WEBVTT mathematics/linear-algebra/hovasapian
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Hello and welcome to Linear Algebra, welcome to educator.com.
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This is the first lesson of Linear Algebra course, here at Educator.com.
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It is a complete Linear Algebra course from beginning to end.
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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.
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It is the study of something called Linear Mappings or Linear transformations, also known as linear functions between vector spaces.
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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.
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So, don't worry about what these terms mean vector space, linear mapping, transformation, things like that, we will get to that eventually.
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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.
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In one form or another, we are always going to be solving some set of linear equations.
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So having said that, welcome again, let's get started.
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Okay, so let's just start with something that many of you have seen already, if not, no worries.
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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.
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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.
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And believe it or not, it actually has nothing to do with a straight line.
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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.
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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.
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Okay, so this is sort of a single variable, we have ax=b, something like for example, [inaudible].
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Well, that's okay we will just leave it like that.
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If I can write this, A¹X¹ + A²X² + A³X³ = B, well these answer just different coefficients, 5, 4, 6, (-7).
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These x1, x2 and x3 are the variable, so now instead of just the one variable, some equation up here.
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We have three variables X¹, X², X³, we can have any number of them and B.
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So a solution to something like this is a series of X's that satisfy this particular equation.
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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.
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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.
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Okay, so let's use a specific example, so if I had something like 6X¹ - 3X² + 4X³ = (-13).
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I might have something like...
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...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.
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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.
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note however that we can also have X1 = 3...
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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.
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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.
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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...
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A^11X1^ + A^12X² + ... + A¹N XN = b₁, so this just is our first equation, we have n variable, that's what the X¹ to X^10, and these are just the coefficients in front of those variables X's and this is just some number.
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So this is just one linear equation, now we'll write another one A^21X¹, and I'll explain what these subscripts mean in just a moment + A^22X² + ... + A²NX < font size="-6" > n < /font > = B².
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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.
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And then I am going to write AM¹X¹ + AM²X² + ... + AM^n, I know that's a little small but that's an MN right there, equals B < font size="-6" > m < /font > , so notice we used two subscripts here, like for example we usually the subscripts I, J.
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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 A^11 is the first equation and the second entry J represents that particular column, that particular entry.
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So, A^11 represents the first coefficient in the first equation, if I did something like let's say I had A^32, that would mean the third equation, the second entry, the second coefficient, the coefficient for X².
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That's all this means, so here I have notice X all the way to n, X^n X^n all the way down, oops I forgot an X^n right here, so I have n variables....
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...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.
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We just arrange it like this, so this is a system of linear equations.
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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...
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We want a set of X1^, X², all the way to X^n, such that all of these equations are satisfied simultaneously...
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... 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...
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... simultaneously...
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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.
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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.
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It has to satisfy all of them, that's the whole idea.
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Let's see what we've got here....
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... okay, we are going to use a process called elimination...
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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.
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One solution infinitely manages solutions, no solutions, what are the things that can happen when dealing with linear system.
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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.
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Now of course many of you have dealt with these in algebra.
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You have seen the method of elimination; you have used the method of substitution.
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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.
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Because again in algebra, ultimately when you are solving an equation, you can deal with one variable at a time.
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Lets just jump in and I think the, the technique itself will be self-explanatory...
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...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.
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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.
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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.
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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.
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So let us go ahead and do that.
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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.
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What this ends up being is...
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-3X - 6Y= (-24) and of course this equation we just leave it alone.
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We don't need to make any changes to it.
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3X - 4Y = 4.
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And now we can go ahead and then, the -3X + 3X, that goes away, -6Y - 4Y gives us -10Y, -24 + 4 is -20.
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And when we divide through by -10, we get Y = 2.
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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.
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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.
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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...
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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.
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Both of these equalities will be satisfied, so in this particular case, we have one solution.
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We do this in red.....
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...one solution, okay....
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Now let's try X - 3Y = -7, 2X - 6Y = 7, so let's see what happens here.
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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.
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-2 times X is -2X, -2 times -3Y is +6Y = -2 times -7 gives 14.
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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.
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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.
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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.
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You might want to avail yourself over the mathematical software, okay.
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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.
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We get something like this, 0 = 21; well 0 does not equal 21, okay, so this is no solution.
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We call this an inconsistent system, so any time you see something that is not true, that tells you that there is no solution.
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In other words there is no way for me to pick an X and a Y that will satisfy both of these equalities simultaneously.
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It is not possible, no solution also called inconsistent.
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Okay...
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Example three, okay, now we have got three equations and three unknowns, X, Y and Z.
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Well we deal with these two equations at a time, so let's go ahead, we see an X here, and a 2, 3.
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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.
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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.
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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.
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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.
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That's the best thing to do.
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You will never ever go wrong if you write everything own, and yes I am guilty of that myself.
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Okay, so this becomes, let us write it over here, -2X - 4Y - 2 + 3Z is -6Z, -2 times 6 is -12.
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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.
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And now we have reduced these two, eliminated the X, so now we have an equation in two unknowns,
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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.
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I am going to multiply by a -3 this time.
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When I do that, I end up with -3X...
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-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.
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And then again I leave this third one unchanged, 3X + Y - Z = -2.
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I eliminate those -6Y + 1Y is -5y, and then I get -9 -1 is -10Z, -18 - 2, I get -20.
