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in this episode you’ll learn what augmented matrices are and why they are

useful [Music] okay so let’s begin by considering a set

of m linear equations that involve n unknown quantities the r that are

represented by these n variables here so i’m using indexes for variable um

so x1 x2 all the way to xn so we have n unknown quantities

and we have m linear equations m of them

and we’re going to need some coefficient so we’re going to say a i j represents

the number the real number that is the coefficient of the x j so j

is running between 1 to n in the ith equation so i is the first index here

and that that’s for the ith equation and j is for the j variable

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so that the variable number comes second and then we need some given real number

so b 1 through b m and so we’re going to build an m by n

system of linear equations and here they are

and so you can see everything written out without the

a i j right so here we have a one one a one two a one in all the way to a one n

and this is the x1s the x2s all the way to the xns

and then we have numbers over here b1 b2 all the way to bm so we have m

equations m linear equations and n variables right here

and so this is called the system of simultaneous linear equations so we’re

looking for a solution that satisfies all of them simultaneously

so um before we continue though let me just say that

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this video is part of the series and in the first episode we talked about

linear equations systems of linear equations and so this is continuing this

discussion and so if you haven’t seen that first

video you may want to go and check the playlist in the description

and check out that video below and so let’s continue on now so a solution

to this system is an ordered set of n numbers because we have n variables so

we need n numbers that satisfies each of these m statements so each linear

equation you can think about as a statement a linear system with no solution is

called inconsistent and if there’s at least one solution

then it’s called consistent so a linear system may have one solution

or there may be infinitely many solutions but in any case if there’s at

least one solution it’s called consistent so here’s a m by n linear system of a

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system of linear equations here and here is the um

augmented matrix right here so we’re going to take this first equation right

here and build the first row here and i’m going to use this right here

think of this right here this vertical line right here which some people by the

way might dash so some people put vertical straight

vertical or some people will dash it in but any case these are the constants

over here okay so this is called the augmented

matrix corresponding to this system of linear equations right here

and so we’re going to be looking at as we’re solving linear systems

or systems of linear equations we’re going to be looking at them uh augmented

matrix and seeing what’s happening with them so we’re going to work through

several examples and our goal is to solve the linear system

and to keep an eye on and and just to kind of see what hap what’s happening to

the augmented matrix as we move along now again this video is part of a series

and what’s going to be coming up in the next couple of videos is

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uh we’re going to talk about elementary row operations we’re going to talk about

equivalent systems and then we’re going to talk about gas

elimination and gauss jordan elimination

also so all that’s part of the series so this video right here is just one step

in this process all right so let’s look at our first example here

find all solutions to the system of linear equations

so right here we have two equations in three unknowns

and we’re going to like write down the augmented matrices and we’re also going

to write down the solution to this right here so first off we can write down the

augmented matrix for this so this z over here is um written on the

right-hand side so we need to move it to the left-hand side

and so we can write the augmented matrix for it so we’re going to do something

like that right here so we’re going to say here this is going to be minus 150

and then we have a 50 and then we have a 500 and a 100

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and then we have a minus one and a one for the z’s

and then over here since we move this z over and we just have a zero left

so we’re going to just be looking at zero here and then 200 here

so this would be the augmented matrix for this system here

now let’s go about our business of trying to solve it so what i want to try

to do one approach would be to try to multiply this first eq um

multiply the second equation this is just one approach but i’m going to

multiply the second equation by three and so i’m looking at the x’s right here

so i’m going to multiply by 3. so i’m going to keep this top equation the same

so let’s put that right here so minus 150x and then we’re going to have here

plus 500y minus z equals zero right so it’s the same equation so i’m

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going to multiply by minus three here or sorry by

by three so i’m going to get 150 x and then plus three hundred y

and plus three z and that’s going to be equal to six hundred so

you know i just multiplied the second equation by three now what’s our new

augmented matrix right here so this will be 100 minus 150 150 500 300. 1 3

and then 0 600 so not a lot um not interesting not much interesting

happened to the augmented matrix so from this augmented matrix to this augmented

matrix but now something interesting we’re going to do here

is we’re going to add together the x’s and so what we’re going to get is 800y

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and then plus 2z then equals 600 and so we’re going to get here

