WEBVTT
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Hi, this is the first lecture in MIT's course 1806, linear algebra, and I'm Gilbert Strang.
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The text for the course is this book introduction to linear algebra, and the course web page,
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which has got a lot of exercises from the past, the MATLAB code, the syllabus for the course,
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is web.mit.edu slash 1806.
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And this is the first lecture, lecture 1.
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So, and later we'll say, which will give the web address for viewing these videotakes.
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Okay, so what's in the first lecture?
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This is my plan, the fundamental problem of linear algebra, which is to solve a system
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of linear equations.
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So let's start with the case when we have some number of equations, say n equations,
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and n unknowns.
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So an equal number of equations and unknowns, that's the normal, nice case.
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And what I want to do is, with examples, of course, to describe, first, what I call the
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row picture.
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That's the picture of one equation at a time.
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It's the picture you've seen before in two by two equations where lines meet.
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So in a minute you'll see lines meeting.
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The second picture, I put a star beside that, because that's such an important one.
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And maybe new to you is the picture a column at a time.
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And those are the rows and columns of a matrix.
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So the third, the algebra way to look at the problem, is the matrix form, using a matrix
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that I'll call A.
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Okay, so can I do an example?
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The whole semester will be example, and then see what's going on with the example.
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So take an example.
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Two equations, two unknowns.
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So let me take two x minus y equals zero, let's say, and minus x plus say two y equals
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three.
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Okay.
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I can even say right away what's the matrix that is what's the coefficient matrix.
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The matrix that involves these numbers, a matrix is just a rectangular array of numbers.
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Here it's two rows and two columns.
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So two and minus one in the first row, minus one and two in the second row.
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That's the matrix and the unknown, well, we've got two unknowns.
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So we've got a vector with two components, x and y.
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And we've got two right hand sides that go into a vector zero three.
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I couldn't resist writing the matrix form, even before the pictures.
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So I always will think of this as the matrix a, the matrix of coefficients.
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Then there's a vector of unknowns.
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Here we've only got two unknowns.
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Later we'll have any number of unknowns.
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And that vector of unknowns, well, I'll make that extra, extra bold.
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And the right hand side is also a vector that I'll always call b.
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The linear equations are a, x equal b.
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And the idea now is to solve this particular example and then step back to see the bigger
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picture.
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OK, what's the picture for this example?
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The row picture.
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OK, so here comes the row picture.
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So that means I take one row at a time and I'm drawing here the x, y, plane.
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And I'm going to put plot all the points to satisfy that first equation.
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So I'm looking at all the points to satisfy 2x minus y equals zero.
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I'm often, it's often good to start with which point on the horizontal line.
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On this horizontal line, y is zero.
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The x-axis as y is zero and that in this case actually then x is zero.
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So the point, the origin, the point with coordinates zero zero is on the line.
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It solves that equation.
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OK, tell me, well, I guess I have to tell you, another point that solves this same equation.
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Let me suppose x is one.
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So I'll take x to be one.
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Then y should be two, right?
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So there's the point one, two that also solves this equation.
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And then I could put in more points.
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But let me put in all the points at once because they all lie on a straight line.
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This is a linear equation and that word linear has got the letters for line in it.
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That's the equation.
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That's the line of solutions to 2x minus y equals zero.
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My first row, first equation.
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So typically maybe x equals a half, y, or one.
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Well, work.
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And it's sure enough it does.
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OK, that's the first one.
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Now the second one is not going to go through the origin.
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So it's always important that we go through the origin or not.
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In this case, yes, because there's a zero over there.
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In this case, we don't go through the origin because if x and y are zero, we don't get
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three.
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So let me again say suppose y is zero.
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What x do we actually get?
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If y is zero, then I get x is minus three.
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So if y is zero, I go along minus three.
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So there's one point on this second line.
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Now let me say, well, suppose x is minus one.
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Just to take another x.
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If x is minus one, then this is a one.
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And I think y should be a one.
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Because if x is minus one, then I think y should be a one and we'll get that point.
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Is that right?
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If x is minus one, that's a one.
