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  • Okay, this is linear algebra, lecture four.

  • And, the first thing I have to do is something that was on the list for last time, but here

  • it is now.

  • What's the inverse of a product?

  • If I multiply two matrices together and I know their inverses, how do I get the inverse

  • of A times B?

  • So I know what inverses mean for a single matrix A and for a matrix B.

  • What matrix do I multiply by to get the identity if I have A B?

  • Okay, that'll be simple but so basic.

  • Then I'm going to use that to -- I will have a product of matrices and the product that

  • we'll meet will be these elimination matrices and the net result of today's lectures is

  • the big formula for elimination, so the net result of today's lecture is this great way

  • to look at Gaussian elimination.

  • We know that we get from A to U by elimination.

  • We know the steps -- but now we get the right way to look at it, A equals L U.

  • So that's the high point for today.

  • Okay.

  • Can I take the easy part, the first step first?

  • So, suppose A is invertible -- and of course it's going to be a big question, when is the

  • matrix invertible?

  • But let's say A is invertible and B is invertible, then what matrix gives me the inverse of A B?

  • So that's the direct question.

  • What's the inverse of A B?

  • Do I multiply those separate inverses?

  • Yes. I multiply the two matrices A inverse and B inverse, but what order do I multiply?

  • In reverse order.

  • And you see why.

  • So the right thing to put here is B inverse A inverse.

  • That's the inverse I'm after.

  • We can just check that A B times that matrix gives the identity.

  • Okay.

  • So why -- once again, it's this fact that I can move parentheses around.

  • I can just erase them all and do the multiplications any way I want to.

  • So what's the right multiplication to do first?

  • B times B inverse.

  • This product here I is the identity.

  • Then A times the identity is the identity and then finally A times A inverse gives the

  • identity.

  • So forgive the dumb example in the book.

  • Why do you, do the inverse things in reverse order?

  • It's just like -- you take off your shoes, you take off your socks, then the good way

  • to invert that process is socks back on first, then shoes.

  • Sorry, okay.

  • I'm sorry that's on the tape.

  • And, of course, on the other side we should really just check -- on the other side I have

  • B inverse,

  • A inverse. That does multiply A B, and this time it's these guys that give the identity,

  • squeeze down, they give the identity, we're in shape.

  • Okay. So there's the inverse.

  • Good. While we're at it, let me do a transpose, because the next lecture has got a lot to

  • -- involves transposes.

  • So how do I -- if I transpose a matrix, I'm talking about square, invertible matrices

  • right now.

  • If I transpose one, what's its inverse?

  • Well, the nice formula is -- let's see.

  • Let me start from A, A inverse equal the identity.

  • And let me transpose both sides.

  • That will bring a transpose into the picture.

  • So if I transpose the identity matrix, what do I have?

  • The identity, right?

  • If I exchange rows and columns, the identity is a symmetric matrix.

  • It doesn't know the difference.

  • If I transpose these guys, that product, then again it turns out that I have to reverse

  • the order.

  • I can transpose them separately, but when I multiply, those transposes come in the opposite

  • order.

  • So it's A inverse transpose times A transpose giving the identity.

  • So that's -- this equation is -- just comes directly from that

  • one. But this equation tells me what I wanted to know, namely what is the inverse of this

  • guy A transpose?

  • What's the inverse of that -- if I transpose a matrix, what'ss the inverse of the result?

  • And this equation tells me that here it is.

  • This is the inverse of A transpose.

  • Inverse of A transpose.

  • Of A transpose.

  • So I'll put a big circle around that, because that's the answer, that's the best answer

  • we could hope for.

  • That if you want to know the inverse of A transpose and you know the inverse of A, then

  • you just transpose that.

  • So in a -- to put it another way, transposing and inversing you can do in either order for

  • a single matrix.

  • Okay. So these are like basic facts that we can now use, all right -- so now I put it

  • to use.

  • I put it to use by thinking -- we're really completing, the subject of elimination.

