15. Since det A = 6 and the cofactors of the given matrix are 1112,32C == 12102,22C −=−=− 13111,23C −==−− 21000,32C =−= 22306,22C ==− 23309,23C =−=−− 3100,1C == 31000,1C == 33303,11C ==− 200adj 260193A ⎡⎤⎢⎥=⎢⎥⎢⎥−−⎣⎦ and 11/3001adj 1/310.det 1/63/21/2A A A−⎡⎤⎢⎥==⎢⎥⎢⎥−−⎣⎦16. Since det A = –9 and the cofactors of the given matrix are 11319,03C −==− 12010,03C =−= 13030,00C −== 21246,03C =−=− 22143,03C == 23120,00C =−= 312414,31C ==− 32141,01C =−=− 33123,03C ==−− 9614adj 031003A −−⎡⎤⎢⎥=−⎢⎥⎢⎥−⎣⎦ and 112/314/91adj 01/31/9.det 001/3A A A−−⎡⎤⎢⎥==−⎢⎥⎢⎥⎣⎦17. Let ab A cd ⎡⎤=⎢⎥⎣⎦. Then the cofactors of A are 11,C d d == 12,C c c =−=− 21C b b =−=−, and 22C a a ==. Thus adj d b A ca −⎡⎤=⎢⎥−⎣⎦. Since det A = ad – bc , Theorem 8 gives that 111adj det d b A A c a A ad bc −−⎡⎤==⎢⎥−−⎣⎦. This result is identical to that of Theorem 4 in Section 2.2. 18. Each cofactor of A is an integer since it is a sum of products of entries in A . Hence all entries in adj Awill be integers. Since det A = 1, the inverse formula in Theorem 8 shows that all the entries in 1A − will be integers. 19. The parallelogram is determined by the columns of 5624A ⎡⎤=⎢⎥⎣⎦, so the area of the parallelogram is |det A | = |8| = 8. 20. The parallelogram is determined by the columns of 1435A −⎡⎤=⎢⎥−⎣⎦, so the area of the parallelogram is |det A | = |–7| = 7.21. First translate one vertex to the origin. For example, subtract (–1, 0) from each vertex to get a newparallelogram with vertices (0, 0),(1, 5),(2, –4), and (3, 1). This parallelogram has the same area as theoriginal, and is determined by the columns of 1254A ⎡⎤=⎢⎥−⎣⎦, so the area of the parallelogram is |det A | = |–14| = 14. 22. First translate one vertex to the origin. For example, subtract (0, –2) from each vertex to get a newparallelogram with vertices (0, 0),(6, 1),(–3, 3), and (3, 4). This parallelogram has the same area asthe original, and is determined by the columns of 6313A −⎡⎤=⎢⎥⎣⎦, so the area of the parallelogram is |det A | = |21| = 21.。
MIT线性代数Problem set 6
Total: 100 points
Section 4.3. Problem 4: Write down E = �Ax − b�2 as a sum of four squares– the last one is (C + 4D − 20)2 . Find the derivative equations ∂E/∂C = 0 and ∂E/∂D = 0. Divide by 2 to obtain the normal equations AT Ax � = AT b. Solution (4 points)
⎜1⎟ ⎜ 8
⎟
⎟ , b =
⎜ ⎟ ,
AT A = (26), AT b = (112). A =
⎜ ⎝
⎝
8
⎠
3⎠
4 20 Thus, solving AT Ax = AT b, we arrive at D = 56/13. Here is the diagram analogous to figure 4.9a.
the sum of the first equation and two times the third equation. One gets 0 = 20.
Hence, these equations are not simultaneously solvable.
Computing, we get
⎜ C + E − 1
⎟ ⎟ .
Ax − b =
⎜ ⎝C
− D − 3⎠
C −E−4 Computing, we find that the first entry of AT (Ax − b) is 4C − 8. This is zero when C = 2, the average of the entries of b. Plugging in the point (0, 0), we get C + D(0) + E (0) = C = 2 as desired. Section 4.3. Problem 29: Usually there will be exactly one hyperplane in Rn that contains the n given points x = 0, a1 , . . . , an−1 . (Example for n=3: There will be exactly one plane containing 0, a1 , a2 unless .) What is the test to have n exactly one hyperplane in R ? Solution (12 points) The sentence in paranthesis can be completed a couple of different ways. One could write “There will be exactly one plane containing 0, a1 , a2 unless these three points are colinear”. Another acceptable answer is “There will be exactly one plane containing 0, a1 , a2 unless the vectors a1 and a2 are linearly dependent”. In general, 0, a1 , . . . , an−1 will be contained in an unique hyperplane unless all of the points 0, a1 , . . . , an−1 are contained in an n − 2 dimensional subspace. Said another way, 0, a1 , . . . , an−1 will be contained in an unique hyperplane unless the vectors a1 , . . . , an−1 are linearly dependent. Section 4.4. Problem 10: Orthonormal vectors are automatically linearly inde pendent. (a) Vector proof: When c1 q1 + c2 q2 + c3 q3 = 0, what dot product leads to c1 = 0? Similarly c2 = 0 and c3 = 0. Thus, the q ’s are independent. (b) Matrix proof: Show that Qx = 0 leads to x = 0. Since Q may be rectangular, you can use QT but not Q−1 .
第十届哈佛大学-麻省理工数学竞赛代数题及解答
Note that y 2 ≥ 0, so x3 ≥ −1 and x ≥ −1. Let the circle be defined by (x − 4)2 + y 2 = c
dy 2 for some c ≥ 0. Now differentiate the equations with respect to x, obtaining 2y d x = 3x from the dy dy given and 2y dx = −2x + 8 from the circle. For tangency, the two expressions dx must be equal if they are well-defined, and this is almost always the case. Thus, −2x0 + 8 = 3x2 0 so x0 = −2 or x0 = 4/3, but only the latter corresponds to a point on y 2 = x3 + 1. Otherwise, y0 = 0, and this gives the trivial solution x0 = −1.
∞
1 1 + + ··· 3! 4!
=3−
1 1 1 + + + ··· 0! 1! 2!
= 3 − e.
Alternatively, but with considerably less motivation, we can induce telescoping by adding and subtracting e − 2 = 1/2! + 1/3! + · · · , obtaining 2−e+ n(n + 1) + 1 (n + 1)2 − n =2−e+ n · (n + 1) · (n + 1)! n · (n + 1) · (n + 1)! n=1 n=1 1 1 − = 3 − e. n · n! (n + 1) · (n + 1)! n=1
MIT数学分析考试试题3
�
Problem 3 Let f be a continuous function on [a, b]. Explain whether each of the following statements is always true, with brief but precise reasoning. �b (1) The function g (x) = x f (y )dy is well defined. (2) The function g is continuous. (3) The function g is decreasing. (4) The function g is uniformly continuous.
