#1




لمن يريد تعلم اساسيات الجبر
موقع
http://www.wtamu.edu/academic/anns/m...ebra/index.htm لمن يريد تعلم اساسيات الجبر اخوكم / محمد ابوزيد
__________________
**تستطيع ان تنجح في حياتك و لو كان كل الناس يعتقدون انك غير ناجح، ولكنك لا تنجح ابدا اذا كنت تعتقد في نفسك انك غير ناجح** 
#2




رد: لمن يريد تعلم اساسيات الجبر
رااااائع جدا استاذي مبسط كثير للي يبغى الاساسيات ، اسمح لي بترك الموضوع قليلا ثم تثبيته ...

#3




رد: لمن يريد تعلم اساسيات الجبر
بآرك الله فيك ~

#4




رد: لمن يريد تعلم اساسيات الجبر
[SIZE="5"][FONT="Courier New"]محاضرة فى الجبر الخطى
ويمكنك الاستمتاع بالترجمة بالضغط على cc واختيار اللغة العربية ولكن لا اعرف كيف يتم تثبيتها [youtube]http://www.youtube.com/watch?v=ZK3O402wf1c&feature=player_embedded#![/youtube] اخوكم / محمد ابوزيد
__________________
**تستطيع ان تنجح في حياتك و لو كان كل الناس يعتقدون انك غير ناجح، ولكنك لا تنجح ابدا اذا كنت تعتقد في نفسك انك غير ناجح** 
#5




رد: لمن يريد تعلم اساسيات الجبر
اشكرك اخى العزيز مهند على تثبيت الموضوع
ملاحظة: فى الفيديو اشر بالماوس على العلامة الحمراء واختار cc واضغط مرتين ثم اختار الاختيار الثانى ثم اللغة العربية وتمتع بالترجمة اخوكم / محمد ابوزيد
__________________
**تستطيع ان تنجح في حياتك و لو كان كل الناس يعتقدون انك غير ناجح، ولكنك لا تنجح ابدا اذا كنت تعتقد في نفسك انك غير ناجح** 
#6