  • Actually, -- the thing about elimination is it's the right way to understand what the

  • matrix has got.

  • This A equal L U is the most basic factorization of a matrix.

  • I always worry that you will think this course is all elimination.

  • It's just row operations.

  • And please don't.

  • We'll be beyond that, but it's the right algebra to do first.

  • Okay. So, now I'm coming near the end of it, but I want to get it in a decent form.

  • So my decent form is matrix form.

  • I have a matrix A, let's suppose it's a good matrix, I can do elimination, no row exchanges

  • -- So no row exchanges for now.

  • Pivots all fine, nothing zero in the pivot position.

  • I get to the very end, which is U.

  • So I get from A to U.

  • And I want to know what's the connection?

  • How is A related to U?

  • And this is going to tell me that there's a matrix L that connects them.

  • Okay.

  • Can I do it for a two by two first?

  • Okay. Two by two, elimination.

  • Okay, so I'll do it under here.

  • Okay. So let my matrix A be -- We'll keep it simple, say two and an eight, so we know

  • that the first pivot is a two, and the multiplier's going to be a four and then let me put a one

  • here and what number do I not want to put there?

  • Four. I don't want a four there, because in that case, the second pivot would not -- we

  • wouldn't have a second pivot.

  • The matrix would be singular, general screw-up. Okay.

  • So let me put some other number here like seven.

  • Okay.

  • Okay. Now I want to operate on that with my elementary matrix.

  • So what's the elementary matrix?

  • Strictly speaking, it's E21, because it's the guy that's going to produce a zero in

  • that position.

  • And it's going to produce U in one shot, because it's just a two by two matrix.

  • So two one and I'm going to take four of those away from those, produce that zero and leave

  • a three there.

  • And that's U.

  • And what's the matrix that did it?

  • Quick review, then.

  • What's the elimination elementary matrix E21 -- it's one zero, thanks.

  • And -- negative four one,

  • right. Good.

  • Okay. So that -- you see the difference between this and what I'm shooting for.

  • I'm shooting for A on one side and the other matrices on the other side of the equation.

  • Okay.

  • So I can do that right away.

  • Now here's going to be my A equals L U.

  • And you won't have any trouble telling me what -- so A is still two one eight seven.

  • L is what you're going to tell me and U is still two one zero three.

  • Okay. So what's L in this case?

  • Well, first -- so how is L related to this E guy?

  • It's the inverse, because I want to multiply through by the inverse of this, which will

  • put the identity here, and the inverse will show up there and I'll call it L.

  • So what is the inverse of this?

  • Remember those elimination matrices are easy to invert.

  • The inverse matrix for this one is 1 0 4 1, it has the plus sign because it adds back

  • what this removes.

  • Okay. Do you want -- if we did the numbers right, we must -- this should be correct.

  • Okay. And of course it is.

  • That says the first row's right, four times the first row plus the second row is eight

  • seven. Good. Okay.

  • That's simple, two by two.

  • But it already shows the form that we're headed for.

  • It shows -- so what's the L stand for?

  • Why the letter L?

  • If U stood for upper triangular, then of course L stands for lower triangular.

  • And actually, it has ones on the diagonal, where this thing has the pivots on the diagonal.

  • Oh, sometimes we may want to separate out the pivots, so can I just mention that sometimes

  • we could also write this as -- we could have this one zero four one -- I'll just show you

  • how I would divide out this matrix of pivots -- two three.

  • There's a diagonal matrix.

  • And I just -- whatever is left is here.

  • Now what's left?

  • If I divide this first row by two to pull out the two, then I have a one and a one half.

  • And if I divide the second row by three to pull out the three, then I have a one.

  • So if this is L U, this is maybe called L D or pivot U.

  • And now it's a little more balanced, because we have ones on the diagonal here and here.

  • And the diagonal matrix in the middle.

  • So both of those...

  • Matlab would produce either one.