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If y is a one, that's a two.
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And the one and the two make three.
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And that points on the equation.
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OK, now I should just draw the line.
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Connecting those two points, that will give me the whole line.
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And if I've done this reasonably well, I think it's going to happen to go through, well,
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it's happened.
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It was a range to go through that point.
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So I think that the second line is this one.
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And this is the all important point that lies on both lines.
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So we just check that point, which is the point x equal one.
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And y was two, right?
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That's the point there.
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And that, I believe, solves both equations.
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Let's just check this.
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If x is one, I have a minus one plus four equals three.
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OK, apologies for drawing this picture that you've seen before.
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But this, seeing the row picture, first of all, for n equal two, two equations and two
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unknowns, it's the right place to start.
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OK, so we've got the solution, the point that lies on both lines.
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Now can I come to the column picture?
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Pay attention.
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This is the key point.
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So the column picture.
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I'm now going to look at the columns of the matrix.
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I'm going to look at this part and this part.
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I'm going to say that the x part is really, so it's really x times, you see, I'm putting
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the two, I'm kind of getting the two equations at one.
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That part, and then I have a y, and in the first equation it's multiplying a minus one,
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and in the second equation, a two, and on the right hand side, zero and three.
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You see the columns of the matrix, the columns of a are here, and the right hand side b is
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there.
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And now what is the equation asking for?
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It's asking us to find somehow to combine that vector and this one in the right amount
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to get that one.
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It's asking us to find the right linear combination.
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This is called a linear combination, and it's the most fundamental operation in the whole
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course.
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So linear combination of the columns.
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That's what we're seeing on the left side.
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Again, I don't want to write down a big definition.
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You can see what it is.
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There's column one, there's column two.
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I multiply by some numbers, and I add.
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That's a combination, a linear combination, and I want to make those numbers the right
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numbers to produce zero three.
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Now I want to draw a picture that represents what this is algebra, what's the geometry,
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what's the picture that goes with it.
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So again, these vectors have two components, so I better draw a picture like that.
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So can I put down these columns?
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I'll draw these columns as they are, and then I'll do a combination of them.
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So the first column is over two and down one, right?
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So there's the first column.
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The first column, column one.
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It's the vector two minus one.
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The second column is, see, I go over minus one is the first component and up two, it's
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here.
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There's column two.
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So this, again, you see what its components are.
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Its components are minus one two.
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Good, that's this guy.
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Now I have to take a combination.
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What combination shall I take?
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Why not the right combination?
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What the hell?
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Okay.
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The combination I'm going to take is the right one to produce zero three, and then we'll
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see it happen in the picture.
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So the right combination is to take x is one of those and two of these.
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It's because we already know that that's the right x and y, so why not take the correct
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combination here and see it happen.
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Okay, so how do I picture this linear combination?
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So I start with this vector, that's already here.
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So that's one of column one, that's one times column one, right there.
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And now I want to add on, so I'm going to hook the next vector on to the front of the
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arrow, we'll start the next vector and it'll go this way.
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So let's see.
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Can I do it right?
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If I added on one of these vectors it would go left one and up two, so it would go left
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one and up two, so it would probably get us to there.
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Maybe I'll do dotted line for that.
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Okay?
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That's one of column two, tucked on to the end, but I wanted to tuck on two of column two.
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So the second one will go up, but left one and up two also, it'll probably end there,
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and there's another one.
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So what I put in here is two of column two, added on.
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And where did I end up?
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What are the coordinates of this result?
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What do I get when I take one of this plus two of that?
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I do get that of course.
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There it is, there it is, x is zero, y is three, that's b, that's the answer we want.
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And how do I do it?
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You see I do it just like the first component, I have a two and a minus two that produces
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a zero.
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And in the second component I have a minus one and a four, they combine to give the three.
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But look at this picture.
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So here's our key picture.
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I combine this column and this column to get, maybe I better to get this guy.
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That was the b, that's the zero three.
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Okay.
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So that idea of linear combinations is crucial.
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And also, do we want to think about this question?
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Sure, why not?