Problem 2 �1 1 If f : [0, 1] −→ [0, ∞) is increasing and f ( 2 ) > 1, show that 0 f (x)dx > 1 2. Note the question originally said f : [0, ∞] −→ [0, 1] but was corrected during the test. Solution. Since f is increasing, it is Riemann integrable on [0, 1] and hence on any subinterval. Then 1 � 1 � 2 � 1 1 f ≥0+ f= f+ 1 2 0 0 2
1 1 Solution. We can decompose α = α1 + α2 where α1 = 1 2 x and α2 = 2 H (x − 2 ) 1 . Both are monotonic increasing so by a is the Heaviside function with mp at 2 result from class or Rudin, f is RiemannStieltjes intregrable with respect to α if it is RiemannStieltjes integrable separately with respect to α1 and α2 . Again by results from class or Rudin, f is continuous at 1 2 so is RiemannStieltjes integrable with respect to α2 and since it is increasing it is Riemann integrable, which is to say RiemannStieltjes integrable with respect to x and hence α1 = 1 2 x. Thus indeed an increasing f is RiemannStieltjes intregable with respect to α if it is continuous 1 at 2 . �
MIT线性代数试题concep_ques
复习中的一些概念问题第一章1.1 哪些向量是v = (3, 1)和w = (4, 3)的线性组合?1.2 比较v = (3, 1)和w = (4, 3)的点积和它们的模。
哪个更大？满足什么样的不等式？1.3 问题1.2中v和w的角度的余弦是多少?x轴和v的角度的余弦是多少？第二章2.1 矩阵A乘以列向量x = (2, 1)后得到A的什么样的组合？A有多少行和列？2.2 如果Ax = b，b是否是矩阵A中向量的线性组合？b是否位于A的向量空间中？2.3 如果A是2阶方阵2166，它的主元素是多少？2.4 如果A是矩阵0111怎样进行消元？怎样求出转置矩阵P？2.5 如果A是矩阵2163，找到b和c使得Ax = b无解，Ax = c有唯一解。
2.6 当3阶矩阵L乘以什么样的3行的矩阵时使得A的第3行加上第2行的5倍，第2行加上第1行的2倍？2.7 当3阶矩阵E乘以什么样的3行的矩阵时使得A的第2行减去第1行的2倍，第3行减去第2行的5倍？E和问题2.6中的L有何关系?2.8 如果A是4×3阶矩阵，B是3×7阶矩阵，AB中有多少行乘以列？有多少列乘以行？进行多少次乘法运算（两者都有）？2.9 假设A = I U0I是2阶分块方阵，它的逆矩阵是什么？2.10 通过[A I]怎样求出A的逆矩阵？求解方程Ax = I的列，解x是多少？2.11 通过消元怎样怎样判断方阵A是否可逆？2.12 假设经过消元使得A变为U（下三角形）和L（上三角形）的乘积，为什么A的最后一列等于L的最后一列乘以U？2.13 可逆方阵怎样进行分解（通过行变换进行消元）？2.14 AB的逆的转置是什么？2.15 怎样证明置换矩阵的逆还是置换矩阵？对转置矩阵呢？第3章3.1 可逆的n阶方阵的列空间是什么？它的零空间呢？3.2 如果A的每一列都是第一列的倍数，那么A的列空间是什么？3.3 R n中的一些向量是子空间的两个充分条件是什么？3.4 如果矩阵A的行最简形的第一行都是1，怎样判断其它行都是0？它的零空间是什么？3.5 假设A的零空间中只有零向量，方程Ax = b的解是什么形式?3.6 通过行最简形怎样判断原矩阵的秩？3.7 假设A的第4列是第1、2、3列之和，找到零空间中的一个向量。
国开电大本科《理工英语4》机考第三大题阅读理解选择总题库[珍藏版]
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国开电大本科《理工英语4》机考总题库[珍藏版]说明：资料整理于2023年8月，适用于国开电大一平台期末机考考试。
第三大题阅读理解选择题首字母AAccording to experts,the advantage of modified food is that it helps in preventing the occurrence of allergies and also has along shelf life.However,a major portion of the population is against the consumption of genetically modified food and is concerned about its disadvantages.If you are also confused about the consumption of genetically modified food,read through the pros and cons given below and decide for yourself.Advantages of Genetically Modified Foods●One of the most prominent advantages of genetically modified food is that it helps in controlling the occurrence of certain diseases.Certain foods cause allergy to people.Their genetic modification alters the DNA system of these foods,thereby making them non-allergic.●Another major advantage of such type of foods is that they grow faster than their traditionally grown counterparts.As a result,there is greater productivity and more food.●A person need not bother about any additional costs,such as buying chemicals and pesticides,when it comes to genetically engineered food.The food is naturally pest-resistant.●Farming these foods can be a great way to fight world hunger.Since these crops grow faster or more effectively,increased production would help countries wherein crops may not normally prosper because of less than desirable environmental conditions.●The increased shelf life of these food products helps in reducing the quantity of rotten food.Disadvantages of Genetically Modified Foods●The biggest disadvantage of genetically modified food is that they have harmful effects on the human body.The consumption of these genetically engineered foods can lead to development of diseases that are immune to antibiotics.●Chances of developing cancer are high in people who regularly consume engineered food.●Since it is an unnatural way of producing foods,there is an increased health hazard such as allergens,transfer of antibiotic resistance markers and unknown effects.●Since the production of engineered food involves infusing animal genes in the crops and tampering with nature,it might not be good for consumption in the long run.特别提醒：本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！[单选题]◆How does genetically modified food control the occurrence of certain diseases?A.By cross-pollinating.B.By conventional crossbreeding.C.By altering the DNA system of these foods.[答案]C[单选题]◆Do farmers need to buy any chemicals and pesticides if they plant genetically modified food?A.Yes,they need to.Because the food is not resist to pests.B.No,they don’t need to.Because the food is naturally pest-resistant.C.No,they don’t need to.Because the chemicals and pesticides are provided by the government.[答案]B[单选题]◆Which statement is not the advantage of genetically modified food?A.It may have harmful effects on human body.B.Genetically modified food grows faster than their traditionally grown counterparts.C.Farming genetically modified food can help fight world hunger.[答案]A[单选题]◆What would happen to a person if he regularly consume genetically modified food?A.He would get cold.B.He would lose some weight.C.He would have cancer.[答案]C[单选题]◆Which of the following would be the best title for this passage?A.Can Genetically Modified Foods Combat World Hunger?B.Genetically Modified Mood：Our Answer Is No.C.Genetically Modified Foods：Pros And Cons[答案]C首字母II am very happy to share my story about breast enlargement in Thailand.Anyone who is not sure about breast enlargement should read this story first.Firstly,I was amazed at the level of care I received at the Thailand private hospital.It was like being in a luxury hotel.The nursing staffs were also better than those who I have ever met in all my years in America.Not only were they professional,but kinder and more patient than Western nurses.The hospital food was excellent,so I was able to choose from a large and varied Menu.In my private room,there was cable T.V.,a fridge,a phone,and impressive views of Bangkok(曼谷).My Translator was able to communicate with me,and the plastic surgeons arranged the whole operation without me lifting a finger.The operation cost me a third of the price of the U.K.With the low cost of the air tickets and a holiday,it worked out at about half UK’s price,because the cost in Thailand is so cheap.Combining a plastic surgery with a holiday in an amazing place is an unforgettable experience.From the moment I arrived in Bangkok airport,I felt relaxed and welcomed by the Plastic Surgery&Holiday team,it made me feel that I had friends here already.特别提醒：本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！[单选题]◆Where did the author have her breast-enlargement operation?A.In Thailand.B.In America.C.In U.K.[答案]A[单选题]◆Was the author satisfied with the nursing staffs for her?A.No,because they were no better than America nurses.B.Yes,because they were more patient than Western nurses.C.No,because the care was very poor.[单选题]◆Which of the following statement is TRUE?A.The author was content with the food in the hospital.B.When the author was in the hospital,she lived in a public room.C.There was cable T.V.,a bridge and a phone in the author’s room.[答案]A[单选题]◆What does the underlined phrase lifting a finger mean in paragraph three?A.Doing something with only one finger.B.Helping with something.C.Carrying something.[答案]B[单选题]◆Why did the author think her experience unforgettable?A.Because she had an operation there.B.Because she was satisfied with everything there.C.Because she was satisfied with the service and price there.[答案]CIf you think3D printing is only good for making flimsy paperweights.then you’re pretty much right.However.a group of audacious Dutch architects have already begun3D printing an entire canal house in Amsterdam.Is the first3D printed house a gimmick?Definitely!Is it an experiment that pushes the possibilities for3D printing technology and architecture?Maybel3D Print Canal House is the brainchild of DUS Architects.“KamerMaker’.Dutch for’room maker”,is their giant printer built inside a shipping container.Just a few weeks ago,KamerMaker began printing plastic furniture and walls at an empty canal-side lot in northern Amsterdam.Smaller versions of the unusual honeycomb walls have been tested on normal desktop printers,and everyone just needed to scale it up.Well,if only it were that easy.The printer did not always manage to spread the plastic evenly,making some of the ribs of the honeycomb structure of the block uneven.reported ARN（Australian Radio Network,to the surprise of no one who has used a 3D printer.But the architects remain optimistic.The first few printed blocks are meant to be test pieces as they improve the process.It takes about a week to print a3-meter high block right now.The project’s leader hopes to eventually get that down to two hours and finish the first of12rooms in a year.and the entire house in less than3years. Ultimately,3D Print Canal House is an architectural research project,one that is very much being conducted in the public eye:you can actually buy tickets to visit the construction site for2、50Euros.The top floors will become more omate.for example.as newer techniques are incorporated.They’ll explore the possibilities of3D printing.like the honeycomb walls or an entire room recycled.redesigned,and built anew.Nobody’s going to move into a3D printed house soon,but,if any on ever does.something might be learned from this experiment by the canal.特别提醒:本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！[单选题]◆What have the Dutch architects begun to use the3D printer to do according to the passage?A.To make heavy paperweights.B.To print the unusual honeycombC.To print an entire canal house.D.To build a shipping container.[答案]B[单选题]◆Where did KamerMaker begin printing plastic furniture and walls several weeks ago?A.It was inside a shipping container in Amsterdam.B.It was at an empty canal-side lot in northern Amsterdam.C.It was in a canal house in Amsterdam.D.It was in a room.[答案]B[单选题]◆What is the weakness of the3D printer mentioned in the passage?A.It cannot always spread the plastic evenly.B.It needs to be shortened.C.It cannot make the ribs of the honeycomb structure.D.It can print anything smoothly.[答案]B[单选题]◆How long will it take for the3D printer to make the canal house in Amsterdam?A.A week.B.Two hours.C.More than3years.D.Less than3years.