  • I'll basically stay with L U.

  • Okay. Now I have to think about bigger than two by two.

  • But right now, this was just like easy exercise.

  • And, to tell the truth, this one was a minus sign and this one was a plus sign.

  • I mean, that's the only difference.

  • But, with three by three, there's a more significant difference.

  • Let me show you how that works.

  • Let me move up to a three by three, let's say some matrix A, okay?

  • Let's imagine it's three by three.

  • I won't write numbers down for now.

  • So what's the first elimination step that I do, the first matrix I multiply it by, what

  • letter will I use for

  • that? It'll be E two one, because it's -- the first step will be to get a zero in that two

  • one position. right?

  • And then the next step will be to get a zero in the three one position.

  • And the final step will be to get a zero in the three two

  • That's what elimination is, and it produced U. position.

  • And again, no row exchanges.

  • I'm taking the nice case, now, the typical case, too -- when I don't have to do any row

  • exchange, all I do is these elimination steps.

  • Okay.

  • Now, suppose I want that stuff over on the right-hand side, as I really do.

  • That's, like, my point here.

  • I can multiply these together to get a matrix E, but I want it over on the right.

  • I want its inverse over there.

  • So what's the right expression now?

  • If I write A and U, what goes there?

  • Okay.

  • So I've got the inverse of this, I've got three matrices in

  • a row now.

  • And it's their inverses that are going to show up, because each one is easy to invert.

  • Question is, what about the whole bunch?

  • How easy is it to invert the whole bunch?

  • So, that's what we know how to do.

  • We know how to invert, we should take the separate inverses, but they go in the opposite

  • order.

  • So what goes here?

  • E three two inverse, right, because I'll multiply from the left by E three two inverse, then

  • I'll pop it up next to U.

  • And then will come E three one inverse.

  • And then this'll be the only guy left standing and that's gone when I do an E two one inverse.

  • So there is L.

  • That's L U.

  • L is product of inverses.

  • Now you still can ask why is this guy preferring inverses?

  • And let me explain why.

  • Let me explain why is this product nicer than this one?

  • This product turns out to be better than this one.

  • Let me take a typical case here.

  • Let me take a typical case.

  • So let me -- I have to do three by three for you to see the improvement.

  • Two by two, it was just one E, no problem.

  • But let me go up to this case.

  • Suppose my matrices E21 -- suppose E21 has a minus two in there.

  • Suppose that -- and now suppose -- oh, I'll even suppose E31 is the identity.

  • I'm going to make the point with just a couple of these.

  • Okay.

  • Now this guy will have -- do something -- now let's suppose minus five one.

  • Okay. There's typical.

  • That's a typical case in which we didn't need an E31. Maybe we already had a zero in that

  • three one position.

  • Okay.

  • Let me see -- is that going to be enough to, show my point?

  • Let me do that multiplication.

  • So if I do that multiplication it's like good practice to

  • multiply these matrices.

  • Tell me what's above the diagonal when I do this multiplication?

  • All zeroes. When I do this multiplication, I'm going to get ones on the diagonal and

  • zeroes above.

  • Because -- what does that say?

  • That says that I'm subtracting rows from lower rows.

  • So nothing is moving upwards as it did last time in Gauss-Jordan. Okay.

  • Now -- so really, what I have to do is check this minus two one zero, now this is -- what's

  • that number?

  • This is the number that I'm really have in mind.

  • That number is ten.

  • And this one is -- what goes here?

  • Row three against column two, it looks like the minus five.

  • Whatit's that ten.

  • How did that ten get in there?

  • I don't like that ten.

  • I mean -- of course, I don't want to erase it, because it's right.

  • But I don't want it there.

  • It's because -- the ten got in there because I subtracted two of row one from row two,

  • and then I subtracted five of that new row two from row three.

  • So doing it in that order, how did row one effect row three?

  • Well, it did, because two of it got removed from row two and then five of those got removed

  • from row three.

