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Lecture Comments (13)

1 answer

Last reply by: Professor Hovasapian
Fri Oct 24, 2014 10:16 PM

Post by David Llewellyn on October 16, 2014

In the last example in the Homogeneous System section, does the fact that the variables x & z and y & w are equivalent reduce it to a system with only two unknowns and would there be any way, not necessarily for homogeneous systems, where you could tell that variables (not the value the variables take in a particular solution) are the same?

1 answer

Last reply by: Professor Hovasapian
Fri Jun 27, 2014 5:56 PM

Post by Abot Bol on June 26, 2014

Hello Hovasapian
in example 4 of linear systems  i got -7 6 and 1 in the last column  however your solution has shown 0 0 1 . Am i wrong or is also ok?

1 answer

Last reply by: Professor Hovasapian
Tue Aug 27, 2013 9:04 PM

Post by Christian Fischer on August 27, 2013

Hi professor, Thank you for the great videos. At the end of this video you wrote "A homogeneous system always has a trivial solution", So if we have the equation from your example

Does a trivial solution then mean "If I do substitution within the three equations (so isolate x in the first equation and insert in the next) then I will always get X=0, Y=0 and Z=0??

And the only situations where homogeneous equations do have one or infinite solutions is when there are more variables than equations.

Thank you and have a great day.

1 answer

Last reply by: Professor Hovasapian
Mon Jul 23, 2012 6:09 PM

Post by Winnie So on July 22, 2012

I think that x1 should be equal to -2r + 3s + t

4 answers

Last reply by: Professor Hovasapian
Fri Oct 11, 2013 8:21 PM

Post by Jason Mannion on October 11, 2011

there is a mistake in this video, in example 1.
the last row has (0,-5,-2,17), but it should be (o,-4,0,12).

Solutions of Linear Systems, Part II

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Solutions of Linear Systems 0:11
    • Solutions of Linear Systems
  • Example I 3:25
    • Solve the Linear System 1
    • Solve the Linear System 2
  • Example II 17:41
    • Solve the Linear System 3
    • Solve the Linear System 4
  • Homogeneous Systems 21:54
    • Homogeneous Systems Overview
    • Theorem and Example

Transcription: Solutions of Linear Systems, Part II

Welcome back to and thank you for joining use for linear algebra.0000

Today we are going to be discussing solutions of linear systems part-2, okay let's go ahead and get started.0005

We closed off the last session by converting the following matrix...0013

... And you will remember of course sometimes I actually leave off the little brackets on the side of the matrices, simply because it's a personal notational preference that's all, as long as you understand which grouping of numbers goes together okay.0020

We had 91, 2, 3, 9, 2, -1, 1, 8) and (3, 0, -1, 3) and we converted that to reduced row echelon form,0034

(2, 1, 0, 02,), (0, 1, 0, -1) and (0, 0, 1, 3), and notice the reduced row echelon form, it has 1 as leading entries and wherever the row has a 1, that is a leading entry.0052

Everything else in that column is a 0, this column of ‘course doesn't matter, because it's the third row here, so it is irrelevant as far as the definition of reduced row echelon is concerned, okay.0072

Now, we did that just for matrices, well we already know that a given matrix represents a linear system of equations, so this system looks like this in terms of the variables....0087

... X + 2Y +3Z = 9, notice X + 2Y + 3Z = 9, and then 2X - Y + Z = 8.0104

And then 3X + 0Y, you don't necessarily have to put this there, I like to put it there simply because for me it keeps thing in order, in line and it just keeps things consistent.0123

-Z = 3 and again what we have done is...0136

... Whenever we transform a matrix from the standard matrix to its reduced row echelon form, we form an equivalent system, so what we produce is the following, this time I am going to actually avoid the 0's.0144

We have X = 2, Y = -1 and Z = 3, so this is our solution.0156

Matrix represents the linear system, converted to reduced row echelon form, we get, that's one of the reasons why we like these 1's in these positions and 0's everywhere else.0172