um you know because we can divide all this by two so let’s say 400 y

plus z equals 300 and so what we can do is we can write the augmented uh

we can write the augmented matrix for this minus 150 500 minus one zero

and here we’re going to have 0 and then 400 for the y

and then one for the z and then 300. so eliminating x

is trying to get a zero is the same thing as trying to get a zero in one of

the rows so you know because when you’re trying to solve linear system of linear

equations you’re trying to eliminate some of the variables

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and what that corresponds to in augmented matrices is trying to get

zeros different places here so we eliminated the x and we noticed

that that was the same thing as obtaining a zero here

okay but now we still have a y and a z left and so what i’m going to do is i’m

going to say let you know let y be t a free variable that ybt

then i can go and find the z so i can go find the z right here

so the z will be 300 and then move the 400 y over so minus 400 y

but we’re saying the y is a t so then we have the z right there

now that we have the y and the z in terms of some free variable now we can

go and find the x so then and we’ll start off with 50x right so

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let’s look at this one right here so we’re going to get 200 and then minus 100y

and the y’s of t and then minus the z and the z is all of this right here 300

minus 400 t so what does all this come out to be so this comes out to be 200

minus 100 t and then minus 300 and then plus 400 t

in other words 50x is equal to minus 100 plus 300 t

so if we divide by 50 we’re going to get minus 2 plus 6t

so now we have the x the y and the z and it’s all in terms of a free variable

so how would we write the solution but it says find off solution so we’ll say

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all solutions so we need an order set and this will be the x

and this is the y which is just a t let’s just write that as a t

and then we have the z 300 minus 400 t so the set of ordered pairs x y z

and then t is a free variable so we’ll say t is a real number

and so this gives us the infinitely many ordered pairs x y z

choose any t you want any real number plug it in we’ll get an x y and z and we

got a solution to this linear system to the system of

linear equations right here okay so um the augmented matrix part really didn’t

have a much of a role to play at this stage um um eventually we’re going to

get to in the next episode um we’re only using um augmented matrices

but right now we’re just trying to figure out what they are and what role

they play and the only thing we saw in this example is that when we eliminated

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the x that gave us a zero here and you know eliminating the x was

important now another strategy might have been to eliminate the z

in which case we would have come up with some other zero in our in our augmented

matrix here okay let’s look at our next example here so in our next example

we’re gonna look at this system right here and this is just a two by two system

here and so what is the augmented matrix right here so augmented matrix right

here will be so i’m looking at the x’s here so i have a one and a two

and looking at the y so i have a minus one and a zero

and then i have a one and a four so this right here

um is the augmented matrix and we see we

have a zero here and that corresponds to no y here which means we can already

solve this for x we can we can solve this for x of course you could have seen

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that from right here so from the second equation we can see that x is just two

so let’s come up over here and solve it so we get two x equals four

we get x equals two and then substituting that into the

second equation we’re going to you know switch the x switch the y and the one so

we’re going to get x minus one equals y and so y is just in fact what y is just

one so the solution is just two one so this system has one unique solution

one point there x is two y is one that’s the only solution and we saw that the y

was eliminated we had a zero here so there’s really nothing more

uh to this example here just trying to see what the augmented matrix is

and solving it and see what role what we can find out from the augmented matrix

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there okay so let’s look at another example

let’s look at this equation here or this system right here

so this system we have two equations in three unknowns

and so let’s see what could happen first off let’s write down the augmented

matrix so we’re looking at the two and the three for our first column

and then we have a one and a minus two and then we have a minus one and a three

and then we have nineteen and seven so that’s our augmented matrix there and

as you can see we don’t have any zeros and so right now we need to try to

eliminate some of the variables here and so what i’m going to do is i’m going

to multiply the first equation by three and so let’s do that so i’m going to get

here a 6x plus 3y and then minus 3z and then 19 times 3 so we get 57.