رد: لمن يريد تعلم اساسيات الجبر
هذا هو نص المحاضرة باللغة الانجليزية مع الوقت الخاص بها
ان شاء الله ساحاول معرفة كيف يتم تثبيت الترجمة 0:05Hi. This is the first lecture inMIT's course 18.06, 0:11linear algebra, and I'm Gilbert Strang.0:17The text for the course is this book, Introduction to Linear Algebra.0:22And the course web page, which has got a lot of exercises0:28from the past, MatLab codes, the syllabus for the0:34course, is web.mit.edu/18.06. And this is the first0:40lecture, lecture one. So, and later we'll give the web0:46address for viewing these, videotapes. Okay, so what's in the0:52first lecture? This is my plan.0:58The fundamental problem of linear algebra, which is to solve a system1:03of linear equations. So let's start with a case when we1:09have some number of equations, say n equations and n unknowns.1:14So an equal number of equations and unknowns.1:20That's the normal, nice case. And what I want to do is1:25 with examples, of course  to describe,1:29first, what I call the Row picture. That's the picture of one equation1:34at a time. It's the picture you've seen before in two by two equations1:39where lines meet. So in a minute,1:44you'll see lines meeting. The second picture, I'll put a star1:49beside that, because that's such an important one.1:53And maybe new to you is the picture  a column at a time.1:58And those are the rows and columns of a matrix.2:02So the third  the algebra way to look at the2:07problem is the matrix form and using a matrix that I'll call A.2:12Okay, so can I do an example? The whole semester will be examples2:17and then see what's going on with the example. So,2:22take an example. Two equations, two unknowns.2:27So let me take 2x y =0, let's say. And x +2y=3.2:36Okay. let me  I can even say right away  what's the matrix,2:44that is, what's the coefficient matrix? The matrix that involves2:53these numbers  a matrix is just a rectangular array2:59of numbers. Here it's two rows and two columns, so 2 and 3:04minus 1 in the first row minus 1 and 2 in the second row,3:09that's the matrix. And the righthand  the,3:13unknown  well, we've got two unknowns.3:18So we've got a vector, with two components, x and x,3:24and we've got two righthand sides that go into a vector 0 3.3:30I couldn't resist writing the matrix form, right  even3:35before the pictures. So I always will think of this as3:41the matrix A, the matrix of coefficients, then there's a vector3:45of unknowns. Here we've only got two unknowns. Later we'll have any3:49number of unknowns. And that vector of unknowns,3:54well I'll often  I'll make that x  extra bold.3:58A and the righthand side is also a vector that I'll always call b.4:05So linear equations are A x equal b and the idea now is to solve this4:11particular example and then step back to see the bigger picture.4:17Okay, what's the picture for this example, the Row picture?4:24Okay, so here comes the Row picture. So that means I take one row at a4:31time and I'm drawing here the xy plane and I'm going to plot all the4:38points that satisfy that first equation. So I'm looking at all the4:46points that satisfy 2xy =0. It's often good to start with which4:53point on the horizontal line  on this horizontal line,5:00y is zero. The x axis has y as zero and that  in this case,5:05actually, then x is zero. So the point, the origin 5:10the point with coordinates (0, ) is on the line. It solves that5:16equation. Okay, tell me in  well, I guess I have to tell you5:21another point that solves this same equation. Let me suppose x is one,5:27so I'll take x to be one. Then y should be two,5:33right? So there's the point one two that also solves this equation.5:39And I could put in more points. But, but let me put in all the5:44points at once, because they all lie on a straight5:48line. This is a linear equation and that word linear got the letters for5:52line in it. That's the equation  this is the line that ... of5:56solutions to 2xy=0 my first row, first equation.6:00So typically, maybe, x equal a half, y equal one will6:07work. And sure enough it does. Okay, that's the first one. Now6:13the second one is not going to go through the origin. It's6:19always important. Do we go through the origin or not?6:25In this case, yes, because there's a zero over there.6:30In this case we don't go through the origin, because if x and y are6:35zero, we don't get three. So, let me again say suppose y is6:40zero, what x do we actually get? If y is zero, then I get x is minus6:46three. So if y is zero, I go along minus three. So there's6:52one point on this second line. Now let me say, well, suppose x is6:58minus one  just to take another x. If x is minus one, then this is a7:05one and I think y should be a one, because if x is minus one,7:10then I think y should be a one and we'll get that point.7:15Is that right? If x is minus one, that's a one. If y is a one, that's7:19a two and the one and the two make three and that point's7:24on the equation. Okay. Now, I should just draw the7:28line, right, connecting those two points at  that will give me the7:32whole line. And if I've done this reasonably well,7:36I think it's going to happen to go through  well,7:39not happen  it was arranged to go through that point.7:43So I think that the second line is this one, and this is the7:49allimportant point that lies on both lines. Shall we just check7:54that that point which is the point x