[答案]A[单选题]◆Which of the following statements is WRONG according to the passage?A.This canal house is being built in the public eye.B.The top floors of the canal house will be more decorated.C.Some one is going to move into this canal house soon.D.One can visit the construction site for2.50Euros[答案]CImagine you sitting on your front garden admiring the shining sun when your neighbor comes with that annoying car with the noisy sound and the exhaust pipe blowing out the horrible black smoke.What are you going to do about it？In our modem days,too many people depend on their cars to get to work or to drop off their children to school.However,is it safe to have an increasing number of cars on our roads？Research shows that cars create serious pollution.Exhaust from all combustion engines produce harmful effects on the health of both car users and all innocent walkers.Cities have become islands of toxic chemicals from the unrestrained use of vehicles burning fossil fuels.The harmful health effects of car exhaust are pervasive and difficult to measure.For example,people with asthma would suffer with attacks due to the pollution.The world wild life also suffers from the cars’toxic emissions.No matter how much manufacturers try to make cars environmentally friendly it will damage our earth in one way or another.Additionally cars are too dangerous for our community.Careless drivers put themselves and others in danger.Cars are critical to walkers especially to children if the driver is not looking on the road or somehow distracted Have you known someone whose toes or feet have been run over by cars while crossing the street？Lastly.cars create social problems,which could lead to poor health.A few drivers suffer“Road Rage”and put themselves.loved ones and others in a threatening situation.Some cars create disturbing noises,which disturbs people living near highways,freeways,etc.not to get enough sleep.People depend on their cars so much that they forgot to exercise their bodies and end up being obese or having a heart problem.As a result of the excessive number of cars on our roads,everyone’s life is in danger.In order to create an environmentally friendly and safe world,it is necessary for each country to limit the permits in number to control regional air pollution.Second,car manufacturers should try to improve the efficiency of vehicles,such as finding solutions to emissions of combustion engines,developing new power sources such as new fuels,natural gas.Third,try to improve efficiency of traffic by setting up dedicated bus lanes and giving priority to car-pools and vehicles with3or more passengers.Besides,traffic can be scheduled;for example,commercial traffic atnight;large companies can shift working hours and decentralize administrative operations.特别提醒：本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！[单选题]◆Exhaust from cars is________.A.harmlessB.pollutiveC.not serious[答案]B[单选题]◆In the author’s view,it is________to make cars be friendly to environment.A.impossibleB.feasibleC.hopeful[答案]A[单选题]◆________will be the major victims in the car accident.A.WalkersB.ChildrenC.Car-drivers[答案]B[单选题]◆Cars caused some social problems EXCEPT________.A.obesityB.a quick temperC.employment[答案]C[单选题]◆Which of the following is NOT TRUE as to handling car problems？A.Reducing the number of cars.ing a car in the restricted areaC.Finding alternative fuels.[答案]BIn the grass,a baby duck was hatched(孵化)out in a college biology lab with a backward left foot.At the time,it probably wouldn’t have survived,if Mike Gary didn’t recommend a proper operation to cut off that bad foot.Months after the healing process,engineers at Nova copy Company produced a3D printed model that would be used to create a prosthetic(修复的)foot for the duck.Considering that ABS plastics were not flexible enough for this type of exercise,the Nova copy company made a three-dimension ally modeled foot made of silicone,with the hope to create a permanent prosthesis(假肢)for the duck.It took just13hours to produce.This3D prosthesis demonstrated the3D technologies can be used not only for human application,but also in foreign subjects such as animals and other medical treatments.特别提醒：本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！[单选题]◆Where was the baby duck hatched out?A.In a college biology lab.B.In a back yard.C.In Nova copy Company.[答案]A[单选题]◆What was wrong with the baby duck?A.It didn’t survived.B.It was born with a backward left foot.C.Mike Gary cut off that bad foot.[答案]B[单选题]◆Could ABS plastics be used to produce a3D printed model that would be used to create a prosthetic foot for the duck?A.No,because ABS plastics were not flexible enough.B.Yes,but it could only be produced by Nova copy Company.C.Yes,because it was made of silicone.[答案]A[单选题]◆How long did it take to create a permanent prosthesis for the duck?A.10hours.B.13hours.C.30hours.[答案]B[单选题]◆What did the last sentence of this passage indicate?A.3D technologies can be used only for human application.B.3D technologies can be used only for foreigners.C.3D technologies can be used to human,animals and other medical treatments.[答案]C首字母MMy hero is a person who has given me so many things,who has taught me about everything,and who has shown me how beautiful life is.And gradually as I get older and I grow up,my hero has opened my eyes to the fact that life is not always beautiful and happy like a fairy tale,but life is much more than that.My hero has explained to me that life can be very hard.Everyone must struggle hard to make all their dreams come true and to be a successful person.Maybe,because my her olives so close to me,I haven’t realized that this person has inspired me so much.My only true heroism y mother.She was born in Jakarta,on the1st of July.My mother is the youngest child in her family.She has two sisters,who always love and take care of her.Maybe,because of that,she becomes a cheerful,kind,friendly,and love able person.She wants everybody to feel comfortable when they are around her.My mother likes to study and to read everything,from books to newspapers.Everyone really likes to talk and chat with her,so she has many friends.My mother believes all you need is to be a hard-working person who always wants to study hard,so that you can be a successful person.She says that many people around her have succeeded because they have studied and worked hard.特别提醒：本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！[单选题]◆If a person wants to succeed in his life,he________.A.should listen to a fairy tale everydayB.must dream a good dream everydayC.has to work hard all his life[答案]C[单选题]◆The underlined word“inspired”in Paragraph2may have the same meaning as“________”.A.encouragedB.showedC.cared[答案]A[单选题]◆The writer’s mother becomes a love able person becauseA.she is the youngest in the familyB.her two sisters love her very muchC.she feels comfortable in her family[答案]B[单选题]◆From the third passage,We can infer(推测)that the writer________.A.is a friendly person like his motherB.has many friends to talk withC.knows the meaning of success[答案]C[单选题]◆Which of the following is the best title for the reading?A.Study well!B.What a beautiful life!C.My mother,my hero.[答案]C首字母RRoby Mini is just40centimeters tall,has a large round screen for ahead and has been designed to help senior citizens living alone.Manufactured by robotics company Shanghai PT Info,the brightly colored robot,which costs 5,000yuan($790),was launched on Aug14to help entertain and monitor elderly people.In Shanghai last year,the number of registered residents at least60years old was4million.That was nearly30 percent of the population.By2018,the figure is projected to rise to5million.With many living alone,robots such as Roby Mini can help playa crucial role in caring for the elderly.“It will change the way people interact with digital products,”Hong says.Founded in May,Shanghai PT Info had already done the groundwork.In fact,the research team spent three years developing the speech recognition system before the firm was officially launched.With360-degree wheels hidden under its body,Roby Mini can follow a person through face recognition and strike up a conversation,tell jokes or provide information.The robot also works as a computer terminal,supplying weather information,ordering groceries online and even booking taxis.Roby Mini can be connected to other gadgets as well,such as smart wristbands or magnetic devices on doors.Roby Mini can also connect an elderly owner to a doctor through its high-definition screen.Already Shanghai PT Info is talking to three community hospitals in Shanghai’s Min hang district.Since the robot is linked to the Internet,it can support long-distance audio and video calls.Mass production started this month and Shanghai PT Info plans to manufacture200,000units this year.But the future is not easy for Shanghai PT Info.Earlier this year,Flying wings Intelligent Robot Technology(Shanghai)introduced a machine with sensors that will ring an alarm when an elderly person falls down.Tianjin Shangjiu Sheng yuan Technology Group also developed a robot for seniors in2014.Owners can turn on the lights and air-conditioning units by pressing buttons on the machine.It can also remind elderly owners to take pills,and inform them of hospital checkups and family birthdays.特别提醒：本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！◆Roby Mini is the name of_________.A.a robotics companyB.a robotC.a senior citizen living alone[答案]B[单选题]◆Roby Mini is designed to_________.A.help look after children living alone in senior high schoolB.help entertain and monitor senior citizensC.change the way people interact with digital products[答案]B[单选题]◆According to this passage,the number of registered residents at least60years old will be___.A.4million next yearB.30percent of the populationC.increase to5million by2018[答案]C[单选题]◆The writer thinks“the future is not easy for Shanghai PT Info”because_________.A.there are too many elder people in ShanghaiB.Roby Mini can not ring an alarm when an elderly person falls down.C.the company will face tough competition in the marketplace[答案]C[单选题]◆The writer thinks“the future is not easy for Shanghai PT Info”because_________.A.there are too many elder people in ShanghaiB.the company will face tough competition in the marketplaceC.Roby Mini can not ring an alarm when an elderly person falls down[答案]B[单选题]◆The best title for this passage would be“_________”.A.How to Use a RobotB.Elder People Increase in ShanghaiC.Robot Gives Help to Older People[答案]C首字母TThere are three kinds of goals:short-term,medium-range and long-term goals.Short-term goals are those that usually deal with current activities,which we can apply on a daily basis.Such goals can be achieved in a week or less,or two weeks,or possibly months.It should be remembered that just as a building is no stronger than its foundation,our long-term goals cannot amount to very much without the achievement of solid short-term goals.Upon completing our short-term goals,we should date the occasion and then add new short-term goals that will build on those that have been completed.The intermediate goals build on the foundation of the short-term goals.They might deal with just one term of school or the entire school year,or they could even extend for several years.Any time you move a step at a time,you should never allow yourself to become discouraged or overwhelmed.As you complete each step,you will enforce the belief in your ability to grow and succeed.And as your list of completion dates grows,your motivation and desire will increase.Long-term goals may be related to our dreams of the future.They might cover five years or more.Life is not a static thing.We should never allow a long-term goal to limit us or our course of action.