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Lecture Comments (59)

0 answers

Post by John Lins on January 30, 2017

Could you please explain what would be the solution to this question below?
Show that this pair of augmented matrices are row equivalent, assuming ad-be# 0:
(a b |e) ~ (1 0 |(de-bf) / (ad-bc))
c d |f    (0 1 |(af-ce) / (ad-bc)

1 answer

Last reply by: Professor Hovasapian
Wed Jan 18, 2017 8:13 PM

Post by Mohsin Alibrahim on January 10, 2017

Hi Professor H

In ex 5, why didn't you multiply the first equation by (-5) and eliminate the x variables the x immediately instead of what you did when worked on 1,2 and 1,4 separately.

1 answer

Last reply by: Professor Hovasapian
Tue Nov 29, 2016 2:40 AM

Post by manu vats on November 25, 2016

which textbook should i use for this course

1 answer

Last reply by: Professor Hovasapian
Sun Nov 6, 2016 4:37 PM

Post by El Einstein on November 5, 2016

Would you classify "infinite # of solutions" as Consistent or Inconsistent? Im slightly confused.
I'm assuming that as long as the system of linear equations has at least one solution, it will be considered as Consistent. Is this a correct assumption?

3 answers

Last reply by: cary pope
Fri Nov 18, 2016 9:27 PM

Post by cary pope on October 15, 2016

Professor Hovasapian, You are so good at explaining science and mathematical concepts. Your videos are well formed and extremely useful. The first lessons of yours I watched was on multivariable calculus. I was way ahead of my class and fell in love with vector calculus, with your videos and explanations being a major factor. I feel lucky, I'm taking linear algebra and chemistry at the moment and was pleasantly surprised to find out that you have courses for each on educator.com.  I just wanted to mention how much you're helping me and how good you are. I love how you simplify things and include tricks to visualize the process of what you are actually doing with the mathematics, your multivariable calculus course was brilliant. I'm enjoying your chemistry and linear algebra coases as well. Any chances of you making a course on complex analysis? I'm taking that next semester!
Thank you for making these effective videos that are on target and to the purpose!

1 answer

Last reply by: Professor Hovasapian
Sat Sep 10, 2016 2:17 AM

Post by Kaye Lim on September 9, 2016

For a case of 2 unknowns with 3 equation, after solving this, let's say we get 1 solution (x,y). How would the graph look like? Does it mean 3 lines meet at a point (x,y) in the solution?

-Is it possible for these 3 lines to meet at 2 points?

1 answer

Last reply by: Professor Hovasapian
Fri Jun 17, 2016 6:24 PM

Post by Emily Lewis on June 16, 2016

Why did you decide to use elimination for all of these examples when substitution would have been much easier for some of them?

1 answer

Last reply by: Professor Hovasapian
Tue Apr 7, 2015 10:50 PM

Post by Micheal Bingham on March 31, 2015

You have an absolutely beautiful verbiage, I know that this does not pertain to the lecture but do you have any advice how one could become as articulate as you?

1 answer

Last reply by: Professor Hovasapian
Mon Mar 2, 2015 6:37 PM

Post by julius mogyorossy on March 1, 2015

Dr. do I need your course to take the college Algebra CLEP test?

1 answer

Last reply by: Professor Hovasapian
Mon Feb 2, 2015 4:02 PM

Post by Danial Shadmany on February 1, 2015

Hi Professor,

I was just wondering if you will cover singular value decomposition in this course. I didn't see it in the syllabus and wondered if you'd still cover it in this course.


1 answer

Last reply by: Professor Hovasapian
Fri Nov 21, 2014 9:29 PM

Post by Eric Liu on November 21, 2014

Hi Professor Hovasapian,

I know your Calculus AB course is going to be released sometime in the coming months, but I was wondering if you will be doing a Calculus BC course for educator.com as well?


2 answers

Last reply by: Miguel Villarreal
Mon Jun 2, 2014 9:44 AM

Post by Miguel Villarreal on June 2, 2014

@31:32 Example VI 3 equations and 2 unknowns because y=4 if the other y=x another variable would we solve for both or does become an inconsistent system because they are different?