equal one and y was two,8:00right? That's the point there and that, I believe, solves8:05both equations. Let's just check this.8:12If x is one, I have a minus one plus four equals three,8:19okay. Apologies for drawing this picture that you've seen before.8:27But this  seeing the row picture  first of all, for n equal 2,8:32two equations and two unknowns, it's the right place to start. Okay.8:36So we've got the solution. The point that lies on both lines.8:40Now can I come to the column picture? Pay attention,8:44this is the key point. So the column picture.8:48I'm now going to look at the columns of the matrix.8:52I'm going to look at this part and this part. I'm going to say that8:57the x part is really x times  you see, I'm putting the two  I'm9:02kind of getting the two equations at once 9:07that part and then I have a y and in the first equation it's multiplying9:12a minus one and in the second equation a two,9:17and on the righthand side, zero and three. You see, the9:23columns of the matrix, the columns of A are here and the9:28righthand side b is there. And now what is the equation asking9:34for? It's asking us to find  somehow to combine that vector and9:42this one in the right amounts to get that one. It's asking us to find9:49the right linear combination  this is called a linear combination.9:56And it's the most fundamental operation in the whole course.10:00It's a linear combination of the columns. That's what we're seeing10:04on the left side. Again, I don't want to write down a10:08big definition. You can see what it is.10:12There's column one, there's column two.10:15I multiply by some numbers and I add. That's a combination 10:21a linear combination and I want to make those numbers the right numbers10:26to produce zero three. Okay. Now I want to draw a picture10:32that, represents what this  this is algebra.10:37What's the geometry, what's the picture that goes with it?10:42Okay. So again, these vectors have two components,10:46so I better draw a picture like that. So can I put down these columns?10:51I'll draw these columns as they are, and then I'll do a combination of10:55them. So the first column is over two and down one, right?10:59So there's the first column. The first column. Column one.11:06It's the vector two minus one. The second column is 11:12minus one is the first component and up two. It's here.11:19There's column two. So this, again, you see what its11:26components are. Its components are minus one,11:32two. Good. That's this guy. Now I have to take a combination.11:39What combination shall I take? Why not the right combination, what11:45the hell? Okay. So the combination I'm going11:51to take is the right one to produce zero three and then we'll see it11:57happen in the picture. So the right combination is to take12:03x as one of those and two of these. It's because we already know that12:08that's the right x and y, so why not take the correct12:13combination here and see it happen? Okay, so how do I picture this12:17linear combination? So I start with this vector that's12:22already here  so that's one of column one,12:26that's one times column one, right there. And now I want to add on 12:30so I'm going to hook the next vector onto the front of the arrow will12:35start the next vector and it will go this way. So let's see,12:39can I do it right? If I added on one of these vectors,12:44it would go left one and up two, so we'd go left one and up two, so12:48it would probably get us to there. Maybe I'll do dotted line for that.12:53Okay? That's one of column two tucked onto the end,12:57but I wanted to tuck on two of column two. So that 13:01the second one  we'll go up left one and up two also.13:05It'll probably end there. And there's another one. So what13:09I've put in here is two of column two.13:13Added on. And where did I end up? What are the coordinates of this13:21result? What do I get when I take one of this plus two of that?13:30I do get that, of course. There it is, x is zero, y is three, that's b.13:38That's the answer we wanted. And how do I do it? You see I do13:43it just like the first component. I have a two and a minus two that13:47produces a zero, and in the second component I have a13:52minus one and a four, they combine to give the three.13:57But look at this picture. So here's our key picture.14:02I combine this column and this column to get this guy.14:08That was the b. That's the zero three. Okay. So that idea of14:15linear combination is crucial, and also  do we want to think14:22about this question? Sure, why not. What are all the combinations?14:29If I took  can I go back to xs and ys? This is a question for14:36really  it's going to come up over and over, but why don't we see it14:42once now? If I took all the xs and all the ys, all the combinations,14:49what would be all the results? And, actually,14:55the result would be that I could get any righthand side at all.15:00The combinations of this and this would fill the whole plane.15:06You can tuck that away. We'll, explore it further. But this idea15:12of what linear combination gives b and what do all the linear15:17combinations give, what are all the possible,15:22achievable righthand sides be  that's going to be basic. Okay.15:27Can I move to three equations and three unknowns?15:32Because it's easy to picture the two by two case.15:37Let me do a three by three example. Okay, I'll sort of start it