It's because it just gives you the answer, X is one thing, Y is another and Z is another okay.0182

And in this particular case we have a solution, so it is a unique solution, for this particular system, for this particular, well, for this particular system, there is no other, it is unique, okay.0189

Let's take a look at another example, let's actually go ahead and exquisitely give the linear system this time, and then we will just deal with the matrix, so we have, let me, let me write this system up here.0206

We have X + Y + 2Z - 5W = 3, we have 2X + 5Y - Z -9W = -3, we have 2X again.0220

+ Y - Z + 3W = -11, and we have X - 3Y, excuse me, + 2Z + 7W = -5.0245

And again I think the biggest problem that you are going to run across with linear lagebra, working with matrices is just keeping everything in order, you have a bunch of letters floating around, you have a bunch of numbers floating around.0264

The biggest problem you really going to run into is, believe it or not, just planar arithmetic, keeping everything straight.0274

Okay let's go ahead and just take the coefficients, make it a matrix augmented with this, you know series of numbers right here, turn it into reduced echelon form, and we will have our solution and see what it is.0280

Our matrix looks like this, we just take the coefficients, we have (1, 1, 2, -5) and we have 3, let me go ahead and put a little, just to show you that these are the coefficients of the variables and these are the solution set.0292

We have (2, 5, -1, -9), we have (2, 1, -1, 3) and we have (1, -3, 2 and 7) and then (-3, -11, -5), (-3, -11, -5).0311

This matrix is the one is the one that we want to convert into reduced row echelon form, okay.0330

We notice, remember our process, we go ahead and we find the first row that actually, or the first column that has non-zero entries, which is this one, and then we find the first row that has a non-zero entry, which is up here.0339

this becomes our pivot, once we find our pivot, we divide everything in that row by that number, to turn it into a 1.0352

But in this case it's already a 1, so it's not a problem, so in order to get rid of, the next step is to get rid of this number, this number, this number, to turn everything in that column into a 0, and of ‘course we do that by.0358

We are going to multiply the first equation by -2 added to the second row, and then we are going to multiply the first row by -2 added to this row, and then the third thing we're are going to do is multiply this by -1, and add it to this row.0371

And when we do that, we will have converted this column to 0, and then we move over on to the next column, if you want to check up the process, you can go back to the next lesson and review it, what we did it very, carefully.0386

Each individual pivot, each sub matrix and so forth, I'll go ahead and do just the first conversion, once I have done these three, we end up with the following matrix, of ‘course the original one doesn't change.0398

We have (1, 1, 2, -5, 3) we have (0, 3, -5, 1, -9), (0, -1, -5, 13, -17), and we have (0, -5, -2, 17).0412

I think that's 17, yes and -11, so these are all 0's and now I submit this is the entire matrix, our first set of conversions.0437

Now we notice this, so we can leave that alone, now we just deal with this, so this is al 0's we move to this column, so this particular column, 3 happens to be the first number that we run across going down.0448

This becomes our pivot, so our next step would be to divide this entire row by 3, to turn this into a 1q, and then do the same process.0463

Multiply this by 1, add it to this row, multiply this row by 5, add it to this row, and so on until we get reduced row echelon form.0471

Now I am going to go ahead and skip the process, and I am just going to give you the reduced row echelon form, and then I am going to take a couple of minutes to talk about how it is that we do that and the use of mathematical software to make this much easier for you.0480

What we end up with is the following matrix, let me just draw a little arrow here, and put in RRE, our reduce row echelon form.0493

What you end up with is (1, 0, 0, 2, -5, 0, 1, 0, -3, 2) excuse me for this little marks that show up, i don't want to confuse the numbers here.0502

And then we have (0, 0, 1, -2, 3) and the final row is going to be a row of all 0's which is fine, it doesn't matter.0522

Here is our reduced row echelon, notice (1, 1, 1,), these are all entries, 0's, the columns, so this is what we end up with.0533