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What are all the combinations?
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If I took, can I go back to x is and y?
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This is a question for really, it's going to come up over and over.
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But why don't we see it once now?
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If I took all the x's and all the y's, all the combinations, what would be all the results?
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And actually the result would be that I could get any right hand side at all.
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The combinations of this and this would fill the whole plane.
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You can tuck that away, we'll explore it further.
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But this idea of what linear combination gives b and what do all the linear combinations
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give?
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What are all the possible achievable right hand sides b?
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That's going to be basic.
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Okay, can I move to three equations and three unknowns?
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Because it's easy to picture the two by two case.
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Let me do a three by three example.
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Okay, I'll start at the same way.
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So maybe two x minus y.
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And maybe I'll take no z's as a zero.
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And maybe a minus x and a two y.
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And maybe a minus z is a.
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Oh, let me make that a minus one.
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And just for variety, let me take minus three z minus three y's.
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I should keep the y's in that line.
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And four z's is say four.
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Okay.
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That's three equations.
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I mean three dimensions x, y, z.
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And I don't have a solution yet.
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So I want to understand the equations and then solve them.
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Okay.
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So how do I, how do you understand them?
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The row picture is one way.
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The column picture is another very important way.
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Just let's remember the matrix form here.
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That's easy.
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The matrix form, what's our matrix A?
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The matrix A is this right hand side, the two and the minus one and the zero from the
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first row.
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The minus one and the two and the minus one from the second row.
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The zero, the minus three and the four from the third row.
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So it's a three by three matrix, three equations, three unknowns.
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And what's our right hand side?
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Of course it's the vector zero minus one four.
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Okay.
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So that's the way, well that's the shorthand to write out the three equations.
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But it's the picture that I'm looking for today.
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Okay.
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So the row picture.
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All right.
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So I'm in three dimensions x, y and z.
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And I want to take those equations one at a time and ask and make a picture of all the
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points that satisfy, let's take equation number two.
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If I make a picture of all the points that satisfy all the x, y, z points, let's solve
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this equation.
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So first of all, the origin is not one of them.
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X, y, z being zero, zero, zero would not solve that equation.
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What are some points that do solve the equation?
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Let's see.
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Maybe if x is one, y and z could be zero.
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That would work, right?
253
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So there's one point.
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I'm looking at this second equation here just to start with.
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Let's see.
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Also, I guess if z could be one, x and y could be zero.
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So that would just go straight up that axis.
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And probably I want a third point here that may take x to be zero.
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Let's say x to be zero, z to be zero, then y would be minus half, right?
260
00:19:02.880 --> 00:19:06.080
So there's a third point somewhere.
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Oh, okay.
262
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Let's see.
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00:19:11.200 --> 00:19:15.600
I want to put in all the points that satisfy that equation.
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00:19:15.600 --> 00:19:20.600
Do you know what that bunch of points will be?
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It's a plane.
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00:19:21.920 --> 00:19:27.920
If we have a linear equation, then fortunately, the graph of the thing, the plot of all the
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points that solve it, are a plane.
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00:19:31.720 --> 00:19:40.560
So these three points determine a plane, but your lecture is not rim-brant.
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And the art is going to be the weak point here.
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So I'm just going to draw a plane, right?
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There's a plane somewhere.
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That's my plane.
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00:19:53.760 --> 00:19:57.800
That plane is all the points that solve this guy.
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And what about this one?
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2x minus y plus zero z.
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So z actually can be anything.
277
00:20:07.600 --> 00:20:11.200
Again, it's going to be another plane.
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Each row in a three by three problem gives us a plane in three dimensions.
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So this one is going to be some other plane.
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Maybe it, maybe I'll try to draw it like this.
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And those two planes meet in a line.
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00:20:28.000 --> 00:20:35.320
So if I have two equations, just the first two equations in three dimensions, those give
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00:20:35.320 --> 00:20:36.320
me a line.
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00:20:36.320 --> 00:20:38.080
The line where those two planes meet.
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And now the third guy is a third plane.
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And it goes somewhere.