特别提醒：本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！[单选题]◆Short-term goals are the ones that________.A.motivate you over the long haulB.keep you engaged on a daily basisC.might be measured for many years[答案]B[单选题]◆The medium-range goals are built upon________.A.belief and successB.the motivation and desireC.the completion of the short-term goals[答案]C[单选题]◆Once we set ourselves long-term goals,________.A.we should stick to them until we complete themB.we may change our goals as we have new ideas and opportunitiesC.we had better wait for the exciting news of success[答案]B[单选题]◆What does the underlined word“static”mean in the last paragraph?A.movingB.developingC.not moving[答案]C[单选题]◆Which of the following statements is TRUE according to the passage?A.Short-term goals all help in achieving your long-term goals.B.Any time you move a step at a time,you may become discouraged.C.Life is a static thing,thus never allowing a long-term goal to limit us.[答案]A首字母WWhen computer servers operate a complex program,they can get very hot.Cooling the servers can be costly.So researchers asked what would happen if the heat created by the servers could be captured and used?Data centers of large Internet companies such as Google and Microsoft have thousands of computer servers.As these servers process information,they create large amounts of heat,so they need huge cooling systems,These systems send the heat into the air.The Dutch company Nerdalize thinks paying for electrieity to operate the servers and then paying again to cool them is a waste of energy.So it developed a device called the e-Radiator.It is a computer server that also works as a heating source.Boaz Leupe is the chief executive officer of Nerdalize,He says the e-Radiator saves money because companies don’t have to pay to cool their servers.“The kilowatt(千瓦)hour you are using is used twice-once to heat the home and once to compute the client’s task without the cooling overhead.”He says five homeowners in the Netherlands are testing the heating device in their homes.“We reimburse the electricity the server uses,and that we can do because of the computer clients on the other side,and,in that way,homeowners actually get heating for free,and computer users don’t have to pay for the overhead of the data center.”Jan Visser is one of the participants in the year-long experiment.He says the amount of heat produced by the e-Radiator into room depends on the work being done by the computer server.He says it can not be used as theprimary source of heat.But he is ready to try it.He says if it provides enough warmth,he will be able to use his home’s heating system less,which will save him money.Nerdalize says e-Radiators create heat temperatures of up to55℃.It says the devices could save users up to$440in heating costs a year.特别提醒：本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！[单选题]◆Why did researchers ask what would happen if the heat created by the servers could be captured and used?A.Because computers can get very hot.puter servers is a complex program,C.Cooling computer servers costs a lot of money.[答案]C[单选题]◆The computer servers in large Internet companies such as Google and Microsoft can________.A.create large amounts of informationB.create large quantity of heatC.manage the huge cooling systems[答案]B[单选题]◆Nerdalize developed the e-Radiator because_________.A.e-Radiator cost less energyB.e-Radiator can cool the servers for freeC.e-Radiator can create cold air for free[答案]B[单选题]◆According to Jan Visser,the amount of heat produced by the e-Radiator_________.A.depends on the work being processed by PCB.can not be used as the home’s heating systemC.can not be used as the main source of heat[答案]C[单选题]◆The best title for this passage would be“_________”.A.The New Way to Heat HomesB.The New Way to Cool HomesC.Hot Computers Could Be Used for Free[答案]AWHY BUILD A SPACE ELEVATOR?The space elevator will reduce the cost of getting from Earth to space.It will also allow us to take very large payloads into space very easily,very safely.Because of that,we can build cities on the moon.We can build space stations.We can build large solar arrays in space to collect energy from the sun and beam it down to Earth.WHY NOT JUST USE ROCKETS INSTEAD?Rockets are very expensive.The shuttle costs S10,000a pound to put a payload into low-Earth orbit,and if you want to go to the moon or Mars,it’s hundreds of thousands of dollars per pound or more.Rockets are also very limited in the amount of payload they can carry.and they’re risky.With a space elevator,the cost is a fraction of that-it may go down to100th or1,000th of current costs-and there are no rocket engines that might explode.WHAT WOULD IT BE LIKE TO RIDE THE ELEVATOR?You’d go to an ocean platform,you’d climb into a module,and you’d feel it start moving.You’d see the Earth fall away.In just a half hour or so,you’d pass up through the clouds,and you’d start to see the curve of the horizon.Another half an hour to an hour later,you’d basically be in space.You’d see stars,even in the middle of the day.Eventually you’d be weightless.It’d be a very smooth ride:there wouldn’t be any shaking.ARE THERE POTENTIAL DRAWBACKS TO OPENING SPACE THIS WAY?Everything we do has the potential for creating some bad.If we mine an asteroid,the asteroid’s going to look ugly.just like a mine here.But I think the benefits of opening up space far outweigh the damage that we can see-benefits in terms of gaining energy from space to replace oil.additional capabilities in telecommunications. manufacturing in space.additional real estate and exploration.特别提醒：本题共5个小题,需要拉动答题框右边的滚动条使5个试题显示出来做答！！[单选题]◆If a space elevator is built,what can’t be built by us?A.Cities on the moon.B.Space stations.C.Skyscrapers in space.D.Expressways in space.[答案]C[单选题]◆The main reason of building a space elevator instead of using rockets is that________.A.the technology needed in building a space elevator is easier than that of rocketsB.the safety of a space elevator is better than that of rocketsC.the cost of building a space elevator is a fraction of using rocketsD.a space elevator is much more expensive than a rocket[答案]C[单选题]◆Which of the following statements is NOT correct,when you are riding the space elevator?A.You’d feel the Earth fall away.B.You’d see the horizon in just half an hour.C.You can’t see the stars in the middle of the day.D.You’d be weightless.[答案]C[单选题]◆What does the word“outweigh”in Paragraph4mean?A.Be less important than.B.Be greater than.C.Be lighter than.D.Be less valuable than[答案]B[单选题]◆Which of the following can summarize the main idea of this passage?A.Why do we need to build a space elevator?B.What can we do after building the space elevator?C.Why does building a space elevator cost less than using rockets?D.How can we build a space elevator?[答案]AWould a robot serving you coffee in bed make waking up easier on weekday mornings?Could a household robot help an elderly relative who is living alone?How would you like to climb into a robotic car and eat breakfast with the kids while you’re all driven to school and work?These scenarios may sound like science fiction,but experts say they’re a lot closer to becoming reality than you probably think.Brown University roboticist Chad Jenkins expects a near-term robot revolution that will echo the computing revolution of recent decades.And he says it will be driven by enabling robots to learn more like humans do-by watching others demonstrate behaviors and by asking questions.。
第二届哈佛大学-麻省理工数学竞赛代数题及解答
Harvard Math TournamentFebruary 27, 19991. If a@b = , for how many real values of a does a@1 = 0?2. For what single digit n does 91 divide the 9-digit number 12345n789?3. Alex is stuck on a platform floating over an abyss at 1 ft/s. An evil physicist has arranged for the platform to fall in (taking Alex with it) after traveling 100ft. One minute after the platform was launched, Edward arrives with a second platform capable of floating all the way across the abyss. He calculates for 5 seconds, then launches the second platform in such a way as to maximize the time that one end of Alex’s platform is between the two ends of the new platform, thus giving Alex as much time as possible to switch. If both platforms are 5 ft long and move with constant velocity once launched, what is the speed of the second platform (in ft/s)?4. Find all possible values of d where a2 6ad + 8d2 = 0, a = 0.5. You are trapped in a room with only one exit, a long hallway with a series of doors and land mines. To get out you must open all the doors and disarm all the mines. In the room is a panel with 3 buttons, which conveniently contains an instruction manual. The red button arms a mine, the yellow button disarms two mines and closes a door, and the green button opens two doors. Initially 3 doors are closed and 3 mines are armed. The manual warns that attempting to disarm two mines or open two doors when only one is armed/closed will reset the system to its initial state. What is the minimum number of buttons you must push to get out?6. Carl and Bob can demolish a building in 6 days, Anne and Bob can do it in 3, Anne and Carl in 5. How many days does it take all of them working together if Carl gets injured at the end of the rst day and can’t come back? Express your answer as a fraction in lowest terms.7. Matt has somewhere between 1000 and 2000 pieces of paper he’s trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries 2, 3, 4, 5, 6, 7, and 8 piles but ends up with one sheet left over each time. How many piles does he need?8. If f (x) is a monic quartic polynomial such that f ( 1) = 1, f (2) = 4, f ( 3) = 9, andf (4) = 16, nd f (1).9. How many ways are there to cover a 3 × 8 rectangle with 12 identical dominoes?10. Pyramid EARLY is placed in (x y z) coordinates so that E = (10 10 0), A = (10 10 0), R = ( 10 10 0), L = ( 10 10 0), and Y = (0 0 10). Tunnels are drilled through the pyramid in such a way that one can move from (x y z) to any of the 9 points (x y z 1), (x ± 1 y z 1), (x y ± 1 z 1),(x ± 1 y ± 1 z 1). Sean starts at Y and moves randomly down to the base of thepyramid, choosing each of the possible paths with probabilityeach time. What is the probability9that he ends up at the point (8 9 0)?1Algebra SolutionsHarvard-MIT Math TournamentFebruary 27, 1999Problem A1 [3 points]If a@b = a a 3b 3 , for how many real values of a does a@1 = 0?e S x o t l r u a t neo ion:us If ss ,in 0c e t h th en a t a ma k e 1s ,he 0 d o e r n 2o f +th a o g in =a l 0 .ex pres Thus si o a n 1,o r a wh i is c h a i r s oo a t nof a 2 + a + 1. But this quadratic has no real roots, in particular its roots are − 1±2√−3 . Thereforethere are no such real values of a, so the answer is 0.Problem A2 [3 points]For what single digit n does 91 divide the 9-digit number 12345n789?Solution 1: 123450789 leaves a remainder of 7 when divided by 91, and 1000 leaves a remainder of 90, or -1, so adding 7 multiples of 1000 will give us a multiple of 91.Solution 2: For those who don’t like long division, there is a quicker way. First notice that 91 = 7 13, and 7 11 13 = 1001. Observe that 12345n789 = 123 1001000+45 n 1001 123 1001+123 45n+789 It follows that 91 will divide 12345n789 i ff 91 divides 123 45n + 789 = 462n. The number 462 is divisible by 7 and leaves a remainder of 7 when divided by 13.Problem A3 [4 points]Alex is stuck on a platform floating over an abyss at 1 ft/s. An evil physicist has arranged for the platform to fall in (taking Alex with it) after traveling 100ft. One minute after the platform was launched, Edward arrives with a second platform capable of floating all the way across the abyss. He calculates for 5 seconds, then launches the second platform in such a way as to maximize the time that one end of Alex’s platform is between the two ends of the new platform, thus giving Alex as much time as possible to switch. If both platforms are 5 ft long and move with constant velocity once launched, what is the speed of the second platform (in ft/s)?Solution: The slower the second platform is moving, the longer it will stay next to the first platform. However, it needs to be moving fast enough to reach the first platform before it’s too late. Let v be the velocity of the second platform. It starts 65 feet behind the first platform, so it reaches the s．i pdel edhhlines up with the front of the first platform at the instant that the first platform has travelled 100ft, which occurs after 100 seconds. Since the second platform is launched 65 seconds later and has to travel 105 feet, its speed is 105/35 = 3ft/s.12Find all possible values of a dwhere a 2 − 6ad + 8d 2 = 0, a = 0. Solution: Dividing a 2 − 6ad + 8d 2 = 0 by a 2 , we get 1 − 6a d + 8(ad )2 = 0. The roots of this quadratic are 21 , 41 .Problem A5 [5 points]You are trapped in a room with only one exit, a long hallway with a series of doors and land mines. To get out you must open all the doors and disarm all the mines. In the room is a panel with 3 buttons, which conveniently contains an instruction manual. The red button arms a mine, the yellow button disarms two mines and closes a door, and the green button opens two doors. Initially 3 doors are closed and 3 mines are armed. The manual warns that attempting to disarm two mines or open two doors when only one is armed/closed will reset the system to its initial state. What is the minimum number of buttons you must push to get out?Solution: Clearly we do not want to reset the system at any time. After pressing the red button r times, the yellow button y times, and the green button g times, there will be 3 + r − 2y armed mines and 3 + y − 2g closed doors, so we want the values of r , y , and g that make both of these quantities 0 while minimizing r + y + g. From the number of doors we see that y must be odd, from the number of mines we see y = (3 + r)/2 ≥ 3/2, so y ≥ 3. Then g = (3 + y)/2 ≥ 3, and r = 2y − 3 ≥ 3, so r + y + g ≥ 9. Call the red, yellow, and green buttons 1, 2, and 3 respectively for notational convenience, then a sequence of buttons that would get you out is 123123123. Another possibility is 111222333, and of course there are others. Therefore the answer is 9.Problem A6 [5 points]Carl and Bob can demolish a building in 6 days, Anne and Bob can do it in 3, Anne and Carl in5. How many days does it take all of them working together if Carl gets injured at the end of the first day and can’t come back? Express your answer as a fraction in lowest terms.Solution: Let a be the portion of the work that Anne does in one day, similarly b for Bob and c for Carl. Then what we are given is the system of equations b + c = 1/6, a + b = 1/3, and a + c = 1/5. Thus in the first day they complete a + b + c = 21(1/6 + 1/3 + 1/5) = 7/20, leaving 13/20 for Anne and Bob to complete. This takes 113//320= 39/20 days, for a total of 2059 .Matt has somewhere between 1000 and 2000 pieces of paper he’s trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries 2, 3, 4, 5, 6, 7, and 8 piles but ends up with one sheet left over each time. How many piles does he need?Solution: The number of sheets will leave a remainder of 1 when divided by the least common multiple of 2, 3, 4, 5, 6, 7, and 8, which is 8 · 3 · 5 · 7 = 840. Since the number of sheets is between 1000 and 2000, the only possibility is 1681. The number of piles must be a divisor of 1681 = 412 , hence it must be 41.23If f (x) is a monic quartic polynomial such that f (−1) = −1, f (2) = −4, f (−3) = −9, and f (4) = −16, find f (1).Solution: The given data tells us that the roots of f (x) + x 2 are -1, 2, -3, and 4. Combining with the fact that f is monic and quartic we get f (x) + x 2 = (x + 1)(x − 2)(x + 3)(x − 4). Hence f (1) = (2)(−1)(4)(−3) − 1 = 23.Problem A9 [7 points]How many ways are there to cover a 3 × 8 rectangle with 12 identical dominoes?Solution: Trivially there is 1 way to tile a 3 × 0 rectangle, and it is not hard to see there are 3 waysto tile a 3 × 2. Let T n be the number of tilings of a 3 × n rectangle, where n is even. From the diagram below we see the recursion T n = 3T n −2 + 2(T n −4 + T n −6 + . . . + T 2 + T 0). Given that, we can just calculate T 4 = 11, T 6 = 41, and T 8 is 153.Problem A10 [8 points]Pyramid EARLY is placed in (x,y , z) coordinates so that E = (10, 10, 0), A = (10, −10, 0), R = (−10, −10, 0), L = (−10, 10, 0), and Y = (0, 0, 10). Tunnels are drilled through the pyramid in such a way that one can move from (x,y , z) to any of the 9 points (x,y , z − 1), (x ± 1, y , z − 1), (x, y ± 1, z − 1),(x ± 1, y ± 1, z − 1). Sean starts at Y and moves randomly down to the base of the pyramid, choosing each of the possible paths with probability 91each time. What is the probability that he ends up at the point (8, 9, 0)?. . . . . . etc...Solution 1: Start by figuring out the probabilities of ending up at each point on the way down the pyramid. Obviously we start at the top vertex with probability 1, and each point on the next level down with probability 1/9. Since each probability after n steps will be some integer over 9n, we will look only at those numerators. The third level down has probabilities as shown below. Think of this as what you would see if you looked at the pyramid from above, and peeled offthe top two layers.1 2 3 2 1246423 6 9 6 3246421 2 3 2 134What we can observe here is not only the symmetry along vertical, horizontal, and diagonal axes, but also that each number is the product of the numbers at the ends of its row and column (e.g. 6 = 2 · 3). This comes from the notion of independence of events, i.e. that if we east and then south, we end up in the same place as if we had moved south and then east. Since we are only looking for the probability of ending up at (8, 9, 0), we need only know that this is true for the top two rows of the square of probabilities, which depend only on the top two rows of the previous layer. This will follow from the calculation of the top row of each square, which we can do via an algorithm similar to Pascal’s triangle. In the diagram below, each element is the sum of the 3 above it.11 1 11 2 3 2 11 3 6 7 6 3 11 4 10 16 19 16 10 4 11 5 15 30 45 51 45 30 15 5 1Now observe that the first 3 numbers in row n , where the top is row 0, are 1, n , n (n2+1) . This factis easily proved by induction on n , so the details are left to the reader. Now we can calculate the top two rows of each square via another induction argument, or by independence, to establish that the second row is always n times the first row. Therefore the probability of ending up at the point (8,9,0) is 910550 .Solution 2: At each move, the ① and y coordinates can each increase by 1, decrease by 1, or stay the same. The y coordinate must increase 9 times and stay the same 1 times, the ① coordinate can either increase 8 times and stay the same 1 time or decrease 1 time and increase 9 times. Now we consider every possible case. First consider the cases where the ① coordinate decreases once. If the ① coordinate decreases while the y coordinate increases, then we have 8 moves that are the sameand 2 that are di fferent, which can be done in 8!10! = 90 ways. If the ① coordinate decreases whilethe y coordinate stays the same, then we have 9 moves that are the same and 1 other, which can bedone in 9!10! = 10 ways. Now consider the cases where the ① coordinate stays the same twice. If the y coordinate stays the same while the ① coordinate increases, then we have 7 moves that are thesame, 2 that are the same, and 1 other, which can be done in 71!20!!= 360 ways. If the y coordinate stays the same while the ① coordinate stays the same, then we have 8 moves that are the same and 24 that are di fferent, which can be done in 8!10! = 90 ways. Therefore there are 360 + 90 + 90 + 10 = 550paths to (8,9,0), out of 91 0 possible paths to the bottom, so the probability of ending up at the point (8,9,0) is 910550 .。
麻省理工学院 线性代数
Lemma 1.4. The matrix representative [T(f,d),(g,e)]C,B is A(f,d),(g,e).
