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0:00OK, when the camera says, we'll start.
0:08You want to give me a signal? OK, this is lecture eight in linear
0:13algebra, and this is the lecture where we completely solve
0:19linear equations. So A X equal B.
0:25That's our goal. I- i- i- If it has a solution.
0:30It certainly can happen that there is no solution.
0:35We have to identify that possibility by elimination.
0:39And then if there is a solution we want to find out is there only one
0:43solution or are -- is there a whole family of solutions,
0:48and then find them all. OK.
0:51Can I use as an example the same matrix that I had last time when we
0:58were looking for the null space. So the, the matrix has rows one two
1:05two two, two four six eight, and the third row -- you remember
1:11the main point was the third row, three six eight ten, is the sum of
1:17row one plus row two. In other words,
1:22if I add those left-hand sides, I get the third left-hand side.
1:27So you can tell me right away what elimination is going to discover
1:32about the right-hand sides. What's -- there is a condition on B1,
1:37B2, and B3 for this system to have a solution.
1:43Most s- cases -- if I took these numbers to be one,
1:48five, and seventeen, there would not be a solution.
1:52In fact, if I took those first numbers to be one and five,
1:58w- w- what is the only B3 that would be OK?
2:03Six. If the left-hand --
2:06if the -- if these left-hand sides add up to that,
2:10then B -- I need B1 plus B2 to equal B3.
2:14Let's, let's just see how elimination discovers that.
2:18But, but we can see it coming, right?
2:20That, that if, if -- let me say it in other words.
2:25If some combination on the left-hand side gives all zeros then the same
2:30combination on the right-hand side must give zero.
2:34OK. So let's, uh,
2:38let me take that example and write down instead of copying out all the
2:43plus signs, let me write down the matrix.
2:48One two two two, two four six eight,
2:54and that six three eight ten, where the third row is the sum of
2:59the first two rows. Now how do we deal with
3:04the right-hand side? That's -- wi- we want to do the same
3:07thing to the right-hand side that we're doing to these rows on the
3:11left side, so we just tack on the right-hand
3:15side as another vector, another column.
3:19So this is the augmented matrix. It's, it's the matrix A with the
3:31vector B tacked on. In Matlab, that's all you
3:36would need to type. OK.
3:38So we do elimination on that. Can we just do elimination quickly?
3:42The first pivot is fine, I subtract two of this away from
3:48this, three of this away from this, so I have one two two two B1.
3:54Two of those away will give me zero zero two and four, and that
4:00was B2 minus two B1. I, I have to do the same thing to
4:05that third, that last column. And then three of these away from
4:11this gave me zero zero two four B3 minus three B1s.
4:17So that's the, that's elimination with the first
4:22column completed. We move on.
4:26There's the first pivot still. Here is the second pivot.
4:30We're always remembering, now, these are then going to be the
4:37pivot columns. And let me get the final result --
4:44well, let me -- can I, can I, do it by eraser?
4:53We're capable of subtracting this row from this row,
4:59just by -- that'll knock this out completely and give me the row of
5:05zeros, and on the right-hand side, when I subtract this away from this,
5:11what do I have? I think I have B3 minus a B2,
5:18and I had minus three B1s. This is going to, it's going
5:23to be a minus a B1. Oh yethat's exactly
5:27what I expect. So now the -- what's
5:31the last equation? The last equation,
5:35th- th- th- this represented by this zero row, that last equation is,
5:40says zero equals B3 minus B2 minus B1.
5:45So that's the condition for solvability.
5:50That's the condition on the right-hand side that we expected.
5:53It says that B1 plus B2 has to match B3, and if our numbers happen to
5:58have been one, five, and six --
6:02so let me take, suppose B is one five six.
6:06That's an OK B. And when I do this elimination,
6:10what will I have? The B1 will still be a one.
6:14B2 would be five minus two, this would be a three.
6:19Five -- my six minus five minus one, this will be -- this is the main
6:23point -- this will be a zero, thanks.
6:27OK. So the last equation is OK now.