the15:41same way, say maybe 2xy and maybe I'll take15:48no zs as a zero and maybe a x+2y and maybe a z as a 15:56oh, let me make that a minus one and, just for variety let me take,16:053z, 3ys, I should keep the ys in that line, and 4zs is, say, 4.16:13Okay. That's three equations. I'm in three dimensions, x, y,16:19z. And, I don't have a solution yet. So I want to understand the16:24equations and then solve them. Okay. So how do I you understand16:30them? The row picture one way. The column picture is another very16:35important way. Just let's remember the matrix form,16:41here, because that's easy. The matrix form  what's our matrix A?16:46Our matrix A is this righthand side, the two and the minus one and16:51the zero from the first row, the minus one and the two and the16:57minus one from the second row, the zero, the minus three and the17:02four from the third row. So it's a three by three matrix.17:07Three equations, three unknowns. And what's our righthand side?17:12Of course, it's the vector, zero minus one, four.17:17Okay. So that's the way, well, that's the shorthand to write17:22out the three equations. But it's the picture that I'm17:28looking for today. Okay, so the row picture.17:34All right, so I'm in three dimensions, x,17:41y and z. And I want to take those equations one at a time and ask 17:48and make a picture of all the points that satisfy 17:55let's take equation number two. If I make a picture of all the17:59points that satisfy  all the x, y, z points that solve18:03this equation  well, first of all,18:07the origin is not one of them. x, y, z  it being 0, 0, 0 would18:11not solve that equation. So what are some points that do18:16solve the equation? Let's see, maybe if x is one,18:21y and z could be zero. That would work, right? So there's one point.18:26I'm looking at this second equation, here, just, to start18:31with. Let's see.Also, I guess,18:37if z could be one, x and y could be zero,18:43so that would just go straight up that axis. And,18:49probably I'd want a third point here. Let me take x to be zero,18:55z to be zero, then y would be minus a half, right?19:02So there's a third point, somewhere  oh my  okay.19:07Let's see. I want to put in all the points that satisfy that19:12equation. Do you know what that bunch of points will be?19:18It's a plane. If we have a linear equation, then,19:23fortunately, the graph of the thing, the plot of all the points that19:28solve it are a plane. These three points determine a19:33plane, but your lecturer is not Rembrandt and the art is going to be19:38the weak point here. So I'm just going to draw a plane,19:43right? There's a plane somewhere.19:48That's my plane. That plane is all the points that solves this guy.19:53Then, what about this one? Two x minus y plus zero z.19:58So z actually can be anything. Again, it's going to be another20:03plane. Each row in a three by three problem20:09gives us a plane in three dimensions. So this one is going to be some20:14other plane  maybe I'll try to draw it like this.20:20And those two planes meet in a line. So if I have two equations,20:26just the first two equations in three dimensions,20:32those give me a line. The line where those two planes20:39meet. And now, the third guy is a third plane.20:45And it goes somewhere. Okay, those three things meet in a point.20:51Now I don't know where that point is, frankly. But 20:55linear algebra will find it. The main point is that the three21:00planes, because they're not parallel, they're not special.21:04They do meet in one point and that's the solution.21:09But, maybe you can see that this row picture is getting a little hard to21:15see. The row picture was a cinch when we looked at two lines meeting.21:20When we look at three planes meeting, it's not so clear and in21:26four dimensions probably a little less clear.21:32So, can I quit on the row picture? Or quit on the row picture before21:37I've successfully found the point where the three planes meet?21:43All I really want to see is that the row picture consists of three21:48planes and, if everything works right,21:54three planes meet in one point and that's a solution.21:59Now, you can tell I prefer the column picture.22:04Okay, so let me take the column picture. That's x times 22:09so there were two xs in the first equation minus one x is,22:13and no xs in the third. It's just the first column of that.22:19And how many ys are there? There's minus one in the first equations,22:24two in the second and maybe minus three in the third.22:29Just the second column of my matrix. And z times no zs minus one zs and22:35four zs. And it's those three columns,22:40right, that I have to combine to produce the righthand side,22:45which is zero minus one four. Okay. So what have we got on this22:49lefthand side? A linear combination.22:54It's a linear combination now of three vectors,22:58and they happen to be  each one is a three dimensional23:03vector, so we want to know what combination23:07of those three vectors produces that one. Shall I try to draw the column23:11picture, then? So, since these vectors have three23:15components  so it's some multiple  let me draw in the first23:20column as before  x is two and y is minus one.23:25Maybe there is the first column. y  the second column has maybe a23:32minus one and a two and the y is a minus three, somewhere,23:39there possibly, column two. And the third column has 