Now I want to take a couple of minutes to talk to you about the use of mathematical software.0544

The techniques for this, what's this called, George, Gaussian elimination, excuse me, turning it into reduced row echelon form, dealing with the matrices.0550

A lot of the techniques that we are going to be developing are going to be computationally intensive, so it's not a problem if you want to do them by hand, I think it’s certainly a good way to get comfortable with the material, but at some point when the matrices start to get big, even like a 3 by 3 or 4 by 4.0560

You definitely want to start using mathematical software to make your life easier.0575

As long as you understand the process originally, it's not a problem after that, now there are several math software’s that are available, for example one of them is maple.0579

It's one of the oldest, it's the one that I use personally, it's the one that I was trained by, there are something called Mathcad, it's very popular in the engineering field.0590

There is something called Mathematica, also a very powerful and these are all symbolic manipulation software, not altogether different than what's available on ATI 89, so if you have a ATI 89, you can also do matrices on there, because it handles symbolically as supposed to numerically like older calculators used to do.0601

And there's also something specific called mat lab and it stands for matrix laboratory, this is specifically for linear algebra, it's also very powerful programming language.0621

Anyone of these is fine, very simple to use, you'll, you just plug in the matrix, and then you have some pull down menus or some commands, and you say reduced row echelon or find the inverse, find the transpose.0633

You can go from here to here, without having to go through all the intense computation, let the computer do that for you, it’s the math that's important, not necessarily the computation, especially since like I said before.0645

Arithmetic makes mistakes when you are dealing with 100s and 100's of numbers, they are going to happen, so if that I throw that out there, feel free to make use of any of these software and they are reasonable inexpensive.0659

Okay, so let's get back to our solution, so we have this final reduced row echelon form here, and what is actually, let me actually go to another page and rewrite it.0673

I have (1, 0, 0, 2, -5, 0, 1, 0, -3, 2, 0, 0, 1, -2, 3, 0, 0, 0, 0, 0) , by the way when you are doing entries for matrix, notice I tend to go in rows.0687

You can also go in columns, what you don't want to do is just sort of start going randomly, putting numbers here and there to fill the main.0709

believe it or not, that actually creates problem, so be as systematic as possible in all things you do mathematically.0715

This is the matrix, it's equivalent to the following system, well this is X, so we have X + 2 and let's say that this variable is W, so we have X, we have Y, we have Z, we have W and we have the solutions right here.0721

W have X + 2W = -5, we have Y...0737

... -3W = 2, we have the Z - 2W = 3, again it's in the W column, these are the solutions.0747

Now, notice what we have, this system right here, we have an X Y and Z, and in each case W shows up, so W becomes a free parameter.0760

That means I can choose any value I want for W, and the I can solve for X, Y and Z, so what we here is...0770

... Infinite solutions...0779

... You can stop here if you want, let me go ahead and show you what this looks like explicitly in terms of X, Y, Z, in other words I am going to solve for X and Y and Z, because W is a free parameter.0783

I am going to give it, I am going to put it down here, I am just going to call it R, you can give it anything you want, okay.0793

XZ is going to equal to -5...0803

... -2R, Y is going to equal 2 + 3R and Z = 3 + 2R.0809

And these are explicit representations with all of the variables on one side and everything else on the other, this is an implicitly, again it's an implicit relation because you can always solve it for one of the variables, that's all implicit means.0825

these are your solutions, infinite number of solutions, also notice something very, you already notice this, of ‘course the column that doesn't have the leading entry, the column where there are multiple entries, those are the columns that are going to be your free parameters.0839

In this particular system, there is only one column like that, all of the other columns had leading entries, and they were all 0's, if you end up having two or three or four columns, you are going to end up having two or three or four different parameters.0857

Speaking of which, let's go ahead and take a look at 1, okay I'll just do the matrix, and then we will do the row, reduced row echelon form, save us some time.0873