Thank you

0 answers

Post by MOHAMMED ALHUMAIDI on December 7, 2013


When i try to download lecture slides it is show without any  answer on it??


1 answer

Last reply by: Professor Hovasapian
Mon Sep 23, 2013 4:17 PM

Post by Jawad Mustafa on September 22, 2013

Thank you very much for offering this course

Jawad Mustafa
Amman - Jordan

3 answers

Last reply by: Professor Hovasapian
Wed Dec 30, 2015 12:03 AM

Post by Manfred Berger on May 29, 2013

I've been thinking about example 4 for a bit, and it seems to me that the reason you're not getting a single point as your solution there is you're intersecting 2 planes, which in turn leads to a line as your result.

1 answer

Last reply by: Professor Hovasapian
Fri Apr 12, 2013 4:33 PM

Post by Rishabh Jain on April 12, 2013

I love your lecture SIR !!! AMAZINGGGGGGGGG !!!!!!!!!!

1 answer

Last reply by: Professor Hovasapian
Thu Oct 11, 2012 3:29 PM

Post by Aniket Dhawan on October 11, 2012

You are a great teacher. I really liked your explanations,they helped me a lot.

Thankyou professor

3 answers

Last reply by: Aniket Dhawan
Thu Oct 11, 2012 5:09 AM

Post by Suhaib Hasan on October 4, 2012

Your comment about induction and math was great.

2 answers

Last reply by: Rob Lee
Thu Sep 27, 2012 6:18 PM

Post by robert lee on September 26, 2012

Question: at 36:02. I am slightly confused with your graph, I understand the no solution case, where the lines are parallel and never meet.

The one solution case meets at one point (in other words if it was three or more lines, they should all intersect at the same point then?)

Now the infinite case, I do not understand at all, what does it mean when the line is on top of another line? Doesn't this just mean that they meet at one point? So it would be just like the one solution case then... How is this possible?

In my imagination, an infinite solution would be more like a graph of a sin and cos equation, cause then they would intersect at multiple points, but then this would not be linear??

Can you please clarify?

Thank you.

1 answer

Last reply by: Professor Hovasapian
Fri Sep 21, 2012 2:34 PM

Post by Erdem Balikci on September 20, 2012

Very well done! Clear and smooth!!

0 answers

Post by Maimouna Louche on June 17, 2012

Thanks I get it now, it looked scary for nothing.

0 answers

Post by Maimouna Louche on June 15, 2012

I will be taking this class soon. Man it look hard :( I will be fine, I have Educator now! ^^

0 answers

Post by Real Schiran on February 29, 2012

All understood. This lecture is very clear. Thanks

1 answer

Last reply by: Constantin Ficiu
Thu Oct 24, 2013 2:34 PM

Post by amir szeinberg on February 18, 2012

I have a problem re-entering the lecture in the middle, say in the 20th minute, and I have to strart from the begining. Is there any thing I could do about it?
Thanks in advance,

0 answers

Post by thomas kotch on December 18, 2011

He is great!

0 answers

Post by Senghuot Lim on December 18, 2011


0 answers

Post by Arthur Bookstein on October 12, 2011

Very well explained.

1 answer

Last reply by: Constantin Ficiu
Thu Oct 24, 2013 2:30 PM

Post by Jason Mannion on October 4, 2011

I subscribed to this site because I was having great difficulties in my university Linear Algebra class. So I have only watched the "Linear Systems" lecture, and from that I can conclude a few things: 1)Dr. Hovasapian actually has the ability to organize a course, 2) He knows how to present material in a educational manner, and 3) in the one video I learned more than I did in several weeks of my university lectures.
(the problem is simply that my professor lacks any teaching abilities, and cannot organize the material. We started our course with complex numbers, and then went straight into vectors and subspaces, without any explanation as to what a linear system was, how to solve one, or even how to do basic matrix arithmetic! In fact, we never see any matrices in class!)