6:35And I can proceed to solve the two equations that are really there with
6:40four unknowns. OK, I, I, I want to do that,
6:45so this, this B is OK. It allows a solution.
6:50We're going to be, naturally, interested to keep track
6:56what are the conditions on B that, that, are re- that, that make
7:02the equation solvable. So let me put it,
7:07let, let me write down what we already see before I continue
7:11to solve it. Let me first -- solvability,
7:17solvability. So which --
7:25so this is the condition on the right-hand sides.
7:32And what is that condition? This is solvability always
7:36of A X equal B. So A X equal B is solvable --
7:42well, actually, we had an answer in the language of the column space.
7:51Can you remind me what that answer is?
7:53That, that was like our answer from earlier lecture.
7:56B had to be in the column space. Solvable if --
8:03when -- exactly when B is in the column space of A.
8:09Right? That just says that B has to be a
8:15combination of the columns, and of course that's exactly what
8:19the equation is looking for. So that, that --
8:23now I want to answer it -- the same answer but in different
8:27language. U- u- u- s- another way to answer
8:34this, if, if some, if a combination of the rows of A
8:45gives the zero row, if -- a- and this was an example
8:53where it happened, some combination of the rows of A
8:58produced the zero row -- then what's the requirement on B?
9:02That -- since we're going to do the same thing to both sides of all
9:07equations, the same combination of the
9:13components of B has to give zero. Right?
9:16That's -- so if there's a combination of the rows that gives
9:24the zero row, then the same combination of the
9:32entries of B must give zero. And this isn't the zero row,
9:38that's the zero number. OK.
9:42The, th- th- this is another way of saying -- a- and it is not immediate,
9:47right, that these two statements are equivalent.
9:53But somehow they must be, because they're both equivalent to
9:56the solvability of the system. OK.
9:59So we've got this, this sort of -- like question zero
10:04is, does the system have a solution? OK, I'll come back to
10:09discuss that further. Let's go forward when it does.
10:16When there is a solution. And so what's our job now?
10:21Abstractly we sit back and we say, OK, there's a solution, finished.
10:25It exists. But we want to construct it.
10:28So what's the algorithm, the str- the, the sequence of steps
10:33to find the solution? That's what I --
10:38and, and of course the quiz and the final, I'm going to give you a
10:42system A X equal B and I'm going to ask you for the solution,
10:45i- if there is one. And s- so,
10:49so this algorithm that you want to follow.
10:53OK, let's see. So what's the --
11:01so now to find the complete solution to A X equal B.
11:11OK. Let me start by finding one solution,
11:16one particular solution. I'm expecting that I can,
11:23because my system of equations now, that last equation is zero equals
11:29zero, so that's all fine. I really have two equations --
11:35actually I've got four unknowns, so I'm expecting to find not only a
11:40solution but a whole bunch of them. But let's just find one.
11:45So step one, a particular solution, X particular.
11:52How do I find one particular solution?
11:56Well, let me tell you how I, how I find it.
12:00So this is -- since, since there are lots of
12:02solutions, you could have your own way to find
12:05a particular one. But thi- this is a pretty
12:08natural way. Set all free variables to zero.
12:19Since those free variables are the guys that can be anything,
12:25the most convenient choice is zero. And then solve A X equals B for the
12:33pivot variables. So what does that mean
12:41in this example? Which are the free variables?
12:45Which, which are the variables that we can assign freely and then,
12:49and then there's one and only one way to find the pivot variables?
12:53They're X2 and -- so X2 is zero, because that's in a
12:58column without a pivot, the second column has no pivot.
13:03And the -- what's the other one? The fourth, X4 is zero.
13:10Because that, those are the, the free ones.
13:14Those are in the columns with no pivots.
13:18So you see what my -- so when I knock --
13:22when X2 and X4 are zero, I'm left with the -- th- w- what am
13:27I left with here? I'm just left with --
13:32see, now I'm not using the two free columns.