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so multiplying everything by 3 that’s a z here

and then i’m keeping the second equation the same and so plus three z

equals seven and so the augmented matrix for this would just be

um nothing useful so i’m not even going to write it down but

you know what happens now when we add these up here

so then the next system i write down is 6x plus 3y

minus 3z it will be equal to 57. now let’s add them together so when we

add them together we’re going to get 9x plus y and then the z’s add up to 0

and then we’re going to get equal to what 64. so this augmented matrix here

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would be what 6 9 3 1 -3 0 and then 57 and 64. so we have eliminated the

z right there but we still have an x and a y here we still have an x and a y

and so what we need to do is we need to come up with a free variable again

so here we’re going to go let x be equal to t and then so the y would be what

so letting x be t so the y would be 64 minus 64 minus 9t

then we can go and find the z so z would be equal to

and so we’re going to say 2x plus y minus 19 right so that’s coming from

right here this is z just switch sides here the z and the 19. so there’s the z

and now if we substitute in so 2x so x is a t

and then plus a y which is 64 minus 9 t and then still minus 19

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and so what we get from all that is z is 45 minus 7 t

and so now we can write the solution because we have the x we have the y and

we have the z let’s go over here and write all

solutions so all solutions are given by the x which is a t and then the y

and then the z so that triple those triples where t is a real number

so this gives us a set of triples t being free you can choose any real

number you want and substitute it in you’ll get an x y z

and that x y z will satisfy both equations okay so

just curious though what were to happen if i were to say

let y be equal to t some of you may be curious what happens if we let y be t

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well then we would go and find the x and now that will involve some fractions

so i chose the x to be t that way it’d be easier to find the y

but if you wanted to say y is t then the x would be 64 minus t over 9

just by solving this right here for x and then you have to go put this fraction

and that t and go and find the z and so if we do

that let’s do that right here real quick so then we’ll find the z so z

would be equal to 2 x 64 minus t over nine

and then plus a t and then minus 19 and so then we would have to put all that

together and in the end we would get 7 over 9 t minus 43 over 9.

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and so that would be the solution set there using those

x y z using this x this y and this z right here

and so we could write a solution set out using that those those x y and z’s here

okay so this i like a little bit better just because we get no fractions here i

didn’t put any fractions over here so if you can avoid if you could choose one

without fractions i typically like to choose that one right there

all right so let’s look at the next example

so this is an another two by two system here

so in this example here what will be the augmented matrix right here

so we’re looking at one and minus minus two

and we’re looking at minus three and six and we’re looking at two and minus four

so that’s the augmented matrix right there

now when i’m trying to solve this linear

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system i want to try to eliminate one of the variables here

so i’m going to keep the i’m going to multiply this top equation by minus 2

and let’s see what we get up here so i’m going to get minus 2x plus 6y

equals minus 4 and then i’m going to keep the second equation the same

and then in fact what i notice is that those equations are exactly the same so

what that says is that i actually just have all zeroes so i can just write the

one equation down here in other words the this system right here the

augmented matrix is minus 2 0 6 0 and then minus 4 0

and so for this system because they’re actually the same line

so one of the rows an augmented matrix just is all zeros here

and so when you see that then you know you’re going to have some free variables

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here um so let’s see here what if we choose

let’s move it over here let’s let’s see what if we choose y to be equal to t

then what can we find for the x so then the x is 2 plus 3t

right so if we put the we put the y is t right here and then

the x then the x is two plus three t okay good and so then the solutions are

the set of ordered pairs x plus uh sorry x comma y and t is free

you know t is freight he could be anything we want for example t could be zero

and if t was zero what what point would we have we’d have two zero right

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so how do we write dot dot dot in this case we’ll write it we’ll write it like

this right two zeros part of the solution you can plug into zero to either one

but there’s a lot more so how do you describe all of them here’s here’s a way

one way in fact what if we choose say x is t

what if we say x is t then what will happen so then the y will turn out to be

six minus t over three right let’s see here where we going

so let’s solve this for for uh so x is t so let’s move the t over

and so what do we get here let’s look at that let’s say t minus two over three

right if we we’re going to divide by the three so we’re going to say 2 minus t

so we’ll just say 3y here and that will be two

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minus t and then just simply divide by three okay so good so then so

right we just need the x and the y and so we can say here this will be just

two-thirds minus one-third t and so we can say the solutions are

let’s move me out of the way the solutions are

ordered pairs the x is the t this time and the y is two thirds minus one third t

where t is a arbitrary real number so this would be how i write the solution

set out this time now i said over here two zeros in in the

solution 2 0 is also in the solution over here isn’t it

what t would we choose well how about 2 if we choose t is two then what will

this one out here will be two thirds minus two thirds would be zero also

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so no matter if you represent the solution this way or if you represent