0 answers

Post by Travis Torres on October 1, 2011

Very well done. I'll admit I was a little confused by the initial lecture on Linear Systems, but the examples themselves were very informative and helpful.

Linear Systems

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
  • Linear Systems 1:20
    • Introduction to Linear Systems
  • Examples 10:35
    • Example 1
    • Example 2
    • Example 3
    • Example 4
    • Example 5
    • Example 6
  • Number of Solutions 35:08
    • One Solution, No Solution, Infinitely Many Solutions
  • Method of Elimination 36:57
    • Method of Elimination

Transcription: Linear Systems

Hello and welcome to Linear Algebra, welcome to educator.com.0000

This is the first lesson of Linear Algebra course, here at Educator.com.0004

It is a complete Linear Algebra course from beginning to end.0010

So a Linear Algebra, I am going to introduce just a couple of terms right now , just to give you an idea of what it is that you are going to expect in this course.0014

It is the study of something called Linear Mappings or Linear transformations, also known as linear functions between vector spaces.0023

And this is a profoundly important part of mathematics, because linear functions are the heart and soul of Science and Mathematics, everything that you sort of enjoy in your world today consist of essentially a study of linear systems.0032

So, don't worry about what these terms mean vector space, linear mapping, transformation, things like that, we will get to that eventually.0047

Today's topic, our first topic is going to be linear systems and it's going to be the most ubiquitous of the topics, because we are going to use linear systems as our fundamental technique to deal with all of the other mathematical structures that we deal with.0055

In one form or another, we are always going to be solving some set of linear equations.0070

So having said that, welcome again, let's get started.0075

Okay, so let's just start with something that many of you have seen already, if not, no worries.0081

If we have something like AX=B, this is a linear equation, one reason that linear is used, the term linear is because this is the equation of a straight line.0091

However as it turns out, although we use the term linear, because it comes from the straight line later on in the course, we are actually going to get the precise definition of what we mean by linear.0104

And believe it or not, it actually has nothing to do with a straight line.0114

It just so happens that the equation, this AX=B, which can be represented by a straight line on a sheet of paper on a two dimensional surface.0117

It had, happens to be a straight line so we call it linear, but its, but the idea of linearity is actually a deeper algebraic property about how this function actually behaves when we start moving from space to space.0129

Okay, so this is sort of a single variable, we have ax=b, something like for example, [inaudible].0143

Well, that's okay we will just leave it like that.0153

If I can write this, A1X1 + A2X2 + A3X3 = B, well these answer just different coefficients, 5, 4, 6, (-7).0155

These x1, x2 and x3 are the variable, so now instead of just the one variable, some equation up here.0175

We have three variables X1, X2, X3, we can have any number of them and B.0183

So a solution to something like this is a series of X's that satisfy this particular equation.0189

That's all what's going on here, linear equation, you know this linear essentially is when this exponent up here is A, that pretty much is what we are used to see when we deal with linear equations.0197

But again linearity is a deeper algebraic property, which we will explore a little bit later in the class, and that's when linear algebra becomes very, very exciting.0210

Okay, so let's use a specific example, so if I had something like 6X1 - 3X2 + 4X3 = (-13).0220

I might have something like...0234

...X1 = 2, X2 = 3 and X3 = (-4), well this 2, this 3, this (-4) for X1, X2 and X3 is a solution to this linear equation.0239

That's it, we are just looking, it is that, that's all we were looking for, we are looking for variable that satisfy this equality, that's all that's happening here.0256

note however that we can also have X1 = 3...0264

X2 = 1 and X3 = (-7). So if we put 3, 1, (-7) in for X1, X2 and X3 respectively, we also get this equality (-13), so as it turns out these particular variables don't necessarily have to be unique.0271

Several, sometimes they can be unique, other times a whole bunch of, set of numbers can actually satisfy that equality, so we want to find as many of the solutions that satisfy that equality, okay.0287

Now let's generalize this some more and talk about a system of equations, so I am going to go ahead and represent this symbolically, so see we have...0302

A11X1+ A12X2 + ... + A1N XN = b1, so this just is our first equation, we have n variable, that's what the X1 to X10, and these are just the coefficients in front of those variables X's and this is just some number.0315