13:36I'm only using the pivot columns. So I'm really left with X1 --
13:41the first equation is just X1 and two X3s should be the right-hand
13:47side, which we picked to be a one. And the second equation is two X3s,
13:53as it happened, turned out to be, uh, three.
14:00I bd- I just write it again here with the X2 and the X4 knocked out,
14:05since we're set them to zero. And you see that we- we're back in
14:10the n- normal case of having back -- where back substitution'll do it.
14:15So X3 is three halves, and then we go back up and X1 is one
14:21minus two X3. That's probably minus two.
14:27Good. So now we have the solution,
14:33X particular is the vector minus two zero three halves zero.
14:42OK, good. That's one particular solution,
14:48and we should and could plug it into the original system.
14:53Th- really if -- on the quiz, please,
14:56it's a good thing to do. Just a- so we did all this,
15:00these, row operations, but this is supposed to solve the original
15:05system, and I think it does. OK.
15:09So that's X particular which we've got.
15:13So i- i- i- i- that's like what's new today.
15:18The particular solution comes -- first you check that you have zero
15:22equals zero, so you're OK on the last equations.
15:27And then you set the free variables to zero, solve for the pivot
15:31variables, and you've got a particular solution,
15:35the particular solution that has zero free variables.
15:40OK. Now -- but that's only one solution,
15:44and now I'm looking for all. So how do I find the rest?
15:50The point is I can add on to -- add on X -- anything out of the null
15:58space. We know how to find the vectors in
16:03the null space, because we did it last time,
16:06but I'll rema- remind you what we got.
16:09And then I'll add. So the final result will be that the
16:17complete solution -- this is now the complete guy --
16:23the complete solution is this one particular solution plus any,
16:29any vector, all different vectors out of the null space.
16:35X N, OK. Well why, why this pattern,
16:40because this pattern shows up through all of mathematics,
16:44because it shows up everywhere we have linear equations.
16:47Let me just put here the, the reason.
16:50A X P, so that's X particular, so what does A X particular give?
17:00That gives the correct right-hand side B.
17:04And what does A times an X in the null space give?
17:09Zero. So I add, and I put in parentheses.
17:17So X P plus X N is B plus zero, which is B.
17:24So -- oh, what am I saying? Let me just say it in words.
17:29If I have one solution, one solution, I can add on anything
17:33in the null space, because anything in the null space
17:38has a zero right-hand side, and, and I still have the correct
17:43right-hand side B. So that's my system.
17:47That's my complete solution. Now let me write out what that will
17:51be for this example. So, so this example,
17:56in this example, X per- X general, X complete, the complete solution,
18:02is X particular, which is minus two zero three halves
18:08zero, with those zeroes in the free variable, plus --
18:15you remember there were the special solutions in the null space that had
18:18a one in the free variables -- or, or, one and zero in the free
18:21variables, and then we filled in to find the
18:24others? I've forgotten what they were,
18:28but maybe it was that. That was a special solution,
18:33and then there was another special solution that had that free variable
18:39zero and this free variable equal one, and I have to fill those in.
18:44Let's see, can I remember how those fill in?
18:47Maybe this was a minus two and this was a two, possibly?
18:52I think probably that's right. I'm not -- yeah.
18:58Two min- hmm. Does that look write to you?
19:04I would have to remember what are my equations.
19:07Ca- uh- can I, rather than go way back to that
19:11board, let me remember the first equation was two X3 plus two X4
19:15equaling zero now, because I'm looking for the guys in
19:19the null space. So I set X4 to be one and the second
19:24equation, that I didn't copy again, gave me minus two for this and then --
19:30yeso I think that's right. Two minus four and two
19:36gives zero, check. OK.
19:40Those were the special solutions. What do we do to get the
19:45complete solution? How do I get the complete
19:49solution now? I multiply this by anything,
19:54C1, say, and I multiply this by anything -- I take any combination.
19:59Y- remember that's how we described the null space?
20:03The null space consists of all combinations of --
20:08so this is X N -- all combinations of the special solutions.