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00:20:51.240 --> 00:20:54.560
OK, those three things meet in a point.
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00:20:54.560 --> 00:20:57.880
Now I don't know where that point is, frankly.
289
00:20:57.880 --> 00:21:01.080
But the enrages will find it.
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00:21:01.080 --> 00:21:09.200
The main point is that there is, that the three planes, because they're not parallel,
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they're not special.
292
00:21:11.280 --> 00:21:14.680
They do meet in one point, and that's the solution.
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00:21:14.680 --> 00:21:22.240
But maybe you can see that this row picture is getting a little hard to see.
294
00:21:22.240 --> 00:21:28.360
The row picture was a cinch when we looked at two lines meeting.
295
00:21:28.360 --> 00:21:32.360
When we look at three planes meeting, it's not so clear.
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00:21:32.360 --> 00:21:37.760
And in four dimensions, probably a little less clear.
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00:21:37.760 --> 00:21:41.840
So can I quit on the row picture?
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00:21:41.840 --> 00:21:48.960
Quit on the row picture before I've successfully found the point where the three planes meet.
299
00:21:48.960 --> 00:21:56.360
All I really want to see is that the row picture consists of three planes.
300
00:21:56.360 --> 00:22:02.360
And if everything works right, three planes meet in one point, and that's the solution.
301
00:22:02.360 --> 00:22:07.800
Now you can tell I prefer the column picture.
302
00:22:07.800 --> 00:22:10.160
OK, so let me take the column picture.
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00:22:10.160 --> 00:22:17.960
But x times, so there were two x's in the first equation, minus one x's, I'll just take,
304
00:22:17.960 --> 00:22:19.840
and no x's in the third.
305
00:22:19.840 --> 00:22:22.160
It's just the first column of that.
306
00:22:22.160 --> 00:22:24.480
And how many y's are there?
307
00:22:24.480 --> 00:22:29.440
There's minus one in the first equation, two in the second, and maybe minus three in
308
00:22:29.440 --> 00:22:30.680
the third.
309
00:22:30.680 --> 00:22:33.040
Just the second column of my matrix.
310
00:22:33.040 --> 00:22:41.360
And z times, no z, minus one z, and four z.
311
00:22:41.360 --> 00:22:49.920
And it's those three columns that I have to combine to produce the right-hand side,
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00:22:49.920 --> 00:22:56.240
which is zero minus one four.
313
00:22:56.240 --> 00:23:01.160
OK, so what have we got on this left-hand side?
314
00:23:01.160 --> 00:23:02.800
A linear combination.
315
00:23:02.800 --> 00:23:07.480
It's a linear combination now of three vectors, and they happen to be.
316
00:23:07.480 --> 00:23:09.920
Each one is a three-dimensional vector.
317
00:23:09.920 --> 00:23:15.560
So we want to know what combination of those three vectors produces that one.
318
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So I try to draw the column picture then.
319
00:23:19.240 --> 00:23:24.280
So since these vectors have three components, so it's some multiple, let me draw in the
320
00:23:24.280 --> 00:23:26.240
first column as before.
321
00:23:26.240 --> 00:23:35.960
So x is two and y is minus one, maybe there's the first column.
322
00:23:35.960 --> 00:23:43.640
y, the second column has maybe a minus one and a two in the y's and minus three, somewhere
323
00:23:43.640 --> 00:23:48.000
there possibly, column two.
324
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And the third column has zero minus one four.
325
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So how shall I draw that?
326
00:23:56.720 --> 00:24:00.600
So nothing in this was the first component.
327
00:24:00.600 --> 00:24:05.040
The second component was a minus one, maybe up here.
328
00:24:05.040 --> 00:24:08.320
That's column three.
329
00:24:08.320 --> 00:24:15.760
That's the column zero minus one and four, this guy.
330
00:24:15.760 --> 00:24:19.440
So again, what's my problem?
331
00:24:19.440 --> 00:24:28.240
What this equation is asking me to do is to combine these three vectors with the right
332
00:24:28.240 --> 00:24:31.720
combination to produce this one.