  • So altogether ten of row one got thrown into row three.

  • Now my point is in the reverse direction -- so now I can do it -- below it I'll do the inverses.

  • Okay.

  • And, of course, opposite order.

  • Reverse order.

  • Reverse order.

  • Okay. So now this is going to -- this is the E that goes on the left side.

  • Left of A.

  • Now I'm going to do the inverses in the opposite order, so what's the -- So the opposite order

  • means I put this inverse first.

  • And what is its inverse?

  • What's the inverse of E21? Same thing with a plus sign, right?

  • For the individual matrices, instead of taking away two I add back two of row one to row

  • two, so no problem.

  • And now, in reverse order, I want to invert that.

  • Just right?

  • I'm doing just this, this.

  • So now the inverse is again the same thing, but add in the five.

  • And now I'll do that multiplication and I'll get a happy result.

  • I hope.

  • Did I do it right so far?

  • Yes, okay.

  • Let me do the multiplication.

  • I believe this comes out.

  • So row one of the answer is one zero zero.

  • Oh, I know that all this is going to be left,

  • right? Then I have two one zero.

  • So I get two one zero there, right?

  • And what's the third row?

  • What's the third row in this product?

  • Just read it out to me, the third row?

  • 0 5 1

  • Because one way to say isthis is saying take one of the

  • last row and there it is.

  • And this is my matrix L.

  • And it's the one that goes on the left of U.

  • It goes into -- what do I mean here?

  • Maybe rather than saying left of A, left of U, let me right down again what I mean.

  • E A is U, whereas A is L U.

  • Okay.

  • Let me make the point now in words.

  • The order that the matrices come for L is the right order.

  • The two and the five don't sort of interfere to produce this ten one. In the

  • right order, the multipliers just sit in the matrix L.

  • That's the point -- that if I want to know L, I have no work to do.

  • I just keep a record of what those multipliers were, and that gives me L.

  • So I'll draw the -- let me say it.

  • So this is the A=L U.

  • So if no row exchanges, the multipliers that those numbers that we multiplied rows by and

  • subtracted, when we did an elimination step -- the multipliers go directly into L.

  • Okay.

  • So L is -- this is the way, to look at elimination.

  • You go through the elimination steps, and actually if you do it right, you can throw

  • away A as you create L U.

  • If you think about it, those steps of elimination, as when you've finished with row two of A,

  • you've created a new row two of U, which you have to save, and you've created the multipliers

  • that you used -- which you have to save, and then you can forget A.

  • So because it's all there in L and U.

  • So that's -- this moment is maybe the new insight in elimination that comes from matrix

  • -- doing it in matrix form.

  • So it was -- the product of Es is -- we can't see what that product of Es is.

  • The matrix E is not a particularly attractive one.

  • What's great is when we put them on the other side -- their inverses in the opposite order,

  • there the L comes out just right. Okay.

  • Now -- oh gosh, so today's a sort of,

  • like, practical day.

  • Can we think together how expensive is elimination?

  • How many operations do we do?

  • So this is now a kind of new topic which I didn't list as -- on the program, but here

  • it came. Here it comes.

  • How many operations on an n by n matrix A.

  • I mean, it's a very practical question.

  • Can we solve systems of order a thousand, in a second or a minute or a week?

  • Can we solve systems of order a million in a second or an hour or a week?

  • I mean, what's the -- if it's n by n, we often want to take n bigger.

  • I mean, we've put in more information.

  • We make the whole thing is more accurate for the bigger matrix.

  • But it's more expensive, too, and the question is how much more expensive?

  • If I have matrices of order a hundred.

  • Let's say a hundred by a hundred.

  • Let me take n to be a hundred.

  • Say n equal a hundred.

  • How many steps are we doing?

  • How many operations are we actually doing that we -- And let's suppose there aren't

  • any zeroes, because of course if a matrix has got a lot of zeroes in good places, we

  • don't have to do those operations, and,

  • it'll be much faster.