23:46no zero minus one four, so how shall I draw that?23:51So this was the first component. The second component was a minus23:57one. Maybe up here. That's column three,24:02that's the column zero minus one and four.24:07This guy. So, again, what's my problem?24:13What this equation is asking me to do is to combine these three vectors24:19with a right combination to produce this one. Well,24:25you can see what the right combination is, because in24:30this special problem, specially chosen by the lecturer,24:37that righthand side that I'm trying to get is actually one of these24:43columns. So I know how to get that one. So what's the solution?24:49What combination will work? I just want one of these and none of these.24:55So x should be zero, y should be zero and z should be one.25:01That's the combination. One of those is obviously the right25:06one. Column three is actually the same as b in this particular problem.25:11I made it work that way just so we would get an answer,25:17(0,0,1), so somehow that's the point where those three planes met and I25:23couldn't see it before. Of course, I won't always be able25:29to see it from the column picture, either. It's the next lecture,25:35actually, which is about elimination,25:41which is the systematic way that everybody  every bit of software,25:47too  production, largescale software would solve the equations.25:53So the lecture that's coming up. If I was to add that to the syllabus,25:59will be about how to find x, y, z in all cases.26:05Can I just think again, though, about the big picture?26:11By the big picture I mean let's keep this same matrix on the left26:17but imagine that we have a different righthand side.26:23Oh, let me take a different righthand side.26:29So I'll change that righthand side to something that actually is also26:33pretty special. Let me change it to 26:37if I add those first two columns, that would give me a one and a one26:41and a minus three. There's a very special righthand26:45side. I just ****ed it up by adding this one to this one.26:48Now, what's the solution with this new righthand side?26:54The solution with this new righthand side is clear.27:00took one of these and none of those. So actually, it just changed around27:05to this when I took this new righthand side.27:11Okay. So in the row picture, I have three different planes,27:17three new planes meeting now at this point. In the column picture,27:23I have the same three columns, but now I'm combining them27:29to produce this guy, and it turned out that column one27:35plus column two which would be somewhere  there is the right27:41column  one of this and one of this would give me the new b.27:46Okay. So we squeezed in an extra example. But now think about all bs,27:52all righthand sides. Can I solve these equations for27:58every righthand side? Can I ask that question?28:05So that's the algebra question. Can I solve A x=b for every b? Let28:13me write that down. Can I solve A x =b for every28:20righthand side b? I mean, is there a solution?28:26And then, if there is, elimination will give me a way to find it.28:31I really wanted to ask, is there a solution for every righthand side?28:36So now, can I put that in different words  in this linear28:41combination words? So in linear combination words,28:49do the linear combinations of the columns fill three dimensional space?29:00Every b means all the bs in three dimensional space.29:11Do you see that I'm just asking the same question in different words?29:22Solving A x  A x  that's very important.29:28A times x  when I multiply a matrix by a vector,29:34I get a combination of the columns. I'll write that down in a moment.29:40But in my column picture, that's really what I'm doing.29:46I'm taking linear combinations of these three columns and I'm trying29:55to find b. And, actually, the answer for this matrix30:05will be yes. For this matrix A  for these columns, the answer is yes.30:14This matrix  that I chose for an example is a good matrix.30:19A nonsingular matrix. An invertible matrix.30:24Those will be the matrices that we like best. There could be other 30:29and we will see other matrices where the answer becomes,30:34no  oh, actually, you can see when it would become no.30:40What could go wrong? How could it go wrong that out of these 30:45out of three columns and all their combinations 30:51when would I not be able to produce some b off here?30:56When could it go wrong? Do you see that the combinations 31:02let me say when it goes wrong. If these three columns all lie in31:08the same plane, then their combinations will lie in31:14that same plane. So then we're in trouble.31:21If the three columns of my matrix  if those three vectors happen to lie31:27in the same plane  for example, if column three is just the sum of31:33column one and column two, I would be in trouble. That would31:39be a matrix A where the answer would be no, because the combinations 31:45if column three is in the same plane as column one and two,31:50I don't get anything new from that. All the combinations are in the31:55plane and only righthand sides b that I could get would be the ones32:00in that plane. So I could solve it for some32:04righthand sides, when b is in the plane,32:08but most righthand sides would be out of the plane and unreachable.32:13So that would be a singular case. The matrix would be not invertible.32:17There would not be a solution