333
00:24:31.720 --> 00:24:41.440
Well, you can see what the right combination is because in this special problem, especially
334
00:24:41.440 --> 00:24:47.360
chosen by the lecturer, that right hand side that I'm trying to get is actually one of
335
00:24:47.360 --> 00:24:48.840
these columns.
336
00:24:48.840 --> 00:24:50.440
So I know how to get that one.
337
00:24:50.440 --> 00:24:52.280
So what's the solution?
338
00:24:52.280 --> 00:24:54.560
What combination will work?
339
00:24:54.560 --> 00:24:57.680
I just want one of these and none of these.
340
00:24:57.680 --> 00:25:05.600
So x should be zero, y should be zero, and z should be one.
341
00:25:05.600 --> 00:25:06.600
That's the combination.
342
00:25:06.600 --> 00:25:08.520
None of that, none of that.
343
00:25:08.520 --> 00:25:10.640
One of those is obviously the right one.
344
00:25:10.640 --> 00:25:19.840
So this column three is actually the same as b in this particular problem.
345
00:25:19.840 --> 00:25:24.120
I made it work that way just so we would get an answer, zero, zero, one.
346
00:25:24.120 --> 00:25:32.600
So somehow that's the point where those three planes met and I couldn't see it before.
347
00:25:32.600 --> 00:25:37.320
Of course, I won't always be able to see it from the column picture either.
348
00:25:37.320 --> 00:25:46.160
But the next lecture actually, which is about elimination, which is the systematic way that
349
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everybody, every bit of software too, production large scale software would solve the equations.
350
00:26:00.960 --> 00:26:08.080
So the lecture that's coming up, if I was to add that to the syllabus, will be about
351
00:26:08.080 --> 00:26:13.360
how to find x, y, z in all cases.
352
00:26:13.360 --> 00:26:18.800
Can I just think again though about the big picture?
353
00:26:18.800 --> 00:26:25.080
The big by the big picture, I mean, let's keep this same matrix on the left.
354
00:26:25.080 --> 00:26:27.840
But imagine that we have a different right hand side.
355
00:26:27.840 --> 00:26:31.800
Oh, let me take a different right hand side.
356
00:26:31.800 --> 00:26:39.040
So I'll change that right hand side to something that actually is also pretty special.
357
00:26:39.040 --> 00:26:45.160
Let me change it to suppose if I add those first two columns that would give me a one and
358
00:26:45.160 --> 00:26:50.560
a one and a minus three, there's a very special right hand side.
359
00:26:50.560 --> 00:26:55.920
I just cooked it up by adding this one to this one.
360
00:26:55.920 --> 00:27:00.360
Now what's the solution with this new right hand side?
361
00:27:00.360 --> 00:27:03.280
The solution with this new right hand side is clear.
362
00:27:03.280 --> 00:27:09.440
It took, now I took one of these, one of these, and none of those.
363
00:27:09.440 --> 00:27:15.880
So actually, it just changed around to this when I took this new right hand side.
364
00:27:15.880 --> 00:27:17.120
Okay.
365
00:27:17.120 --> 00:27:27.320
So in the row picture, I have three different planes, three new planes meeting now at this
366
00:27:27.320 --> 00:27:28.320
point.
367
00:27:28.320 --> 00:27:31.760
In the column picture, I have the same three columns.
368
00:27:31.760 --> 00:27:35.720
But now I'm combining them to produce this guy.
369
00:27:35.720 --> 00:27:40.840
And it turned out that column one plus column two, which would be somewhere there, there
370
00:27:40.840 --> 00:27:49.920
is the right column, one of this and one of this would give me the new B.
371
00:27:49.920 --> 00:27:50.920
Okay.
372
00:27:50.920 --> 00:27:53.480
So that's like we squeezed in an extra example.
373
00:27:53.480 --> 00:27:59.440
But now think about all B, all right hand side.
374
00:27:59.440 --> 00:28:08.120
Could I get, can I solve these equations for every right hand side?
375
00:28:08.120 --> 00:28:11.920
Can I say that that's a question, so that's the algebra question.