  • But -- so just think for a moment about the first step.

  • So here's our matrix A, hundred by a hundred.

  • And the first step will be -- that column, is got zeroes down

  • here. So it's down to 99 by 99, right?

  • That's really like the first stage of elimination,

  • to get from this hundred

  • by hundred non-zero matrix to this stage where the first pivot is sitting up here and the

  • first row's okay the first column is okay.

  • So, eventually -- how many steps did that take?

  • You see, I'm trying to get an idea.

  • Is the answer proportional to n?

  • Is the total number of steps in elimination, the total number, is it proportional to n

  • -- in which case if I double n from a hundred to two hundred -- does it take me twice as

  • long?

  • Does it square, so it would take me four times as long?

  • Does it cube so it would take me eight times as long?

  • Or is it n factorial, so it would take me a hundred times as long?

  • I think, you know, from a practical point of view, we have to have some idea of the

  • cost, here.

  • So these are the questions that I'm -- let me ask those questions again.

  • Is it proportional -- does it go like n, like n squared, like n cubed -- or some higher

  • power of n?

  • Like n factorial where every step up multiplies by a hundred and then by a hundred and one

  • and then by a hundred and two -- which is it?

  • Okay, so that's the only way I know to answer that is to think through what we actually

  • had to do.

  • Okay.

  • So what was the cost here?

  • Well, let's see.

  • What do I mean by an operation?

  • I guess I mean, well an addition or -- yeah.

  • No big deal.

  • I guess I mean an addition or a subtraction or a multiplication or a division.

  • Okay.

  • And actually, what operation I doing all the time?

  • When I multiply row one by multiplier L and I subtract from row six.

  • What's happening there individually?

  • What's going on?

  • If I multiply -- I do a multiplication by L and then a subtraction.

  • So I guess operation -- Can I count that for the moment as, like, one operation?

  • Or you may want to count them separately.

  • The typical operation is multiply plus a subtract.

  • So if I count those together, my answer's going to come out

  • half as many as if -- I mean, if I count them separately, I'd have a certain

  • number of multiplies, certain number of subtracts.

  • That's really want to do.

  • Okay. How many have I got here?

  • So, I think -- let's see.

  • It's about -- well, how many, roughly?

  • How many operations to get from here to here?

  • Well, maybe one way to look at it is all these numbers had to get changed.

  • The first row didn't get changed, but all the other rows got changed at this step.

  • So this step -- well, I guess maybe -- shall I say it cost about a hundred squared.

  • I mean, if I had changed the first row, then it would have been exactly hundred squared,

  • because -- because that's how many numbers are here.

  • A hundred squared numbers is the total count of the entry, and all but this insignificant

  • first row got changed.

  • So I would say about a hundred squared.

  • Okay.

  • Now, what about the next step?

  • So now the first row is fine.

  • The second row is fine.

  • And I'm changing these zeroes are all fine, so what's up with the second step?

  • And then you're with me.

  • Roughly, what's the cost?

  • If this first step cost a hundred squared, about, operations then this one, which is

  • really working on this guy to produce this, costs about what?

  • How many operations to fix?

  • About ninety-nine squared, or ninety-nine times ninety-eight. But less, right?

  • Less, because our problem's getting smaller.

  • About ninety-nine squared.

  • And then I go down and down and the next one will be ninety-eight squared, the next ninety-seven

  • squared and finally I'm down around one squared or -- where it's just like the little numbers.

  • The big numbers are here.

  • So the number of operations is about n squared plus that was n, right? n was a hundred?

  • n squared for the first step, then n minus one squared, then n minus two squared, finally

  • down to three squared and two squared and even one squared.

  • No way I should have written that -- squeezed that in.

  • Let me try it so the count is n squared plus n minus one squared plus -- all the way down

  • to one squared.

  • That's a pretty decent count.

  • Admittedly, we didn't catch every single tiny operation, but we got the right leading term

  • here.