20:14There were two special solutions because there were two
20:17free variables. And we want to make that count,
20:22com- se- carefully now. B- just while I'm up here.
20:27So there's, that's what the --
20:28that's the kind of answer I'm looking for.
20:30Is there a constant multiplying this guy?
20:33Is there a free constant that multiplies X particular?
20:37No way. Right?
20:40X particular solves A X P equal B. I'm not allowed to multiply
20:45that by three. But A X N,
20:48I'm allowed to multiply X N by three, or add to another X N,
20:51because I keep getting zero on the right.
20:55OK. So, so again, X P is
20:58one particular guy. X N is a whole subspace.
21:03Right? It's one guy plus, plus anything
21:06from a subspace. Let me draw it.
21:09Let me, let me try to -- oh. I want to draw,
21:16I want to graph all this -- graph,-- I want to, I want to
21:21plot all solutions. Now X.
21:28So what dimension am I in? This is a unfortunate point.
21:34How many components does X have? Four.
21:37There are four unknowns. So I have to draw a four dimensional
21:42picture on this MIT cheap blackboard.
21:47OK. So here we go.
21:50X1 -- Einstein could do it, but, this, this is -- th- those
21:58are four perpendicular axes in -- representing four dimensional space.
22:07OK. Where are my solutions?
22:11Do my solutions form a subspace? Duh- duh- does the set of solutions
22:17to A X equal B form a subspace? No way.
22:20What does it actually look like, though?
22:23Ther- a subspace is in this picture. This part is a subspace, right?
22:29That part is some, like, two dimensional,
22:33because I've got two parameters, so it's -- I- I'm thinking of this
22:37null space as a two dimensional subspace inside R4.
22:42Now I have to tell you and will tell you pr- next time,
22:46what does it mean to say a subspace, what's the dimension of a subspace.
22:49But you see what it's going to be. It's the number of free independent
22:55constants that we can choose. So somehow there'll be a two
23:00dimensional subspace, not a line, and not a three
23:05dimensional plane, but only a two dimensional guy.
23:09But it's doesn't go through the origin because it goes
23:13through this point. So there's X particular.
23:16X particular is somewhere here. X particular.
23:20So it's somehow a subspace -- can I try to draw it that way?
23:27It's a two dimensional subspace that goes through X particular and then
23:33onwards by -- so there's X particular,
23:39and I added on X N, and there's X. There's X equal X P plus X N.
23:47But the X N was anywhere in this subspace, so that filled
23:52out a plane. It's, it's a,
23:56it's a subspace -- it's not a subspace,
23:59what am I saying? It's, it's,
24:02it's like a flat thing, it's like a subspace, but it's been
24:05shifted, away from the origin. It doesn't contain zero.
24:12That's the picture, and that's the algorithm.
24:15So the algorithm is just go through elimination and,
24:20find the particular solution, and then find those special
24:24solutions. Y- y- y- you can do that.
24:28Let me take our time here in the lecture to think, about
24:34the bigger picture. So let me think about,
24:41uh -- so this, this is my pattern. Now I want to think --
24:47I want to ask you, uh, about a question -- I, I want to ask
24:53you some questions. So when I mean think bigger,
25:01I mean I'll think about an M by N matrix A of rank R.
25:11OK. What's our definition of rank?
25:16Our current definition of rank is number of pivots.
25:21OK. First of all, how are these
25:24numbers related? Can you tell me a relation
25:27between R and M? If I have M rows in the matrix and R
25:33pivots, then I certainly know, I know, always, what, what relation
25:41do I know between R and M? R is less or equal, right?
25:48Because I've got M rows, I can't have more than M pivots,
25:52I might have M and I might have fewer.
25:55Also, I've got N columns. So what's the relation
26:02between R and N? It's the same,
26:06less or equal, because a column can't have more than one pivot.
26:12So I can't have more than N pivots altogether.
26:16OK, OK. So I have an M by N
26:19matrix of rank R. And I always know R less than or
26:22equal to M, R less than or equal to N.