We have the system (1, 2, 0, -3, 1, 0, 2) 1, 2, 3, 4, 5, 6,, 6 variables, this last column is always the solution.0886

Okay, so we have X, Y, Z, S, T, W, something like that, so this is six variables that we are dealing with here (1, 2, 1, -3, 1, 2, 3).0898

We have (1, 2, 0, -3, 2, 1, 4) and my apologies my 4's always look like 9, (3, 6, 1, -9, 4, 3, 9).0911

This is the particular system that we are dealing with...0928

... Okay we have four equations, 1, 2, 3, 4, and six unknowns, that's our augmented matrix, our solution set.0933

When we go ahead and convert this using our mathematical software, to reduced row echelon form, we get the following, and again remember reduced row echelon is unique, so the reduced row echelon that you get is going to be the same as everybody else's.0941

There's only one reduced row echelon form, so we get (1, 2, 0, -3, 0, -1, 0), we get (0, 0, 1, 0, 0, 2, 1).0954

(0, 0, 0, 0, 1, 1, 2) and we get a row of all 0's, so that’s our reduced row echelon form, let's take a look at the ones with leading entries.0975

We have this one, I have put an arrow for the ones that have leading entries, okay, these are going to be the variable that we actually end up solving for, these are actually not the free parameters.0989

Now, the columns, I’ll circle them, that don't have leading entries, that are just kind of randomly arranged again, this is reduced row echelon form, it's not a problem that satisfies the definition.1001

These are going to be here for your parameters, so column 1, column 3, column 5, and then column 2, column 4 and column 6.1016

Those are going to be your free variables, those can be anything, so let's go ahead and pick some variables, let's actually call them.1027

Well, since we are dealing with this many, let's just go X1, X2, X3, X4, X5 and X6, so once again the ones that are going to be free parameters are 2, 4, and 6, so let me circle those, 2, 4 and 6.1035

Okay, now i should probably move onto other page.1056

Now the linear system that reduced row echelon form represents is the following, X1 + 2X2 -3X4 - X6 = 0.1064

And then we have X3 + 2X6 =1, and then we have X5 + X6 = 2.1080

Okay, now we assign the free parameter's once again we said to the 2, 4, and the 6, so we get something like this X2 = I'll just say R, could be anything.1099

X4 could be anything, so let's just call it S, and X6, that can equal anything, we will call it T.1113

Now, X1, just solve...1121

... That equation, just move this over there, this over there and this over there, and what you end up with is X1 = T + 3S - R,1127

X3, just move this over there, that becomes 1- and we said that X6 is T, so 1 - 2T, and we solve for X5.1142

We just move this over to that side, that's equal to 2 _ X6, which is T...1158

... this is the power of reduced row echelon, again use the mathematical software, and then just literally read the solution right off; this is the implicit expression of the solutions.1170

This is the explicit expression of the solutions; the reduced row echelon had columns with leading entries, those with the variable that you solve for, that's these things at the end.1183

Those are conditional, the columns that had variable entries, the non-leading entries, those are the ones that we assigned free parameters to.1193

They can be anything you choose, so again we are dealing with infinite number of solutions here...1201

... Okay...1214

.. Okay, let's do another one, we will do (1, 2, 3, 4, 5), (1, 3, 5, 7, 11), (1, 0, -1, -2, -6)...1221

.. Put it through our mathematical software, we get a reduced row echelon of (1, 0, -1, -2, 0), we get (0, 1, 2, 3, 0).1239

And then we get (0, 0, 0, 0, and 1), okay let's take a look at this third row, (0, 0, 0, 0, 1).1253

This basically tells me that, so we know that, this is our solution set, and this is our X, X1, X2, X3, X4.1263

This is telling me that....1273

... 0 times X4 is equal to 1, well you know 0 times anything is equal to 0, it's telling me that 0 = 1, that is not true.1277

Again no solution for this system, and reduced row echelon let us know that again unique solution, infinite number of solutions, no solutions, okay.1287