  • And what do those add up to?

  • Okay, so now we're coming to the punch of this, question, this operation count.

  • So the operations on the left side, on the matrix A to finally get to U.

  • And anybody -- so which of these quantities is the right ballpark for that count?

  • If I add a hundred squared to ninety nine squared to ninety eight squared -- ninety

  • seven squared, all the way down to two squared then one squared, what have I got, about?

  • It's just one of these -- let's identify it first.

  • Is it n?

  • Certainly not.

  • Is it n factorial?

  • No.

  • If it was n factorial, we would -- with determinants, it is n factorial.

  • I'll put in a bad mark against determinants, because that -- okay, so what is it?

  • It's n -- well, this is the answer.

  • It's this order -- n cubed.

  • It's like I have n terms, right?

  • I've got n terms in this sum.

  • And the biggest one is n squared.

  • So the worst it could be would be n cubed, but it's not as bad as -- it's n cubed times

  • -- it's about one third of n cubed.

  • That's the magic operation count.

  • Somehow that one third takes account of the fact that the numbers are getting smaller.

  • If they weren't getting smaller, we would have n terms times n squared, but it would

  • be exactly n cubed.

  • But our numbers are getting smaller -- actually, row two and row one moves down to row three.

  • do you remember where does one third come in this -- I'll even allow a mention of calculus.

  • So calculus can be mentioned, integration can be mentioned now in the next minute and

  • not again for weeks.

  • It's not that I don't like 18.01, but18.06 is better.

  • Okay. So, -- so what's -- what's the calculus formula that looks like?

  • It looks like -- if we were in calculus instead of summing stuff, we would integrate.

  • So I would integrate x squared and I would get one third x

  • cubed. So if that was like an integral from one to n, of x squared b x, if the answer

  • would be one third n cubed -- and it's correct for the sum also, because that's, like, the

  • whole point of calculus.

  • The whole point of calculus is -- oh, I don't want to tell you the whole -- I mean, you

  • know the whole point of calculus.

  • Calculus is like sums except it's continuous.

  • Okay. And algebra is discreet.

  • Okay.

  • So the answer is one third n cubed.

  • Now I'll just -- let me say one more thing about operations.

  • What about the right-hand side?

  • This was what it cost on the left side.

  • This is on A.

  • Because this is A that we're working with.

  • But what's the cost on the extra column vector b that we're hanging around here?

  • So b costs a lot less, obviously, because it's just one column.

  • We carry it through elimination and then actually we do back substitution.

  • Let me just tell you the answer there.

  • It's n squared.

  • So the cost for every right hand side is n squared.

  • So let me -- I'll just fit that in here -- for the cost of b turns out to be n squared.

  • So you see if we have, as we often have, a a matrix A and several right-hand sides, then

  • we pay the price on A, the higher price on A to get it split up into L and U to do elimination

  • on A, but then we can process every right-hand side at low cost.

  • Okay.

  • So the -- We really have discussed the most fundamental algorithm for a system of equations.

  • Okay.

  • So, I'm ready to allow row exchanges.

  • I'm ready to allow -- now what happens to this whole -- today's lecture if there are

  • row exchanges?

  • When would there be row exchanges?

  • There are row -- we need to do row exchanges if a zero shows up in the pivot position.

  • So moving then into the final section of this chapter, which is about transposes -- well,

  • we've already seen some transposes, and -- the title of this section is,

  • "Transposes and Permutations."

  • Okay. So can I say, now, where does a permutation come in?

  • Let me talk a little about permutations.

  • So that'll be up here, permutations.

  • So these are the matrices that I need to do row exchanges.

  • And I may have to do two row exchanges.

  • Can you invent a matrix where I would have to do two row exchanges and then would come

  • out fine?

  • Yeah let's just, for the heck of it -- so I'll put it here.

  • Let me do three by threes.

  • Actually, why don't I just plain list all the three by three permutation matrices.