26:25Now I'm specially interested in the case of full rank,
26:30when the rank R is as big as it can be.
26:34Well, I guess I've got two separate possibilities here,
26:39depending on what these numbers M and N are.
26:43So let me talk about the case of full column rank.
26:51And by that I mean -- means R equal N.
27:00And I want to ask you, what's, what does that imply about
27:07our solutions? What does that tell us
27:12about the null space? What does that tell us about,
27:16the, the gen- the complete solution?
27:20OK, so what does that mean? So I, I want to ask you,
27:25well, OK, if the rank is N, what does that mean?
27:31That means there's a pivot in every column.
27:34So how many pivot variables are there?
27:39N. All the columns have
27:42pivots in this case. So how many free variables
27:45are there? None at all.
27:49So no free variables. R equal N, no free variables.
27:56So what does that tell us about what's going to happen then in our,
28:00in our little algorithms? What will, what will be
28:04in the null space? The null space of A
28:09has got what in it? Only the zero vector.
28:14There are no free variables to give other values to.
28:19So the null space is only the zero vector.
28:28And what about our solution to A X equal B?
28:32Solution to A X equal B? What, what's the story on that one?
28:40So now that's coming from today's lecture.
28:45The solution X is -- what's the complete solution?
28:54It's just X particular, right? If, if, if there is an X,
29:00if there is a solution. It's X equal X particular.
29:04There's nothing -- y- you know, there's just one
29:06solution. I- If there's one at all.
29:10So it's unique solution -- unique means only one -- unique
29:20solution if it exists, if it exists. In other words,
29:26I would say -- let me put it a different way.
29:29There're either zero or one solutions.
29:37This is all in this case R equal N. So I'm -- because many,
29:46many applications in reality, the columns will be, will --
29:51w- w- will be what I'll l- later call independent.
29:57And we'll have, nothing to look for in the null
30:01space, and we'll only have particular solutions.
30:06OK. Everybody see that po- possibility?
30:11I- i- i- but I need an example, right?
30:14So let me create an example. What sort of a matrix --
30:19what's the shape of a matrix that has full column rank?
30:24So can I squeeze in a- an, an example here?
30:28If it exists. Let me put in an example,
30:35and a- it's just the right space to put in an example.
30:39Because the example will be like tall and thin.
30:44It will have -- well, I mean, here's an example,
30:51one two six five, three one one one. Brilliant example.
30:55OK. So there's a matrix A,
31:00and what's its rank? What's the rank of that matrix?
31:08How many pivots will I find if I do elimination?
31:13Two, right? Two.
31:15I see a pivot there -- oh- i- i- i- certainly those two
31:20columns are headed off in different directions.
31:25When I do elimination, I'll certainly get another pivot
31:28here, fine, and I can use those to clean out u-
31:32u- below and above. So the -- actually,
31:37tell me what its row reduced -- reduced row echelon form would be.
31:44Can you carry that, that elimination process to the
31:48bitter end? So what do, what does that mean?
31:53I subtract a multiple of this row from these rows.
31:57So I clean up, all zeros there. Then I've got some pivot here.
32:01What do I do with that? I go subtract it below and above,
32:05and then I divide through, and what's the, w- what's
32:09R for that example? Maybe I can --
32:12you'll allow me to put that just here in the next board.
32:15What's the row reduced echelon form, just out of practice, for that
32:20matrix? It's got ones in the pivots.
32:28It's got the identity matrix, a little two by two identity matrix,
32:33and below it all zeros. That's a matrix that really has two
32:38independent rows, and they're the first two,
32:42actually. The first two rows are independent.
32:46They're not in the same direction. But the other rows are combinations
32:50of the first two, so -- so, u- u- uh,
32:54is there always a solution to A X equal B?
32:59Du- tell me what's the picture here? For this matrix A,
33:03this is a case of full column rank. The two columns are
33:08-- give two pivots. There's nothing in the null space.
33:12There's no combination of those columns that gives the zero column
33:18except the zero zero combination. So there's nothing
33:22in the null space. But is there always a solution
33:26to A X equal B? What's up with A X equal B?