Proof. For every i = 1, . . . , d, T(f,d),(g,e)(xi−1, 0) = xi−1g. The coordinate vector of xi−1g with respect to C is the vector of coefficients. Thus [xi−1g]j is the coefficient of xj−1 in xi−1g, i.e., the coefficient of xj−i in g. This is 0 unless 0 ≤ j − i ≤ e, and then it equals bj−i.
polynomials of degree d + e − 1.
Corollary 1.3. Assume either f is nonzero of degree d or g is nonzero of degree e. The polynomials f and g have a common factor of positive degree iff T(f,d),(g,e) has nullity ≥ 1.
⎛ a0 a1 . . . ad 0 . . . 0 . . . 0 ⎞
⎜ 0 a0 . . . ad−1 ad . . . 0 . . . 0 ⎟
英文版院线性代数试卷A
3x + y + z = 4 − x + y − 2 z = −15 − 2 x + 2 y + z = −5
二．(10 points) Let A =
1 2 5 6 and B = . Find X such that 4 A + 2 X = B. 3 4 7 8
六．（15 points）Let the matrix A .
2 3 A= . Compute the inverse 1 1
A −1 and transpose AT of
七． （10 points） Find all values of a such that the matrix
三．（5 points）Find the symmetric matrix A such that X T AX = 5 x 2 + 6 xy + 7 y 2 , where
x . X = y
四 ． （ 10 points ） Prove that u = [1 linearly dependent vectors.
2 0] , v = [− 1 1 1] , w = [0 3 1] are
1 3 − 1 1 五．（10 points） Let A = 2 5 1 5 . Find a matrix P , expressed as a 1 1 1 3
product of elementary matrices, such that PA is a row echelon matrix.
昆 明 理 工 大 学 试 卷（A）
考试科目：线性代数（LinearAlgebra） 考试日期： 2012.6.4 学院： 任课教师：赵宁 专业班级： 上课班级： 学生姓名： 考试座位号： 命题教师：赵宁 学号：
线性代数期末考试考核试卷
4.以下哪个向量组构成一个基？
A. (1, 0, 0), (0, 1, 0), (0, 0, 0)
B. (1, 2, 3), (4, 5, 6), (7, 8, 9)
C. (1, 2, 3), (2, 4, 6), (1, 1, 1)
D. (1, 1, 0), (0, 1, 1), (1, 0, 1)
...
20.（根据实际题目内容填写答案）
二、多选题
1. BCD
2. ABCD
3. ABC
4. AB
5. ABC
...
20.（根据实际题目内容填写答案）
三、填题
1. 1
2.线性无关
3.主
...
10.（根据实际题目内容填写答案）
四、判断题
1. √
2. √
3. √
...
10. ×
五、主观题（参考）
1.向量组线性无关，可以通过计算行列式不为零来证明。一个可以由给定向量组线性表示的向量可以是它们的线性组合，例如\(a\vec{v}_1 + b\vec{v}_2 + c\vec{v}_3\)，其中\(a, b, c\)是适当的系数。
D. (1, 1), (1, -1)
（答题括号：________）
5.在求解线性方程组时，以下哪些情况下可以使用高斯消元法？
A.系数矩阵是方阵
B.系数矩阵是非奇异的
C.方程组中方程的个数等于未知数的个数
D.方程组可能有无穷多解
（答题括号：________）
（以下题目类似，省略以节约空间）
6. ...
A.若A为m×n矩阵，则A的转置为n×m矩阵
B.若A为m×n矩阵，则A的转置为m×n矩阵
线性代数考题-英文
Let[n]:={1,2,···,n}.(1)Draw the tree on vertex set[12]corresponding with the Pr¨u fer sequence(5,5,11,5,2,7,9,2,7,1).(2)Show the complete bipartite graph K n/2 , n/2 is the unique triangle-free graph of order n and sizen2/4 .(3)Show the edge set of a graph can be partitioned into cycles if and only if every vertex has even degree.(4)Show a graph is bipartite if and only if it does contain an odd cycle.(5)Use the contraction-deletion formula to compute the Tutte polyniomials of the following graphs.a.n-cycleb.a graph consists of two vertices joined by k edges.(6)Let G be a simple and connected graph.For a subgraph H of G,use c(H)to denote the numberof components of H.Define two parameters associated with H asσ(H)=c(H)−1andσ∗(H)= |E(H)|−|V(G)|+c(H).The following identity is well knownxσ(H)yσ∗(H),T G(1+x,1+y)=H⊆Gwhere the sum is over all spanning subgraphs H of G.Showa.T G(1,1)is the number of spanning trees of G,b.T G(2,1)is the number of spanning forests of G,c.T G(1,2)is the number of connected spanning subgraphs of G,d.T G(2,2)is the number of spanning subgraphs of G.(7)A parking function of size n is a sequence a=(a1,a2,···,a n)of nonnegative integers such that itsincreasing rearrangement c1≤c2≤···≤c n satisfies c i<i.Show a sequence a=(a1,a2,···,a n)is a parking function if and only if|{i|a i<r}|≥r for every r∈[n].(8)Let G be a connected and simple graph with vertex set V(G)={0,1,2,...,n}and edge set E(G).Forany I⊆V(G)\{0}and v∈I,define outdeg I(v)to be the number of edges from the vertex v to a vertex outside of the subset I in G.A G-parking function is a sequence(b1,b2,···,b n)such that for every I⊆[n]there exists a vertex v∈I satisfying0≤b v<outdeg I(v).Show a sequence(b1,b2,···,b n)is a K n+1-parking function if and only if the sequence is an usual parking function of size n defined in(7), where K n+1is the complete graph on n+1vertices.(9)Find as simple a solution as possiblea.How many subsets of the set[10]contain at least one odd integer?b.How many sequences(a1,a2,···,a12)are there consisting of four0’s and eight1’s,if no two consec-utive terms are both0’s?c.How many functions f:[5]→[5]are at most tow-to-one(i.e.,|{f|f−1(i)≤2for all i∈[5]}|)d.How many permutations of[6]have exactly two cycles?(10)Give a combinatorial proofs of the identityn i =0i n i =n 2n −1,where n is nonnegative integer.(11)Let f (m,n )be the number of paths from (0,0)to (m,n )∈N ×N ,where each step is of the form (1,0),(0,1),or (1,1).Show m ≥0 n ≥0f (m,n )x m y n =(1−x −y −xy )−1.(12)Represent the permutation π=62845731∈S 8by an increasing tree and an increasing binary tree,respectively.(13)Let q be a prime power,and denoted by F q a finite field with q elements and by V n (q )the n -dimensionalvector space F n q ={(a 1,a 2,···,a n )|a i ∈F q }.Show the number of k -dimensional subspaces of V n (q )is n k q= n i =n −k +1(q i −1) k i =1(q i −1).(14)A partition λis a nonincreasing finite sequence λ1≥λ2≥···of positive integers.λi is called the i -th part of λ.Let l (λ)denote the number of parts and |λ|=i λi the sum of the parts.For technical reason,we say that 0has a partition,the emptyset,with l (∅)=0=|∅|.Show λ⊆nk q |λ|= n +k k q,where the sum is over all partitions λsatisfying λ1≤n and l (λ)≤k .(15)Let F (x )=a 1x +a 2x 2+···∈xK [[x ]],where a 1=0(and char K =0),and let k,n ∈Z .a.Show n [x n ] F −1 (x ) k =k [x n −k ] x F (x ) n .b.Show (xe −x ) −1 = n ≥1n n −1x n n !.(16)Let c n be the number of paths from (0,0)to (n,n )∈N ×N ,where each step is of the form (1,−1),(1,1)and N denotes the set of nonnegative integers.Define C (x )= n ≥0c n x n .a.Find the generating function C (x ).b.Show c n =1n +1 2n n.2。
MIT英文版线性代数试卷(1)
has
many solutions .
(b) What is the column space of A? Describe the nullspace of A.
Solution:
The column space is a 3-dimensional space inside a 3-dimensional space , i.e.