So this is just one linear equation, now we'll write another one A21X1, and I'll explain what these subscripts mean in just a moment + A22X2 + ... + A2NXn = B2.0344

Now we have our second equation and then we go down the line, so I am going to put a ... there ... means we are dealing with several equations here.0367

And then I am going to write AM1X1 + AM2X2 + ... + AMn, I know that's a little small but that's an MN right there, equals Bm, so notice we used two subscripts here, like for example we usually the subscripts I, J.0379

And the first subscript represents the row or the equation, so in this case 1, 2,3,4,5 all the way to the nth equation, so A11 is the first equation and the second entry J represents that particular column, that particular entry.0415

So, A11 represents the first coefficient in the first equation, if I did something like let's say I had A32, that would mean the third equation, the second entry, the second coefficient, the coefficient for X2.0437

That's all this means, so here I have notice X all the way to n, Xn Xn all the way down, oops I forgot an Xn right here, so I have n variables....0457

...and I have as many rows M equations and this is exactly what we say when we have n equations and N variables, this many and this many.0473

We just arrange it like this, so this is a system of linear equations.0486

What this means when we are looking for a solution to a system of linear equations as supposed to just one linear equation, we are looking for...0490

We want a set of X1, X2, all the way to Xn, such that all of these equations are satisfied simultaneously...0503

... such that all equalities, I'll say equalities instead of equations, we know we are dealing with equations; we want all of these equalities to satisfied...0521

... simultaneously...0535

In other words we want numbers such that, that holds, that holds, that holds, that holds if one of them doesn't hold, it's not a solution.0540

Let's say you have seven equations, and let's say you found some numbers that satisfy six of them, but they don't satisfy the seventh, that system doesn't have that solution.0547

It has to satisfy all of them, that's the whole idea.0558

Let's see what we've got here....0565

... okay, we are going to use a process called elimination...0571

To solve systems of linear equations, now we are going to start in with the examples to see what kind of situations we can actually come up with.0583

One solution infinitely manages solutions, no solutions, what are the things that can happen when dealing with linear system.0592

How many variables, how many equation and, what's the relationship that exists, just to get a sense of what's going on, just to get us back into the habit of working with these.0598

Now of course many of you have dealt with these in algebra.0605

You have seen the method of elimination; you have used the method of substitution.0608

Essentially elimination is turning one equation, let's say you have two equations and two unknowns, you are going to manipulate one of the equations so that you can eliminate one of the variable.0612

Because again in algebra, ultimately when you are solving an equation, you can deal with one variable at a time.0620

Lets just jump in and I think the, the technique itself will be self-explanatory...0628

...okay, so our first example is X + 2I = 8, 3X - 4Y = 4, we want to find X and Y such that both of these hold simultaneously, okay.0636

In this particular case elimination and it really doesn't matter which variable you eliminate, so a lot of times, it's a question of personal choice.0649

Some people just like one particular variable, often times you look at what look like it's easy to do, that will guide your choice.0658

In this particular case I notice that this coefficient is 1, so chances are if I multiply this by 3, by (-3), this whole equation by (-3) to transform it, and then add it to this equation, the -3X and the 3X will disappear.0665

So let us go ahead and do that.0681

Let us go ahead and multiply everything by (-3) and when I do that, I tend to put a (-3) here, (-3) there to remind me.0684

What this ends up being is...0692

-3X - 6Y= (-24) and of course this equation we just leave it alone.0698

We don't need to make any changes to it.0708

3X - 4Y = 4.0711

And now we can go ahead and then, the -3X + 3X, that goes away, -6Y - 4Y gives us -10Y, -24 + 4 is -20.0717

And when we divide through by -10, we get Y = 2.0731

We are able to find our first variable Y = 2.1219 Now, I can put this Y = 2 back into any one of the original equations, you could put them in these two, it's not a problem.0735

it doesn't, multiplying by a constant doesn't change the nature of the equation, because again you are multiplying, you are retaining the equality, you are doing the same thing to both sides, so Y = 2.0745