33:32I've got four, four equations here,
33:35and only two Xs. So the answer is certainly no.
33:41There's not always a solution. If ther- so I,
33:45I may have zero solutions, and if I make a random choice,
33:49I'll have zero solutions. Or if I make a great particular
33:52choice of the right-hand side, which just happens to be a
33:56combination of those two guys -- like tell me one right-hand side
34:00that would have a solution. Tell me a right-hand side that would
34:04have a solution. Well, zero zero zero zero, OK.
34:09No prize for that one. Tell me another one.
34:13Another right-hand side that has a solution would be four
34:16three seven six. I could add the two columns.
34:20Right? What would be the total complete
34:23solution if the right-hand side was four three seven six?
34:27There would be the particular solution one one,
34:30one of that column plus one of that, and that would be the only
34:34solution. So there would be --
34:38X particular would be one one in the case when the right side is the sum
34:43of those two columns, and that's it. So that would be a case
34:47with one solution. OK.
34:50That, this is the typical setup with full column rank.
34:54Now I go to full row rank. You see the sort of natural symmetry
35:00of this discussion. Full row rank means R equal M.
35:15So this is what I'm interested in now, R equal M.
35:21OK, what's up with that? How many pivots?
35:30M. So what happens when we do
35:35elimination in that case? I'm going to get M pivots.
35:41So every row has a pivot, right? Every row has a pivot.
35:50Then what about solvability? What about this business of --
35:56for which right-hand sides can I solve it?
36:00So that's my question. I can solve A X equal B for which
36:07right-hand sides? For -- do you see what's coming?
36:17I do elimination, I don't get any zero rows.
36:23So there aren't any requirements on B.
36:26I can solve A X equ- equal B for every B.
36:34I can solve A X equal B for every right-hand side.
36:39So this is the existence, exists a solution.
36:47Now tell me, so the, u- u- so every row has a
36:53pivot in it. So how many free variables
36:57are there? How many free variables
37:01in this case? If I had N variables a- to start
37:05with, how many are used up by pivot variables?
37:10R, which is M. So I'm left with,
37:21left with N minus R free variables. OK.
37:31So this case of full row rank I can always solve,
37:35and then this tells me how many variables are free,
37:40and this is of course N minus M. This is N minus M free variables.
37:47Can I do an example? You know, the best way for me to do
37:52an example is just to transpose that example.
37:56So let me take, let me take that matrix that had
38:00column one two six five and make it a row.
38:04And let me take three one one one as the second row.
38:10And let me ask you, this is my matrix A,
38:14what's its rank? What's the rank of that matrix?
38:19Sorry to ask, but not sorry really,
38:23because, because we're just getting the idea of rank.
38:26What's the rank of that matrix? Two, exactly, two.
38:29There will be two pivots. What will the row reduced
38:34echelon form be? Anybody know that one?
38:37Actually, tell me not only -- you have to tell me not only the,
38:41there'll be two pivots but which will be the pivot columns.
38:45Which columns of this matrix will be pivot columns?
38:49So the first column is fine, and then I go on to the next column,
38:53and what do I get? Do I get a second pivot out of,
38:56do- will I get a second pivot in this position?
38:59Yes. So the pivots,
39:02when I get all the way to R, will be there.
39:05And here will be some numbers. I, a- a- a- a- this is the part that
39:14I previously called F. This is the part that --
39:19so the, the here -- the pivot columns in R will be
39:22the identity matrix. There are no zero rows,
39:26no zero rows, because the rank is two.
39:30But there'll be stuff over here. And that will,
39:35enter the special solutions and the null space.
39:41OK. So this is a typical m- matrix with
39:46R equal M smaller than N. Now finally I've got a space here
39:55for R equal M equal N. I'm off in the corner here with the
40:02most important case of all. So what's up with this matrix?
40:07So let me give an example. OK, brilliant example,
40:14one two three one. Tell me what --
40:20how do I describe a matrix that has rank R equal M equal N?