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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.
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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.
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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.
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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.
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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.
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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.
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This will give me a positive value here and a negative value here, this should take care of it.
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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.
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So far so good, 7 times -10 is -70Z and 7 times a -20 is -140.
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Now, when we solve this, the Y's go away and we get +20Z - 70Z for a total of -50Z = -10 - 140 - 150.
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That means Z is equal to 3.
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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.
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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.
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But it's always going to be the basic stuff that is going to slip you up, so Y = -2.
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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.
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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.
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Don't multiply this out and make sure you actually see it like this again.
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Write it all out, + 3 times 3 = 6, we get X - 4 + 9 = 6, get, oops, that is little straight lines here.
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Erase these , if you guys are bothered, okay, X - 4, what is -4 + 9, that's 5 right.
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X + 5 = 6, we get X = 1, and there you have it, you have X = 1, Y = -2, Z = 3.
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Three equations, three unknowns and we have one solution.
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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.
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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.
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Again nothing difficult, just a little long, that's all, okay.
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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.
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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.
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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.
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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.
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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.
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I am going to divide by -3 just to make this a positive, so this becomes....
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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.
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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.
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Let's go ahead and move, solve for one of the variables, so Y = Z -4, so now I have Y = Z - 4.
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And I have this thing I can solve for X, but what do I do with this, as it turns out.
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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.
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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.
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That makes Y = 1 , and now I can go back and solve this equation, so let me just do this one quickly.
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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.
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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.
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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.
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Here we have an infinite number of solutions....
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...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.
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Now let's see what we else we can do here.
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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.
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Okay, this particular system is X + 2Y = 10, 2X - 2Y = -4, 3X + 5Y = 26.
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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.
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I get Y = 4, now notice I have three equations, so this Y = 4, deals with these, this first two.
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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.
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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.
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Now I need to do the first and third.
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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.
00:30:14.000 --> 00:30:22.000
And I have 3X + 5Y = 26.
00:30:22.000 --> 00:30:29.000
Now let's go ahead and do that.
00:30:29.000 --> 00:30:58.000
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.
00:30:58.000 --> 00:31:16.000
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....
00:31:16.000 --> 00:31:23.000
...it worked out the same now, I can take this Y, put it in here and solve for x.
00:31:23.000 --> 00:31:45.000
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.
00:31:45.000 --> 00:32:06.000
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.
00:32:06.000 --> 00:32:15.000
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 system
00:32:15.000 --> 00:32:20.000
This system is equivalent to this system, that's all you are doing.
00:32:20.000 --> 00:32:29.000
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.
00:32:29.000 --> 00:32:36.000
You are good, now you can go back and solve for the X, okay.
00:32:36.000 --> 00:33:04.000
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.
00:33:04.000 --> 00:33:22.000
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.
00:33:22.000 --> 00:33:31.000
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.
00:33:31.000 --> 00:33:54.000
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.
00:33:54.000 --> 00:34:08.000
Now, when I do this, the X's cancel, I am left with -Y = -30 + 20 - 10, I get Y = 10.
00:34:08.000 --> 00:34:11.000
Y = 4, Y = 10.
00:34:11.000 --> 00:34:22.000
there is no way to reconcile these two to make all three equalities satisfied simultaneously, so this is no solution.
00:34:22.000 --> 00:34:34.000
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.
00:34:34.000 --> 00:34:44.000
No solution, these have to be consistent, first and second, this is first and third.
00:34:44.000 --> 00:34:48.000
Again we are looking for this whole thing.
00:34:48.000 --> 00:35:00.000
What we just did here is the equivalent system that we have transformed to is X + 2Y = 10, Y = 4, Y = 10.
00:35:00.000 --> 00:35:10.000
There is your inconsistency okay.
00:35:10.000 --> 00:35:33.000
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.
00:35:33.000 --> 00:35:48.000
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.
00:35:48.000 --> 00:35:54.000
The one solution case was when you had...
00:35:54.000 --> 00:36:03.000
... they meet at a point and the infinitely many solutions is when one line is on top of another line, infinitely many solutions.
00:36:03.000 --> 00:36:13.000
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.
00:36:13.000 --> 00:36:22.000
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.
00:36:22.000 --> 00:36:37.000
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.
00:36:37.000 --> 00:36:44.000
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.
00:36:44.000 --> 00:36:54.000
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.
00:36:54.000 --> 00:37:01.000
Okay, let me just go over one more thing here, the method of elimination, so let's recap.
00:37:01.000 --> 00:37:14.000
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.
00:37:14.000 --> 00:37:21.000
But maybe you prefer it in a different location, it just means switching the order of the equations, nothing strange happening there.
00:37:21.000 --> 00:37:33.000
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.
00:37:33.000 --> 00:37:51.000
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.
00:37:51.000 --> 00:38:25.000
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.
00:38:25.000 --> 00:38:37.000
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.
00:38:37.000 --> 00:38:42.000
But the original equation stays, so this, so that's what we were doing when we do this.
00:38:42.000 --> 00:38:50.000
We are changing a system to an equivalent system, that's what we have to keep in mind when we are doing these eliminations.
00:38:50.000 --> 00:38:56.000
Notice the first equation is unchanged, when we rewrite our entire system.
00:38:56.000 --> 00:39:03.000
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.