376
00:28:11.920 --> 00:28:16.720
Can I solve AX equal B for every B?
377
00:28:16.720 --> 00:28:18.360
Let me write that down.
378
00:28:18.360 --> 00:28:28.080
Can I solve AX equal B for every right hand side B?
379
00:28:28.080 --> 00:28:31.360
I mean, is there a solution?
380
00:28:31.360 --> 00:28:36.400
And then if there is a elimination, it will give me a way to find it.
381
00:28:36.400 --> 00:28:41.760
And I really wanted to ask, is there a solution for every right hand side?
382
00:28:41.760 --> 00:28:48.080
So now, can I put that in different words, in this linear combination words?
383
00:28:48.080 --> 00:29:07.400
So in linear combination words, do the linear combinations of the columns fill three-dimensional
384
00:29:07.400 --> 00:29:10.960
space.
385
00:29:10.960 --> 00:29:17.480
Every B means all the B's in three-dimensional space.
386
00:29:17.480 --> 00:29:25.000
So do you see that I'm just asking the same question in different words?
387
00:29:25.000 --> 00:29:30.000
Solving AX, oh, that's very important.
388
00:29:30.000 --> 00:29:31.400
A times X.
389
00:29:31.400 --> 00:29:39.800
When I multiply a matrix by a vector, I get a combination of the columns.
390
00:29:39.800 --> 00:29:43.600
I'll write that down in a moment.
391
00:29:43.600 --> 00:29:48.640
So that you see, but in my column picture, that's really what I'm doing.
392
00:29:48.640 --> 00:29:56.080
I'm taking linear combinations of these three columns, and I'm trying to find B.
393
00:29:56.080 --> 00:30:00.280
And actually, the answer for this matrix will be yes.
394
00:30:00.280 --> 00:30:11.720
For this matrix, for this matrix A, for these columns, the answer is yes.
395
00:30:11.720 --> 00:30:26.360
This matrix is, this matrix that I chose for an example, is a good matrix, a non-singular
396
00:30:26.360 --> 00:30:28.480
matrix, an invertible matrix.
397
00:30:28.480 --> 00:30:32.080
Those will be the matrices that we like best.
398
00:30:32.080 --> 00:30:39.760
There could be other, and we will see other matrices where the answer becomes no.
399
00:30:39.760 --> 00:30:44.440
Oh, actually, you can see when it would become no.
400
00:30:44.440 --> 00:30:49.320
When could, what could go wrong?
401
00:30:49.320 --> 00:30:57.160
How could it go wrong that out of these three columns and all their combinations, when
402
00:30:57.160 --> 00:31:04.000
would I not be able to produce some B off here?
403
00:31:04.000 --> 00:31:05.800
When could it go wrong?
404
00:31:05.800 --> 00:31:11.960
Do you see that the combinations, let me say when it goes wrong.
405
00:31:11.960 --> 00:31:21.320
If these three columns all lie in the same plane, then their combinations will lie in that
406
00:31:21.320 --> 00:31:23.680
same plane.
407
00:31:23.680 --> 00:31:25.240
So then we're in trouble.
408
00:31:25.240 --> 00:31:33.840
If the three columns of my matrix, if those three vectors happen to lie in the same plane,
409
00:31:33.840 --> 00:31:40.160
for example, if column three is just the sum of column one and column two, I would be in
410
00:31:40.160 --> 00:31:42.160
trouble.
411
00:31:42.160 --> 00:31:46.480
That would be a matrix A where the answer would be no.
412
00:31:46.480 --> 00:31:54.040
Because the combinations, if column three is in the same plane as column one and two,
413
00:31:54.040 --> 00:31:56.040
I don't get anything new from that.
414
00:31:56.040 --> 00:32:02.360
All the combinations are in the plane, and the only right hand side B that I could get
415
00:32:02.360 --> 00:32:05.800
would be the ones in that plane.
416
00:32:05.800 --> 00:32:13.520
So I could solve it for some right hand side when B is in the plane, but most right hand
417
00:32:13.520 --> 00:32:17.040
sides would be out of the plane and unreachable.