40:26So the matrix is square, right, it's a square matrix.
40:31And if I know its rank is -- it's full rank, now.
40:35I don't have to say full column rank or full row rank,
40:39I just say full rank, because the count, column count and
40:43the row count are the same, and the rank is as big as it can be.
40:48And what kind of a matrix have I got?
40:52It's invertible. So that's exactly the
40:57invertible matrices. R equal M equal N means the --
41:02wh- and what's the row echelon form, the, the reduced row echelon form,
41:07for an invertible matrix? For a square, nice, square,
41:11invertible matrix? It's I.
41:15Right. So i- i- i- i- i- i- y- you see that
41:21the, the good matrices are the ones that kind of come out
41:27trivially in R. You reduce them all the way to the
41:31identity matrix. What's the null space for
41:35this, for this matrix? Can I just hammer away with
41:38m- m- m- questions? What's the null space
41:42for this matrix? The null space of that matrix is the
41:47zero vector only. The zero vector only.
41:53What are the conditions to solve A X equal B?
41:57Which right-hand sides B are OK? If I want to solve A X equal B for
42:03this example, so I -- A is this,
42:08B is B1 B2, what are the conditions on B1 and B2?
42:13None at all, right. So this is the case,
42:18this is the case where I can solve -- so I've coming back here, I can --
42:22u- u- since the rank equals M, I can solve for every B.
42:27And since the rank is also N, there's a unique solution.
42:32Let me summarize the whole picture here.
42:37Here's the whole picture. I could have R equal M equal N.
42:44This is the case where this is the identity matrix.
42:49And this is the case where there is one solution.
42:55That's the square invertible chapter two case.
43:00Now we're into chapter three. We could have R equal
43:06M smaller than N. Now that's what we had over there,
43:13and the row echelon form looked like the identity with some zero rows.
43:20And that was the case where there are zero or one solution.
43:29To -- I'm a -- when I say solution I mean to A X
43:33equal B. So this case, there's always one.
43:39This case there's zero or one. And now let me take the case of full
43:45column rank, but some, extra rows.
43:54So now R has -- well, the identity --
44:02i- i- i- i- I'm almost tempted to write the identity matrix and then F,
44:11but that isn't necessarily right. I have, I have -- is that right?
44:18Am I -- sorry, am I, am I getting this correct here?
44:22R -- oh, I'm s- I'm not! My God!
44:25This is the case R equals N, the columns, the columns are,
44:31are OK. That's the case that was on that
44:35board, R equal N, full column rank. Now I want the case where M is
44:40smaller than N and I've got extra columns.
44:45OK. There we go.
44:51So this is now the case of full row rank, and it looks like I F except
44:58that th- th- th- th- I can't be sure that the pivot columns
45:04are the first columns. So the I and the F,
45:09the F could be partly mixed into the I.
45:12Can I, can I write that with just like that?
45:17So the, the, the F could be sort of partly i- i-
45:21into the I if, if, if the first columns weren't the
45:25pivot columns. Now how many solutions in this case?
45:30There's always a solution. This is the existence case.
45:35There's always a solution. We're not getting any zero rows.
45:38There are no zero rows here. So there's always either one or
45:44infinitely many solutions. OK.
45:51Actually, I guess there's always an infinite number,
45:56because we always have some null space to deal with.
46:02Then the final case is where R is smaller than M and smaller than N.
46:07OK. Now that's the case where R is the
46:12identity with some free stuff but ther- with some zero rows too.
46:18And that's the case where there's either no solution,
46:25because we didn't get a zero equals zero for some Bs, or infinitely
46:32many solutions. OK.
46:37Do you, d- u- u- u- w- th- this board really summarizes the lecture,
46:42and this sentence summarizes the lecture.
46:46The rank tells you everything about the number of solutions.
46:53The- that number, the rank R, tells you all the
46:57information except the exact entries in the solutions.
47:01For that you go to the matrix. OK, good.
47:04Have a great weekend, and I'll see you on Monday.