  • There're a nice little group of them.

  • What are all the matrices that exchange no rows at all?

  • Well, I'll include the identity.

  • So that's a permutation matrix that doesn't do anything.

  • Now what's the permutation matrix that exchanges -- what is P12? The permutation matrix that

  • exchanges rows one and two would be -- 0 1 0 -- 1 0 0, right.

  • I just exchanged those rows of the identity and I've got it.

  • Okay. Actually, I'll -- yes.

  • Let me clutter this up.

  • Okay. Give me a complete list of all the row exchange matrices.

  • So what are they?

  • They're all the ways I can take the identity matrix and rearrange its rows.

  • How many will there be?

  • How many three by three permutation matrices?

  • Shall we keep going and get the answer?

  • So tell me some more.

  • STUDENT: Zero one

  • STRANG: ZeroWhat one are you going to do now?

  • STUDENT: I'm going to switch the

  • STRANG: Switch rows one and -- One and three, okay. One and three, leaving two alone.

  • Okay.

  • Now what else? Switch -- what would be the next easy one -- is switch two and

  • three, good. So I'll leave one zero zero alone and I'll switch -- I'll move number three

  • up and number two down.

  • Okay. Those are the ones that just exchange single -- a pair of

  • rows. This guy, this guy and this guy exchanges a pair of rows, but now there are more possibilities.

  • What's left?

  • So tell -- there is another one here.

  • What's that?

  • It's going to move -- it's going to change all rows, right?

  • Where shall we put them?

  • So -- give me a first row.

  • STUDENT: Zero one zero?

  • STRANG: Zero one zero.

  • Okay, now a second row -- say zero zero one and the third guy

  • One zero zero.

  • So that is like a cycle.

  • That puts row two moves up to row one, row three moves up to

  • row two and row one moves down to row three.

  • And there's one more, which is -- let's see.

  • What's left?

  • I'm lost.

  • STUDENT: Is it zero zero one?

  • STRANG: Is it zero zero one? Okay.

  • STUDENT: One zero zero.

  • STRANG: One zero zero, okay.

  • Zero one zero, okay.

  • Great.

  • Six. Six of them.

  • Six P. And they're sort of nice, because what happens if I write, multiply two of them together?

  • If I multiply two of these matrices together, what can you tell me about the answer?

  • It's on the list.

  • If I do some row exchanges and then I do some more row exchanges, then all together I've

  • done row exchanges.

  • So if I multiply -- but, I don't know.

  • And if I invert, then I'm just doing row exchanges to get back again.

  • So the inverses are all there.

  • It's a little family of matrices that -- they've got their own -- if I multiply, I'm still

  • inside this group.

  • If I invert I'm inside this group -- actually, group is the right name for this subject.

  • It's a group of six matrices, and what about the inverses?

  • What's the inverse of this guy, for example?

  • What's the inverse -- if I exchange rows one and two, what's the inverse matrix?

  • Just tell me fast.

  • The inverse of that matrix is -- if I exchange rows one and two, then what I should do to

  • get back to where I started is the same thing.

  • So this thing is its own inverse.

  • That's probably its own inverse.

  • This is probably not -- actually, I think these are inverses of each other.

  • Oh, yeah, actually -- the inverse is the transpose.

  • There's a curious fact about permutations matrices, that the inverses are the transposes.

  • And final moment -- how many are there if I -- how many four by four permutations?

  • So let me take four by four -- how many Ps?

  • Well, okay.

  • Make a good guess.

  • Twenty four, right. Twenty four Ps.

  • Okay. So, we've got these permutation matrices, and in the next lecture, we'll use them.

  • So the next lecture, finishes Chapter 2

  • and moves to Chapter 3.

  • Thank you.

Okay, this is linear algebra, lecture four.

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B1 中級

Lec 4 Aへの因数分解 = LU (Lec 4 Factorization into A = LU)

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    Lin に公開 2021 年 01 月 14 日
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