the solution this way it still comes out to be the same set of ordered pairs

so you could this could have been your written solution or this can in fact

there’s infinitely many ways to parameterize your solution set here

but in terms of augmented matrix this was our first augmented matrix

we obtained this augmented matrix here all zero so we’re going to

get some free variables here all right so let’s look at the next

example and in this example here how many equations and how many unknowns do

we have so we have three equations and we have four unknowns

and so let’s write down the augmented matrix now when i write down the augmented

matrix here there might be two ways just kind of two natural ways to do this

you could say w x y z and order them that way but i’m actually going to use a

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different slightly different way i’m going to say x y z w so i’m going to say

w is the fourth variable here so under that convention i’m going to say my

augmented matrix here is 1 2 2 and then minus 1 1 1 for the y’s

and then for the z’s i’m going to have 2 0 0

and for the w so i’m going to have minus 2 3 3 and then i’m going to have 1 4 6.

and that’s going to assume x y z w is my ordering there

so that may not be the ordering that everyone wants to use but that’s what

i’m going to use here all right so now let’s go about trying

to solve this system here so what i’m going to do is i’m going to multiply

the first equation by let’s see here i’m going to multiply the

first equation by well i’m going to keep the second and third equation the same

but i’m going to multiply the second equation by

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minus 2. so i’m going to say a minus two x and then plus two y minus four z

plus four w equals minus two and then i’m gonna keep the second and third

the same so i’m going to say 2x plus y plus 3w which i’ll put over here

and then that’s equal to 4. and then two x plus y and then plus three w

equals six and so that could be the next system that you write there um

but actually looking at these two right here um

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we actually see that these are the same equation right here right so when i if i

were to look at this augmented matrix right here 2 1 1 -4 0 0 4 3 3 -2 4 6

my augmented matrix right there and then now what i’m going to do is i’m

going to add up the x’s together and let’s see what we get here we’re going

to get a minus 2x plus 2y minus 4z e plus 4w i’m going to keep the same

first equation here and then i’m going to add so i’m going to get 3y minus 4z

and plus 7w equals 2 and then 3y minus 4z plus 7w equals four

so this might be your next system right here

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and what did we do we eliminated the x’s so we got zeros here the more zeros you

can get in your system the better so then we have two three three

minus four minus four minus four four seven seven minus two two and four

so eventually we’re going to realize that we’re having a problem right here

but let’s see here what are we going to do now so now i’m going to [Music]

so when i see this right here this is four and this right here is six so

what’s going to happen here is there’s not going to be this is this

system here is inconsistent so there’s not going to be any x and y and w

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that you can put in to get out 4 and to get the same x y w out and get a six

so this system is inconsistent right here um so can you have seen that from the

augmented matrix we have this row right here is four and

we have this row right here is six and they exactly match um so

you know what we can try to do is subtract them and think about that as one

minus one two minus two and then one and now i’m going to do two minus two

i’m gonna keep the same second same one here

and so now i’m gonna subtract two minus two zero one minus one is zero

zero minus zero three minus three and then four minus six here which is

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minus two and so this last row comes out to be all zero so it says this last row

corresponds to zero x plus zero y plus zero z plus zero w equals in fact

minus two oops it says see that right there and so that right there you can see

that it’s inconsistent right there so you know you may go on and do lots of

work because it may not be so obvious but what happens is eventually you will

get down to this you’ll get down to this row right here

and you’ll get something that makes no sense and so then you have to say the

system is inconsistent there is no solution right here so you know we

eliminated the x’s and you can go on and you know

subtract them right here and you’ll get the row here

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that you know if you if you keep working here we’d get the augmented matrix here

and then subtract them you’re going to get 0 0 0 0

and by subtracting them we’re going to get two minus four so we’ll get minus two

so you know sometimes you can see it right off the bat but more often than not

you cannot tell from the actual beginning system

but as you work down through the system trying to eliminate variable after

variable you can eventually run into something looks like this

that corresponds to an equation that doesn’t make any sense and then from

there you can conclude that the system is inconsistent there’s no solution to

this system right here is inconsistent okay so

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