Lets go ahead and use the first equation, therefore I will go ahead and draw a little line here, we will say X + 2 times 2, which is Y = 8X + 4 = 8X oops...0757

Let us put the X on the left hand side, X = 4, so there you have it, a solution X = 4, Y = 2, if X = 4, if Y = 2, that will solve both of these simultaneously.0780

Both of these equalities will be satisfied, so in this particular case, we have one solution.0796

We do this in red.....0804

...one solution, okay....0817

Now let's try X - 3Y = -7, 2X - 6Y = 7, so let's see what happens here.0826

Well, in this particular case again I notice that I have a 2 and a coefficient of 1, some we have to go ahead and eliminate the X again, so in order to eliminate the X, I need this to be a -2X, so I am going to multiply everything by (-2) of top.0834

-2 times X is -2X, -2 times -3Y is +6Y = -2 times -7 gives 14.0849

I can pretty much guarantee you that in your just, a small digression, the biggest problem in linear algebra is not, as this is not going to be the linear algebra, it is going to be the arithmetic, just keeping track of the negative signs or positive signs and just the arithmetic addition, subtraction, multiplication and division.0861

My recommendation of course is, you can certainly do this by hand, and it is not a problem, but at some point you are going to want to start to use the mathematical software, things like maple, math cad, mathematica, they make life much, much easier.0881

Now, obviously you want to understand what is going on with mathematics, but now some of, as we get into the course, a lot of the computational procedures are going to be kind of tedious in the sense that they are easy, except they are arithmetically heavy, so they are going to take time.0897

You might want to avail yourself over the mathematical software, okay.0912

Let us continue on and then this one doesn't change, so it's 2X - 6Y = 7 and then when we add these, we get +6Y and -6Y, wow these cancel too, so we end up with 0 = 14 + 7 is 21.0917

We get something like this, 0 = 21; well 0 does not equal 21, okay, so this is no solution.0936

We call this an inconsistent system, so any time you see something that is not true, that tells you that there is no solution.0946

In other words there is no way for me to pick an X and a Y that will satisfy both of these equalities simultaneously.0954

It is not possible, no solution also called inconsistent.0960


Example three, okay, now we have got three equations and three unknowns, X, Y and Z.0974

Well we deal with these two equations at a time, so let's go ahead, we see an X here, and a 2, 3.0979

I am going to go ahead and just deal with the first two equations, and I am going to multiply, I am going to go ahead and eliminate the X, so I am going to multiply by -2 here.0986

And again just be very, very systematic in what you do, write everything down, the biggest problems that I had seen with my students is that they want to do things in their head and they want to skip steps.0997

Well, when you are dealing with multiple steps, let us say if you have a seven step problem, and each one of those steps requires may be three or four steps, if you skip a step in each sub portion of the problem, you have skipped about seven steps.1007

I promise there has been a mistake, which there always will be, and when it comes to arithmetic, you are going to have a very hard time finding where you went wrong, so just write everything down.1019

That's the best thing to do.1027

You will never ever go wrong if you write everything own, and yes I am guilty of that myself.1028

Okay, so this becomes, let us write it over here, -2X - 4Y - 2 + 3Z is -6Z, -2 times 6 is -12.1035

And let us bring this equation over unchanged, that is the whole idea, 2X -3Y + 2Z = 14, let us go ahead and, so the X's eliminate, and then we end up with -4Y - 3Y is -7Y, -6Z + 2Z is -4Z, and -12 + 14 = 2, so that's our first equation.1051

And now we have reduced these two, eliminated the X, so now we have an equation in two unknowns,1078

Now, let us deal with the first and the third, so on this particular case, I am going to do this one in blue, I am going to, I want to eliminate the X again, because I eliminated the here, so I am going to eliminate the X here.1086

I am going to multiply by a -3 this time.1100

When I do that, I end up with -3X...1103

-3 time +2 is -6Y, -3 times 3 is -9Z and -3 times 6 is -18, and I am hoping that we are going to confirm my arithmetic here.1112