418
00:32:17.040 --> 00:32:19.960
So that would be a singular case.
419
00:32:19.960 --> 00:32:22.640
The matrix would be not invertible.
420
00:32:22.640 --> 00:32:25.600
There would not be a solution for every B.
421
00:32:25.600 --> 00:32:28.520
The answer would become no for that.
422
00:32:28.520 --> 00:32:31.480
Okay.
423
00:32:31.480 --> 00:32:32.480
I don't know.
424
00:32:32.480 --> 00:32:40.720
So we take this a little shot at thinking about nine dimensions.
425
00:32:40.720 --> 00:32:46.040
Imagine that we have vectors with nine components.
426
00:32:46.040 --> 00:32:50.680
Well, it's going to be hard to visualize those.
427
00:32:50.680 --> 00:32:52.760
I don't pretend to do it.
428
00:32:52.760 --> 00:32:58.160
But somehow pretend you do.
429
00:32:58.160 --> 00:33:06.160
If this was nine equations in nine unknowns, then we would have nine columns, and each one
430
00:33:06.160 --> 00:33:12.840
would be a vector in nine dimensional space, and we would be looking at their linear combinations.
431
00:33:12.840 --> 00:33:18.400
So we would be having the linear combinations of nine vectors in nine dimensional space,
432
00:33:18.400 --> 00:33:24.120
and we would be trying to find the combination that hits the correct right hand side B.
433
00:33:24.120 --> 00:33:29.000
And we might also ask the question, can we always do it?
434
00:33:29.000 --> 00:33:31.600
Can we get every right hand side B?
435
00:33:31.600 --> 00:33:35.480
And certainly it will depend on those nine columns.
436
00:33:35.480 --> 00:33:38.400
Sometimes the answer will be yes.
437
00:33:38.400 --> 00:33:41.480
If I picked a random matrix, it would be yes actually.
438
00:33:41.480 --> 00:33:49.520
If I use MATLAB and just use the random command, picked out a nine by nine matrix, I guarantee
439
00:33:49.520 --> 00:33:51.520
it would be good.
440
00:33:51.520 --> 00:33:55.520
It would be non-sangular, it would be invertible, all beautiful.
441
00:33:55.520 --> 00:34:09.280
But if I choose those columns so that they're not independent, so that the ninth column
442
00:34:09.280 --> 00:34:15.800
is the same as the eighth column, then it contributes nothing new, and there would be right
443
00:34:15.800 --> 00:34:19.440
hand side B that I couldn't get.
444
00:34:19.440 --> 00:34:28.520
You sort of think about nine vectors in nine dimensional space and take their combinations.
445
00:34:28.520 --> 00:34:37.400
That's really the central thought that you get kind of used to in linear algebra, even
446
00:34:37.400 --> 00:34:42.640
so you can't really visualize it, you sort of think you can after a while.
447
00:34:42.640 --> 00:34:50.000
Those nine columns and all their combinations may very well fill out the whole nine dimensional
448
00:34:50.000 --> 00:34:51.720
space.
449
00:34:51.720 --> 00:34:56.200
But if the ninth column happened to be the same as the eighth column and gave nothing new,
450
00:34:56.200 --> 00:35:04.720
then probably what it would fill out would be, here's what they'd even just say this,
451
00:35:04.720 --> 00:35:16.680
it would be a sort of plane, an eight dimensional plane inside nine dimensional space.
452
00:35:16.680 --> 00:35:23.240
And it's those eight dimensional planes inside nine dimensional space that we have to work
453
00:35:23.240 --> 00:35:25.240
with eventually.
454
00:35:25.240 --> 00:35:33.640
For now, let's say with the nice case where the matrices work, we can get every right hand
455
00:35:33.640 --> 00:35:38.840
side B, and here we see how to do it with columns.
456
00:35:38.840 --> 00:35:44.600
Okay, there was one step that I realized, which I realized I was saying in words that I
457
00:35:44.600 --> 00:35:52.600
now want to write in letters, because I'm coming back to the matrix form of the equation,
458
00:35:52.600 --> 00:35:55.960
so let me write it here.