18.06 Linear Algebra
Spring 2010
For information about citing these materials or our Terms of Use, visit: /terms.
⎢ 0 2 4 ⎥ →
⎢ 0 2 4 ⎥ ⎣
⎦
⎣
⎦
⎣
⎦
2 4 14 0 3 6 0 0 0 ⎡
⎤
3 Thus, x =
⎣
⎦ is a solution to Ax = b, and b is in the column space of A. 2
⎡
⎤
MIT OpenCourseWare
1 0 0
4
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ b =
⎢ 3 1 0
⎥ ·
⎢ 0
⎥ =
⎣
⎦
⎣
⎦
5 0 1 0
2. Suppose A is the matrix
⎡ 0 1 2 2
⎤
⎢ ⎥ ⎢ ⎥ A =
⎢ 0 3 8 7
⎥ .
⎣
⎦
0 0 4 2 (a) (16 points) Find all special solutions to Ax = 0 and describe in words the whole nullspace of A. Solution: First, by row reduction ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 0 1 2 2
线性代数 英文习题4
220
Chapter 4. Orthogonalห้องสมุดไป่ตู้ty
Figure 4.6b shows the same problem in 3-dimensional space (b p e space). The vector b is not in the column space of A. That is why we could not solve Ax D b. No line goes
No straight line b D C C Dt goes through those three points. We are asking for two numbers C and D that satisfy three equations. Here are the equations at t D 0; 1; 2 to match the given values b D 6; 0; 0:
The best choice for Abx is p. The smallest possible error is e D b p. The three points at heights .p1; p2; p3/ do lie on a line, because p is in the column space. In fitting a straight line, bx gives the best choice for .C; D/.
浙江工业大学2019.06线性代数期末(英文版)
浙江工业大学2018/2019(二)期末考试试卷《线性代数》2019.06Class Name Student ID Score Please write in English!一、Choose only one correct answer to each question (3 points each, total 21 points)1. Let 0032104321,543210002000001A ⎛⎫⎪⎪⎪= ⎪⎪ ⎪⎝⎭then det()A = ( B )(A) 120; (B) -120; C) 60; (D) 0. 2. Let 12(,,...,)m A a a a =，where m i a ∈R ，if 12,,...,m a a a are linearly dependent, then which of the following statements is Not true? ( C ). (A) The system 0Ax = must have infinitely many solutions;(B) The rank of A must be less than m ;(C) The column vector m a can be expressed linearly by the remain column vectors;(D) The matrix A must be singular.3. If vector ,,αβγ are linearly independent, and ,,αβω are linearly dependent, then ( D )(A) αcan be expressed by ,,;βγω (B) β can be expressed by ,,;αγω (C) γcan be expressed by ,,;αβω (D) ω can be expressed by ,,.αβγ 4. Let 123,,,x x x y and z be vectors in 4R . If 123det(,,,)2,x x x y =213det(,,,)3,x x z x = then 123det(,,,)y z x x x -= ( A )(A) 1; (B) -1; (C) 5-； (D) 5.5. Let A be an n ⨯n matrix, then which of the following statements is not true? ( D ) (A) A and T A have the same determinant; (B) A and T A have the same rank;(C) A and T A have the same eigenvalues; (D) A and T A have the same eigenvectors. 6. Let A be a 3×3 matrix with eigenvalues 0,1,2-. Then which of the matrices that follow is nonsingular? ( C )(A) 2 +A A (B) 2A I - (C) 2+A I (D) 2A -4I 7. Let αbe a nonzero vectors in 4R ， then the dimension of the eigenspace of T αα corresponding the eigenvalue =0λ is equal to ( C )(A) 1; (B) 2; (C) 3; (D) 4.二、Fill in each blank (3 points each, total 24 points)1. Let (1,2,...,2019),T x = then the rank of the matrix ()T adj xx is equal to 0 .2. Let A be a 44⨯ matrix with ||4,A =-then)|2A = -1 . 3. Let 12231426A t-⎛⎫ ⎪= ⎪ ⎪-⎝⎭, if there exists a nonzero matrix 32B ⨯ such that ,AB O = then t = 1 .4. Given vectors 12(1,0,2) and (0,1,0)T T x x ==, if 3x =2,0,1)T - , then23,}x x will be an orthonormal basis for 3R . 5. Let , and a b c be all numbers. If S is the subspace of 4R consisting of all vectors of the form (,2,,)T a b a b c b c +--. Then the dimension of S is 3 .6. Let 2201112()10120xx x p x x x x=-, the coefficient of 3x is 2 .7. Let 12(1,3),(3,1),T T u u ==- then the coordinate of a vector (20,30)T v =respect to12{,}u u is (11,3)T .8. Let ,αβ,n R ∈ if ||||=||||2,αβ=||||3,αβ-=then the scalar projection of α ontoβis -1/4 .三．(7 points) Let 12(),XB A C -=-where A , B , C are 3⨯3 matrices satisfying,AB I = 2.BC I = If 248862002C ⎛⎫⎪= ⎪ ⎪⎝⎭, find X .Solution:,AB I =2.BC I = 2.C A ∴=21=2X A B A C ∴==124431.001⎛⎫ ⎪= ⎪ ⎪⎝⎭四、(10 points) Let A be a 4⨯5 matrix and let U be the reduced row echelon form of A .If 12(2,1,3,2),(1,2,3,1),T Ta a =--=- 10201013020001500000U -⎛⎫⎪- ⎪=⎪⎪⎝⎭.1. Find a basis for N (A ).2. Given that 0x is a solution to ,Ax b = where 0(0,5,3,4)and (3,2,0,2,0)T T b x ==, determine the matrix A and find a basis for the column space of A . Solution:1. N (A ) ((2,3,1,0,0),(1,2,0,5,1)).T Tspan =---2. From U, we see that 31223(1,8,3,1).T a a a =+=- And from 0,Ax b =we get 124322(0,5,3,4).T a a a ++=So 4121(32)(2,1,3,4).2T a b a a =--=--Then 412425(10,10,12,20).T a a a a =--+=--五、(10 points) Let 1234{,,,}x x x x be a basis for 4.R If 1123,y x x x =++2234,y x x x =++ 33444,.y x x y x =+=1. Show that 1234{,,,}y y y y is also a basis for 4.R2. Find the transition matrix from basis 1234{,,,}x x x x to 1234{,,,}y y y y .Proof ：1. Let 10001100,11100111S ⎛⎫⎪⎪= ⎪ ⎪⎝⎭then 12341234(,,,)(,,,).y y y y x x x x S = Since the matrix S is nonsingular, we get 1234{,,,}y y y y is also a basis for 4.R2. The transition matrix from basis 1234{,,,}x x x x to 1234{,,,}y y y y is 11000110001101011S -⎛⎫ ⎪- ⎪= ⎪-⎪--⎝⎭六、(12 points) Let123(,,)A ααα=，where ()121,1,1,(1,1,1),αλαλ=+=+3(1,1,1)αλ=+，and let 2(0,,).b λλ=(1) For what values of λ will the system Ax b = be inconsistent?(2) For what values of λ will the system Ax b = have a unique solution?(3) For what values of λ will the system Ax b = have the infinitely many solutions? And find all the solutions.Solution: 2111111det()(3).111A A λλλλλ+⎛⎫ ⎪=+=+ ⎪ ⎪+⎝⎭， If 0,3, λ≠-then the system Ax b = has a unique solution 1A b -.If 0, λ=then 1110()000 00000A b ⎛⎫⎪→ ⎪ ⎪⎝⎭， it has infinitely many solutions, and all thesolutions are of the form (1,1,0)(1,0,1).T T s t -+-If 3, λ=-then 1129()011 40002A b ⎛-⎫ ⎪→- ⎪ ⎪⎝⎭， it is inconsistent.七、(10 points) Let 210120,00A t ⎛⎫ ⎪= ⎪ ⎪⎝⎭ 23035,005s B ⎛⎫ ⎪= ⎪ ⎪⎝⎭1. For what value of t and s will A be similar to B ?2. Find a matrix S such that 1.A SBS -=Solution:1. det(A)=det(B) and tr(A)=tr(B), so we get 5, 1.t s ==2. Obviously, the eigenvalues of A and B are 1, 3 and 5.For 1,λ= we get the corresponding eigenvectors of A and B are (1,1,0)T -and (1,0,0)T respectively.For 3,λ= we get the corresponding eigenvector of A and B are (1,1,0)T and (1,1,0)T respectively.For 5,λ= we get the corresponding eigenvector of A and B are (0,0,1)T and (4,5,2)T respectively.Let 100110114030,110,015005001002D X Y -⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪=== ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭, Then we have11,.A XDX D Y BY --==So 11,A XY BYX --=and then 1123102.1002S XY -⎛⎫-- ⎪ ⎪==-⎪ ⎪⎝⎭八、(6 points) Let A and B be n ⨯n matrices. Show that(1) If λ is a nonzero eigenvalue of AB , then it is also an eigenvalue of BA. (2) If 0λ= is an eigenvalue of AB , then 0λ= is also an eigenvalue of BA. Proof:(1) Let with ABv v v λθλ=≠≠，0, then 0 and ,Bv BABv Bv λ≠= So λis also an eigenvalue of BA .(2) If 0λ=, then AB is singular , and so does BA . Then 0λ= is also an eigenvalue of BA.。
MIT线性代数Problem set 5
Note that if B has as its columns a basis for the row space of A then the rows of B T will form a basis for the row space of A. Since the row reduced forms of A and B T agree (up to 13 decimal places, but the numbers up there are just rounding error) their rows must span the same space, so the columns of B are indeed a basis for the row space of A. Problem 32. Suppose I give you four nonzero vectors r, n, c, l in R2 . a. What are the conditions for those to be bases for the four fundamental sub spaces C (AT ), N (A), C (A), N (AT ) of a 2 × 2 matrix? b. What is one possible matrix A? Solution (12 points) a. In order for r and n to be bases for N (A) and C (AT ) we must have r · n = 0, as the row space and null space must be orthogonal. Similarly, in order 3
Problem 33. Suppose I give you four nonzero vectors r1 , r2 , n1 , n2 , c1 , c2 , l1 , l2 in R2 . a. What are the conditions for those to be bases for the four fundamental sub spaces C (AT ), N (A), C (A), N (AT ) of a 2 × 2 matrix? b. What is one possible matrix A? Solution (12 points) a. Firstly, by the same kind of dimension considerations as before we need the four sets {r1 , r2 }, {n1 , n2 }, {c1 , c2 } and {l1 , l2 } to each contain linearly independent vectors. (For example, if r1 and r2 are linearly dependent the dim C (AT ) = 1 not 2, and then dim C (AT ) + dim N (A) < 4 which can’t happen.) Secondly, for i = 1, 2 and j = 1, 2 we need ri · nj = 0 and ci · lj = 0. This will imply that the specified row space and nullspace are orthogonal, and that the specified column space and left nullpace are also orthogonal. (When we are given subspaces in terms of bases it suffices to check orthogonality on the basis.) b. One possible such matrix is A= � c1 c2 �� r1 r2 �T .