And then again I leave this third one unchanged, 3X + Y - Z = -2.1127

I eliminate those -6Y + 1Y is -5y, and then I get -9 -1 is -10Z, -18 - 2, I get -20.1141

Now, I have my second equation, and this one was first equation, so now I have two equations and two variables, Y and Z, Y and Z.1155

Now, I can work with these two, so let me go ahead and bring them over and rewrite them, -7Y - 4Z = 2, and -5Y - 10Z = -20.1164

Good, so now we have a little bit of a choice to make, do we eliminate the y or do we eliminate the Z now.1184

It's again, it's a personal choice, I am going to go ahead and eliminate the Y's for no other reasons, and beside and I am just going to work from left to right, not a problem.1192

I am going to multiply, so I need the Y's to disappear and they are both negative, so I thing I am going to multiply the top equation by a -5.1201

And I am going to multiply the bottom equation, I will write that in black, no actually I will keep it in blue, the bottom equation by 7 , 7 here, 7 here.1214

This will give me a positive value here and a negative value here, this should take care of it.1225

Let me multiply the first one, what I get is 35Y right, -5 times -4 is +20Z, -5 times 2 = 1-10., 7 times -5 is -35Y.1231

So far so good, 7 times -10 is -70Z and 7 times a -20 is -140.1252

Now, when we solve this, the Y's go away and we get +20Z - 70Z for a total of -50Z = -10 - 140 - 150.1263

That means Z is equal to 3.1279

Okay, so now that I have Z = 3, I can go back and put it into one of these equations to find Y, so let me go ahead and use the first equation, so let me move over here next.1285

I would write -7Y - 4 times Z which was 3 = 2, I get -7Y -12 = 2, -7Y = 14, Y = -2, notice I didn't skip these steps, I wrote down everything, yes I know its basic algebra.1298

But it's always going to be the basic stuff that is going to slip you up, so Y = -2.1320

I have done my algebra correctly, my arithmetic, that's that one, now that I have a Z and I have a Y, I can go back to any one of my original equations and solve for my X.1325

Okay, I am going to go ahead and take the first one since, because that coefficient is there, so I get X + 2 times Y, which is -2 +, write it out exactly like that.1335

Don't multiply this out and make sure you actually see it like this again.1350

Write it all out, + 3 times 3 = 6, we get X - 4 + 9 = 6, get, oops, that is little straight lines here.1355

Erase these , if you guys are bothered, okay, X - 4, what is -4 + 9, that's 5 right.1371

X + 5 = 6, we get X = 1, and there you have it, you have X = 1, Y = -2, Z = 3.1381

Three equations, three unknowns and we have one solution.1392

Again one solution, notice what we did, we eliminated, we picked two equations, eliminated variable, the first and the third to eliminate the same variable, we dropped it down to now two equations and two unknowns.1401

Now we eliminated the common variable, got down to 1 and the, we worked our way backward, very, very simple, very straight forward, nice and systematic.1413

Again nothing difficult, just a little long, that's all, okay.1421

Let's see what else we have in store here, example four, okay so we have X +2Y - 3Z = -4, 2X + Y - 3Z = 4, notice in this case we have two equations and we have 3 unknowns, so let's see what's going to happen here.1427

Well, this is a coefficient 1, this is 2, so let's multiply this by a -2, let's go ahead and use a blue here so we will do -2 here and -2 there.1444

And let's go, now let's move over in this direction, so we have -2X - 4Y and this is going to +6Z right, equals +8 and then we will leave this one alone, because we want to eliminate the variable 2X.1457

Excuse me, + Y - 3Z = 4, okay let's eliminate those, now we have -4Y + Y, it should be -3Y, 6Z - 3Z is +3Z, 8, 9, 10, 11, 12, that is equal to 12, okay.1478

Now we have -3Y + 3Z = 12, we can simplify this a little bit because every number here, all the coefficients are divisible by 3, so let me go ahead and rewrite this as, let me divide by (-), actually it doesn't really matter.1501