459
00:35:55.960 --> 00:36:05.360
The matrix form of my equation of my system is some matrix A times some vector X equals
460
00:36:05.360 --> 00:36:12.800
some right hand side B. Okay, so this is a multiplication, A times X, matrix times vector,
461
00:36:12.800 --> 00:36:17.360
and I just want to say how do you multiply a matrix by a vector?
462
00:36:17.360 --> 00:36:23.120
Okay, so I'm just going to create a matrix.
463
00:36:23.120 --> 00:36:35.560
Let me take two, five, one, three, and let me take a vector X to be say one and two.
464
00:36:35.560 --> 00:36:41.360
How do I multiply a matrix by a vector?
465
00:36:41.360 --> 00:36:49.000
And we'll just think a little bit about matrix notation and how to do that multiplication.
466
00:36:49.000 --> 00:36:52.400
So let me say how I multiply a matrix by a vector.
467
00:36:52.400 --> 00:36:54.680
Actually there are two ways to do it.
468
00:36:54.680 --> 00:36:57.480
Let me tell you my favorite way.
469
00:36:57.480 --> 00:36:59.760
It's a column again.
470
00:36:59.760 --> 00:37:01.720
It's a column at a time.
471
00:37:01.720 --> 00:37:09.120
For me, this matrix multiplication says I'd say one of that column and two of that column
472
00:37:09.120 --> 00:37:10.520
and add.
473
00:37:10.520 --> 00:37:18.040
So this is in my, the way I would think of it, is one of the first column and two of
474
00:37:18.040 --> 00:37:24.640
the second column and let's just see what we get.
475
00:37:24.640 --> 00:37:30.760
So in the first component, I'm getting a two and a ten, I'm getting a twelve there,
476
00:37:30.760 --> 00:37:35.120
and the second component I'm getting a one and a six, I'm getting a seven.
477
00:37:35.120 --> 00:37:41.040
So that matrix times that vector is twelve seven.
478
00:37:41.040 --> 00:37:44.880
Now you could do that another way.
479
00:37:44.880 --> 00:37:47.160
You could do it a row at a time.
480
00:37:47.160 --> 00:37:52.480
You could do and you would get this twelve, and actually I pretty much did it here.
481
00:37:52.480 --> 00:37:55.920
This way, two, I could take that row times my vector.
482
00:37:55.920 --> 00:38:00.760
This is, this is all the time, this is the idea of a dot product.
483
00:38:00.760 --> 00:38:08.280
This vector times this vector, two times one plus five times two is the twelve.
484
00:38:08.280 --> 00:38:13.480
This vector times this vector, one times one plus three times two is the seven.
485
00:38:13.480 --> 00:38:21.960
So I can do it by rows, and in each row, each row times my x is what I'll later call a
486
00:38:21.960 --> 00:38:26.800
dot product, but I also like to see it by columns.
487
00:38:26.800 --> 00:38:29.960
I see this as a linear combination of a column.
488
00:38:29.960 --> 00:38:31.440
So here's my point.
489
00:38:31.440 --> 00:38:43.520
A times x is a combination of the columns of a.
490
00:38:43.520 --> 00:38:51.760
That's how I hope you'll think of A times x when we need it.
491
00:38:51.760 --> 00:38:57.040
Right now we've got, with small ones, we can always do it in different ways,
492
00:38:57.040 --> 00:39:02.240
but later think of it that way.
493
00:39:02.240 --> 00:39:12.440
So that's the picture for two by two system, and if the right hand side B happened to be
494
00:39:12.440 --> 00:39:20.440
twelve seven, then of course the correct solution would be one two.
495
00:39:20.440 --> 00:39:31.080
So let me come back next time to a systematic way using elimination to find the solution
496
00:39:31.080 --> 00:39:44.080
if there is one to any system of any size, and find out because elimination fails, find
497
00:39:44.080 --> 00:39:46.920
out when there is an solution.
498
00:39:46.920 --> 00:39:53.400
Okay, thanks.