for every b. The answer would become no32:24for that. Okay. I don't know  shall we take just32:30a little shot at thinking about nine dimensions? Imagine that we have32:36vectors with nine components. Well, it's going to be hard to32:42visualize those. I don't pretend to do it.32:48But somehow, pretend you do. Pretend we have  if this was nine32:53equations and nine unknowns, then we would have nine columns,32:59and each one would be a vector in ninedimensional space and we would33:04be looking at their linear combinations.33:09So we would be having the linear combinations of nine vectors in33:13ninedimensional space, and we would be trying to find the33:17combination that hit the correct righthand side b.33:20And we might also ask the question can we always do it?33:24Can we get every righthand side b? And certainly it will depend on33:28those nine columns. Sometimes the answer will be yes 33:32if I picked a random matrix, it would be yes, actually.33:36If I used MatLab and just used the random command,33:40picked out a nine by nine matrix, I guarantee it would be good. It33:44would be nonsingular, it would be invertible,33:48all beautiful. But if I choose those columns so33:54that they're not independent, so that the ninth column is the same34:01as the eighth column, then it contributes nothing new and34:09there would be righthand sides b that I couldn't get.34:16Can you sort of think about nine vectors in ninedimensional space an34:21take their combinations? That's really the central thought 34:25that you get kind of used to in linear algebra.34:30Even though you can't really visualize it, you sort of think you34:34can after a while. Those nine columns and all their34:39combinations may very well fill out the whole ninedimensional space.34:45But if the ninth column happened to be the same as the eighth column and34:50gave nothing new, then probably what it would fill out34:56would be  I hesitate even to say this 35:01it would be a sort of a plane  an eight dimensional plane inside35:06ninedimensional space. And it's those eight dimensional35:11planes inside ninedimensional space that we have to work with eventually.35:16For now, let's stay with a nice case where the matrices work,35:21we can get every righthand side b and here we see how to35:26do it with columns. Okay. There was one step which I35:33realized I was saying in words that I now want to write in letters.35:41Because I'm coming back to the matrix form of the equation,35:48so let me write it here. The matrix form of my equation,35:55of my system is some matrix A times some vector x equals some righthand36:00side b. Okay. So this is a multiplication.36:05A times x. Matrix times vector, and I just want to say how do you36:11multiply a matrix by a vector? Okay, so I'm just going to create a36:16matrix  let me take two five one three 36:21and let me take a vector x to be, say, 1and 2. How do I multiply a36:27matrix by a vector? But just think a little bit about36:33matrix notation and how to do that in multiplication.36:39So let me say how I multiply a matrix by a vector.36:45Actually, there are two ways to do it. Let me tell you my favorite way.36:50It's columns again. It's a column at a time.36:56For me, this matrix multiplication says I take one of that column and37:02two of that column and add. So this is the way I would think of37:07it is one of the first column and two of the second column and let's37:11just see what we get. So in the first component I'm37:15getting a two and a ten. I'm getting a twelve there.37:20In the second component I'm getting a one and a six, I'm37:24getting a seven. So that matrix times that vector is37:29twelve seven. Now, you could do that another way.37:35You could do it a row at a time. And you would get this twelve 37:41and actually I pretty much did it here  this way.37:47Two  I could take that row times my vector. This is the idea of a37:52dot product. This vector times this vector, two times one plus five37:56times two is the twelve. This vector times this vector 38:01one times one plus three times two is the seven.38:05So I can do it by rows, and in each row times my x is what38:11I'll later call a dot product. But I also like to see it by38:17columns. I see this as a linear combination of a column.38:23So here's my point. A times x is a combination of the38:29columns of A. That's how I hope you will think of A times x when we need38:37it. Right now we've got  with small ones, we can always do it38:45in different ways, but later, think of it that way.38:53Okay. So that's the picture for a two by two system.39:01And if the righthand side B happened to be twelve seven,39:07then of course the correct solution would be one two.39:14Okay. So let me come back next time to a systematic way,39:21using elimination, to find the solution, if there is one,39:28to a system of any size and find out  because if elimination fails,39:33find out when there isn't a solution. Okay, thanks. اخوكم / محمد ابوزيد
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#7




رد: لمن يريد تعلم اساسيات الجبر
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#8




رد: لمن يريد تعلم اساسيات الجبر
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#9




رد: لمن يريد تعلم اساسيات الجبر
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#10




رد: لمن يريد تعلم اساسيات الجبر
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