I am going to divide by -3 just to make this a positive, so this becomes....1520

right now let me actually do a little error out, so divide by -3, this becomes Y, this becomes a -Z, and 12 divided by 3 becomes -4, is that correct? Yes, so now we have this equation Y - Z = 4, that's as far as we go.1529

Now let's, what we are going to do is again we need to find the solutions to this, so we need to find the X and the Y and the Z.1554

Let's go ahead and move, solve for one of the variables, so Y = Z -4, so now I have Y = Z - 4.1564

And I have this thing I can solve for X, but what do I do with this, as it turns out.1579

Whenever I have something like this, Z = any real number, so basically when you have a situation like this, you can put in any real number for Z, and whatever number you get, let's say you choose the number 5.1584

If you put 5 in for Z, that means 5 - 4, well let's just do that as an example, so if Z = 5, well 5- 4 = 1.1600

That makes Y = 1 , and now I can go back and solve this equation, so let me just do this one quickly.1611

We get X + 2 times +2 - 15 = -4, 2 - 15 is X -13 = -4, that means X = 4 + 13 should be 9, so X = 9.1620

This is a particular solution, but it's a particular solution based on the fact that I chose Z = 5, so notice any time you have two equations three unknowns, more unknowns than equations, you are going to end up with an infinite number of possibilities depending on how you choose Z.1644

Z can be any real number, once you choose Z you have specified why, and once you know, specified Y, you can go back and you specify X.1660

Here we have an infinite number of solutions....1670

...okay, so an infinite number of solutions is also another possibility, so we have seen one solution, a system that has one solution only, we have seen system that has no solutions , that was inconsistent and now we have seen the system that has an infinite number of solutions, okay.1683

Now let's see what we else we can do here.1698

Just want to be nice and example, happy just to get a, so make sure that every, every all the, all the steps are covered, all the bases are covered, just we know what we are dealing with.1703

Okay, this particular system is X + 2Y = 10, 2X - 2Y = -4, 3X + 5Y = 26.1711

Okay, let's start off by eliminating the X here, so I am going to multiply this by -2, -2 to give us, -2X -4Y = -20, and that of course this one stays the same, 2X -2Y = -4, when I do this I get -6Y = -24, Y -4, okay.1720

I get Y = 4, now notice I have three equations, so this Y = 4, deals with these, this first two.1754

I need all three equations to be handled simultaneously, so now since I can't just stop here and plug back in, it's not going to work.1766

I need to make sure so now I have just done the first and the second, now I am going to do the first and the third, so this is first and second equations.1774

Now I need to do the first and third.1784

So now I am going to, and we do this one in red, this is X, this is 3X, so I am going to multiply by -3, so in this case I have -3X - 6Y = -3 times then actually you cross these out, -3 times that, -30 and make sure my negative signs work here.1789

And I have 3X + 5Y = 26.1814

Now let's go ahead and do that.1822

Okay, 3X's cancel, -6Y + 5Y is a -Y, and -30 + 26 is a -4, divide by -1, so we get Y = 4, okay so notice, our first and second equation we get Y = 4, our first and third equation we get Y = 4, these equations that we come up with, we have transformed this original system.1829

Now our original system has been transformed into X + 2Y = 10, because that's what we are doing, we are just changing equations around, X + 2 = 10, and then we did this one, we got Y = 4 and we go Y = 4, because....1858

...it worked out the same now, I can take this Y, put it in here and solve for x.1876

Let me make this a little clear, select this and we will write an X here, it is definitely a Y, so now I take X + 2 times Y, which is 4 = 10, so I get X + 8 = 10, I get X = 2.1883

And that's my solution, one solution X = 2, Y = 4, so be very, very careful with this, it's just because you end up eliminating and equation or eliminating a variable, in this particular case notice we have three equations and two variable, you can eliminate a variable and end up with a Y = 4, which you can't stop there.1905

You can't, you have to, you have to account for the third equation, so now you do the first and the third, and if there is consistency there, you end up with this system1926

This system is equivalent to this system, that's all you are doing.1935

Every time you make the change, you are creating a new set of equations, you are just, you know, now you are dealing with this system because this and this are the same.1940