第七届哈佛大学-麻省理工数学竞赛代数题及解答
Solution: −2010012/2010013
Let z1, . . . , z5 be the roots of Q(z) = z5 +2004z −1. We can check these are distinct (by using the fact that there’s one in a small neighborhood of each root of z5 + 2004z, or by
2. Find the largest number n such that (2004!)! is divisible by ((n!)!)!. Solution: 6 For positive integers a, b, we have a! | b! ⇔ a! ≤ b! ⇔ a ≤ b. Thus, ((n!)!)! | (2004!)! ⇔ (n!)! ≤ 2004! ⇔ n! ≤ 2004 ⇔ n ≤ 6.
10. There exists a polynomial P of degree 5 with the following property: if z is a complex number such that z5 +2004z = 1, then P (z2) = 0. Calculate the quotient P (1)/P (−1).
Harvard-MIT Mathematics Tournament
February 28, 2004
Individual Round: Algebra Subject Test
1. How many ordered pairs of integers (a,b) satisfy all of the following inequalities?
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5. (11 points) (a) Consider a 120� rotation around the axis x = y = z. Show that the vector i = (1, 0, 0) is
rotated to the vector j = (0, 1, 0). (Similarly j is rotated to k = (0, 0, 1) and k is rotated to i.) How is j − i related to the vector (1, 1, 1) along the axis?
Answer: ⎥ ⎤ .8 .1 . A= .2 .9 The eigenvector with � = 1 is
⎤
1/3 2/3
⎥
This is the steady state starting from
. ⎤
1 0
⎥
.
2 of all students prefer linear algebra! I agree. 3
Answer: A [v1 · · · vn ] = [b1 · · · bn ] or AV = B. Then A = BV −1 if the v � s are independent.
(b) Describe the column space of that matrix A in terms of the given vectors. Answer: The column space of A consists of all linear combinations of b1 , · · · , bn .
�
does the solution
�
x(t) y (t)
�
to this differential equation
lie on a single straight line in R 2 for all t?
Answer: If the initial conditions are a multiple of either eigenvector (2, 1) or (1, 1), the solution is at all times a multiple of that eigenvector.
(c) If a 3 by 3 matrix P projects every vector onto the plane x +2y + z = 0, find three eigenvalues
and three independent eigenvectors of P. No need to compute P .
(a) Write those equations Avi = bi in matrix form. What condition on which vectors allows A
to be determined uniquely? Assuming this condition, find A from V and B.
(d) If the vector b is the sum of the four columns of A, write down the complete solution to
Ax = b.
Answer: ⎡ � ⎡ � ⎡ 1 −2 3 � 1 ⎢ � ⎢ � ⎢ ⎢ + x2 � 1 ⎢ + x4 � −2 ⎢ x=� � 1 ⎣ � 0 ⎣ � 0 ⎣ 1 0 1 �
2. (11 points) This problem finds the curve y = C + D 2 t which gives the best least squares fit
to the points (t, y ) = (0, 6), (1, 4), (2, 0).
(a) Write down the 3 equations that would be satisfied if the curve went through all 3 points. Answer: C + 1D = 6 C + 2D = 4 C + 4D = 0
4. (11 points) (a) Suppose xk is the fraction of MIT students who prefer calculus to linear algebra at year k.
The remaining fraction yk = 1 − xk prefers linear algebra. At year k + 1, 1/5 of those who prefer calculus change their mind (possibly after taking 18.03). Also at year k + 1, 1/10 of those who prefer linear algebra change their mind (possibly because of this exam). � � � � � � xk+1 xk 1 k Create the matrix A to give =A and find the limit of A as k � �. yk 0 yk+1
Solve AT Ax ˆ = AT b : ⎤ 3 7 7 21 ⎥⎤ C D ⎥
� � 1 A A= � 1 2 4 1 � ⎦ � 1 1 1 � � AT b = � 1 2 4 ⎤ 10 14 ⎥ ⎤ C D ⎥
T
1 1 1
=
gives
1 = 14
=
.
(c) What values should y have at times t = 0, 1, 2 so that the best curve is y = 0? Answer: The projection is p = (0, 0, 0) if AT b = 0. In this case, b = values of y = c(2, −3, 1).
(b) Solve these differential equations, starting from x(0) = 1, y (0) = 0 :
dx = 3x − 4y dt dy = 2x − 3y . dt
Answer: ⎤ ⎥ 3 −4 A= . 2 −3 has eigenvalues �1 = 1 and �2 = −1 with eigenvectors x1 = (2, 1) and x2 = (1, 1).
is the resulting output?
�
Answer: Gram-Schmidt will find the unit vector ⎡ 1 1 q1 = � � 2 ⎣ . 14 3 �
But the construction of q2 fails because column 2 = 2 (column 1).
Answer: The plane is perpendicular to the vector (1, 2, 1). This is an eigenvector of P with � = 0. The vectors (−2, 1, 0) and (1, −1, 1) are eigenvectors with � = 0.
(c) What additional condition on which vectors makes A an invertible matrix? Assuming this,
find A−1 from V and B .
Answer: If the b� s are independent, then B is invertible and A−1 = V B −1 .
Answer AAT is m by m but its rank is not greater than n (all columns of AAT are combinations of
A3 = I. What are the eigenvalues of ecause this is three 120� rotations (so 360� ). The eigenvalues satisfy �3 = 1 so � = 1, e2�i/3 , e−2�i/3 = e4�i/3 .
Answer: �
⎡ −1 j − i = � 1 ⎣
0
⎡ 1 is orthogonal to the axis vector � 1 ⎣ .
1
So are k − j and i − k. By symmetry the rotation takes i to j, j to k, k to i. (b) Find the matrix A that produces this rotation (so Av is the rotation of v ). Explain why
3. (11 points) Suppose Avi = bi for the vectors v1 , . . . , vn and b1 , . . . , bn in Rn . Put the v ’s into
the columns of V and put the b’s into the columns of B.
orthonormal eigenvectors.
Answer: ⎡ ⎤ ⎥ 1 2 1 2 3 � 14 28 T A A= 2 4 ⎣= 2 4 6 28 56 3 6 ⎤ ⎥ ⎤ ⎥ 1 1 1 −2 and x2 = � . has �1 = 70 and �2 = 0 with eigenvectors x1 = � 1 5 2 5 ⎤ ⎥ ⎡ � ⎡ ⎥ 1 2 ⎤ 5 10 15 1 2 3 AAT = � 2 4 ⎣ = � 10 20 30 ⎣ has �1 = 70, �2 = 0, �3 = 0 with 2 4 6 3 6 15 30 45 � ⎡ � ⎡ � ⎡ 1 −2 3 1 1 1 x1 = � � 2 ⎣ and x2 = � � 1 ⎣ and x3 = � � 6 ⎣ . 14 3 5 70 −5 0 � (b) If you apply the Gram-Schmidt process (orthonormalization) to the columns of this matrix A, what