They also call this...1299

... An inconsistent system...1303

... Okay...1311

... Want to talk about homogenous systems a little bit, so homogenous system...1315

... Homogeneous systems are linear systems of ‘course...1330

... where each equation...1342

... Is equal to 0, that's it, everything on the right hand side of the equality sign is just a 0, homogenous systems are very important in mathematics, they play a very important role in the theory of differential equations.1349

And of ‘course the theory of differential equations is highly applicable for all fields of engineering and physics, so homogeneous systems, huge field of research.1360

An example would be something like 2X + 3Y - Z = 0, X - 2Y - 2Z = 0, notice everything is 0 on the right hand side, kind of makes it a lot easier to deal with, + 3Y - Z = 0.1370

That's it, this is just an example of a homogenous system, it just means that everything on the right hand side is 0.1393

Now you notice that the homogeneous system always has the trivial solution, which means X, Y and Z , all the variables are 0, so that's called the trivial solution.1400

We are not too concerned with the trivial solution, XY = 0, for all I, X of I meaning X1X2X3, here we have them as X, Y and Z, but if you have more than four or five, you just refer to them as X123456, like we did in the previous example.1412

And they all equals 0 for all I, this upside down A is a symbol which means for all, again just a little bit of formal mathematics.1431

Okay, now we will go ahead and we will give you a theorem, which we won't prove, but which will come in handy, notice here we have three equations and we have four unknowns.1441

Oh no I am sorry, I can't even count now, no it's three equations and three unknowns, so now the theorem says...1457

... A homogeneous system of M...1468

... Equations and N unknowns...1478

... Always has a non-trivial solution...1490

... if M is less than N, in other words if the number of M is the number of equations, so if the number of equations is less than the number of variables, this homogeneous system always has a non-trivial solution, that means there is always at least one solution that is not all 0's.1504

X = 0, Y = 0, Z = 0 and so on, so let's repeat that, a homogenous system of M equations and N unknowns always has a non-trivial solution.1524

It doesn't whether there are infinitely many or one, it just says, there, at least one, at least one exist if M is less than N, if the number of equations is less than the number of unknowns.1534

Lets do an example.1544

Okay, so we have (1, 1, 1, 1, 0), (1, 0, 0, 1, 0), (1, 2, 1, 0, 0).1548

When we subject this to Gaussian elimination, which brings us to reduced row echelon, we end up with the following, (1, 0, 0) (1, 0) (0, 1, 0, -1, 0) and (0, 0, 1).1564

(1, 0)...1584

... Again we are...1588

This is our solution, this is the entire matrix, so here we are talking about three equations, 1, 2, 3, and we have 1, 2, 3, 4 unknowns, this is of ‘course the solution set on the right side of the equality sign.1591

We have more unknowns than equations, and sure enough the reduced row echelon shows us that yes, we have leading entry, leading entry, leading entry.1608

Let's call this X, let's call this Y, let's call this Z, these can be the free parameter, we can call it, we can call the variable W, so we can set W = S.1617

And then solve for this, this has a solution and in fact in this particular case, because we have a free parameter, we have an infinite number of solutions, okay.1629

Let's go ahead and list these out explicitly, what this says is X + W = 0, this says Y - W = 0.1641

This says Z + W = 0, so W's are free parameter, so we will set W = S, we will solve for Z, Z= -W, which is -S.1658

Y = W, which is S and X = -W, which is -S, this is our explicit solution, this is our implicit solution, you might have a favorite, it's up to you.1676

Personally I actually prefer the implicit form, I like to do the math myself, when I see it like this, it's perfectly fine I again, just a personal preference...1691

... You might as well use our erase function here, so again this is implicit, this is explicit, and there you go.1707

Solutions of a homogeneous system Gaussian elimination, reduced row echelon, matrices represent linear systems, linear systems represents matrices.1722

Thank you for joining us at, we will see you next time.1730