You are good, now you can go back and solve for the X, okay.1949

Let's look what we have here, again we have a system of three equations and two unknowns, so we are going to treat it same way, so let's start off by doing the first and second equations, so you write first and second over here, so we are going to multiply this by -2, -2, so we are going to get -2X - 4Y = -20.1956

And this one we leave the 2X - 2Y = -4, when we add the X's cancel, we are left with -6Y = excuse me, and then we are left with Y = 4, again.1984

That's just the same thing that we had before, now we will take care of the first and third equation, this time to multiply again by 3.2002

Let me do this one in blue, -3, -3 and we are left with, so the first equation becomes -3X - 6Y = -30, and then this one becomes 3X + 5Y = 20.2011

Now, when I do this, the X's cancel, I am left with -Y = -30 + 20 - 10, I get Y = 10.2034

Y = 4, Y = 10.2048

there is no way to reconcile these two to make all three equalities satisfied simultaneously, so this is no solution.2051

Again just because you found a solution here, don't stop here, don't stop here and (inaudible) into one of these equations because you just did it for the first two, and certainly don't throw it into the third, because that won't give you anything.2062

No solution, these have to be consistent, first and second, this is first and third.2074

Again we are looking for this whole thing.2084

What we just did here is the equivalent system that we have transformed to is X + 2Y = 10, Y = 4, Y = 10.2088

There is your inconsistency okay.2100

All of these examples that we have done always been the same thing, we see that we either have one solution, unique solution, we have no solution or we have infinitely many solutions, those are the only three possibilities for a linear system, one solution , no solution or infinitely many solutions.2110

Back in algebra, we are dealing with lines, again these are all just equation of lines, the ones and two variables X + Y, well the no solution case, that's when you have parallel lines, they never meet.2133

The one solution case was when you had...2148

... they meet at a point and the infinitely many solutions is when one line is on top of another line, infinitely many solutions.2154

But again, we are using the word linear because we have dealt with lines before we developed a mathematical theory; mathematics tends to work from specific to general.2163

And the process of going to the general, the language that they use to talk about the general is based on the stuff that we have dealt within the specifics.2173

We have dealt with line before we dealt with linear functions, once we actually came up with a precise definition for a linear function, we said let's call it, well the ones who decided to give it a name said let's call it a linear function, a linear map, a linear transformation.2182

It actually has nothing to do with a straight line, it just so happens that the equation for a line happens to be a specific example of a linear function.2197

But linearity itself is a deeper algebraic property which we will explore and which is going to be the very heart of, well linear algebra.2204

Okay, let me just go over one more thing here, the method of elimination, so let's recap.2214

Using the method of elimination we can do three things essentially, we can interchange any two equations and interchange just means switch the order, so if I have the particular equation that has a coefficient of 1 and one of the variable, it's usually an good idea to put that one on top.2221

But maybe you prefer it in a different location, it just means switching the order of the equations, nothing strange happening there.2234

Multiply any equation by a non-zero constant, which is really what we did most of the time here; multiply by -3, -2, 5, 7, whatever you need to do in order to make the elimination of the variables happen.2241

And then third, add a multiple of one equation to another, leaving the one you multiplied by a constant in its original form, so recall when we had X + 2Y, we had X + 2Y = 8, 3X - 4Y = 4.2253

When we multiply the first equation by -3, then add it to equation 2, we ended up with the following equivalent system, so we end up converting this to -3X -6Y and end up -24 and then we brought this one over 3X - 4Y = 4, once we actually found the answer to this, which is say, -10Y = -20.2271

we ended up with a solution, well once we get that solution, that, this is now the new equation, so that's over here, Y = 2.2305

But the original equation stays, so this, so that's what we were doing when we do this.2317

We are changing a system to an equivalent system, that's what we have to keep in mind when we are doing these eliminations.2322

Notice the first equation is unchanged, when we rewrite our entire system.2330

Okay, thank you for joining us here at educator.com, first lesson for linear algebra, we look forward to see you again, take care, bye, bye.2336