So in this video, we are going to solve a system of equations using row operations.
So here is the system of equations that we'll work within this first part of the video.
We're actually going to do three parts, and each part will have a slightly different system
of equations.
The first equation in the system of equations is minus 4 1 plus 7x2 minus 2x x3 equals 2.
Here is the next equation, and the final equations.
There are three equations and then three unknowns.
The unknowns being x1, x2, and x3.
So we're trying to determine, does this system of equations have a solution?
And the way that we're going to go about determining that is we're going to form what's called
the augmented matrix.
So when you form the augmented matrix, for each equation in your system, you have a row.
And the coefficients in the matrix are the coefficients of the unknown variables.
So I've put a minus 4, 7, and negative 2 because those are the coefficients of x1, x2, and
x3, in this first equation.
And it's an augmented matrix because this final column that I'll form right here consists
of the values on the right-hand side of the equation.
So 2, 3, and negative 4 are going to fill in that column.
So I do that for each equation, and when I'm done, I have what I call the augmented matrix.
And then, I'm going to determine if this system of equations has a solution now by performing
row operations on this augmented matrix.
So to perform row operations, there's a few things I can do.
I can always interchange rows.
I can always multiply a row by a number, and I can also add rows to other rows or linear
combinations of other rows.
So the strategy for doing this is we would like to get this augmented matrix into what
we call a reduced row echelon form, which will be more clear when we get to it.
But I'm going to try to get 1's along the diagonal of this matrix, so I can easily pick
out the values of the variable x1, x2, and x3.
So I would like to start off having a 1 right there, so I can do that easily if I switch
equations 1 in 2.
So this thing I've circled here, E1 double arrow E2, that just means interchange the
first and the second rows of this matrix.
So if I do that, I make the second row the first row, and I make the first row the second,
just switch them.
And now, I have a 1 right here like I would like.
So the next thing I would like to do is I would like to get some 0's right here.
Well, I can get a 0 in those entries if I add the second row to a multiple of the first
row, and if I add the third equation to a multiple of the first equation.
For instance, if I let the second equation equal 4 times the first equation plus the
second equation, you can see what happens.
4 times 1 is 4, and if I add that to a negative 4, I get 0.
So I'm going to replace equation 2 with 4 times equation 1 plus equation 2.
And if I do that, I'll get a nice simplification of the augmented matrix, and then do a similar
thing for equation 3.
I would like this to be 0.
Well, I can make this a 0 if I subtract off 2.
The way I can subtract off 2 is by multiplying the first equation by 2, and subtracting it
off.
So that's what I've done right here.
I'm going to replace the third equation with itself minus 2 times the first equation.
So if I do that, the first row, this first equation is going to be unchanged because
I'm not doing anything with equation 1, but the second equation, I'm going to have a 0
here.
The next entry is a negative 1 because remember what I'm doing?
I'm taking 7 minus 4 times a negative 2.
That's a negative 1.
The next entry will be negative 2 minus 4 times 1, which will give us-- I'm sorry, it
will be 4 times this plus this, which gives us 2.
And then finally, here, it's 4 times 3 is 12, plus 2 is 14.
So our strategy was getting a 0 here and we form whatever equation we need to get that
0 here, and then the rest of the values, you compute just the exact same way based on this
equation.
We did the same thing for the last equation.
And equation 3 is going to equal E3 minus 2E1.
So I get a 0 there, I'll get a 1 here, a negative 1 here, and then minus 10 here.
OK, so we're getting there.
We have a 1 here, and we have 0's below it.
So that's what I like to have in my augmented matrix.
Now, what I want to do is I would like a 1 here with 0 here.
Well, how can I do that?
If I replace E3 with E3 plus E2, basically, just add these two equations, and negative
1 plus 1 gives me a 0.
So let's replace the third equation with equation 3 plus equation 2, and then now I have a 0
right there just like I'd like.
Let's go ahead and replace E2 with the negative of E2.
So just multiply this whole row by negative 1.
It turns into there.
I'm not doing anything with E3, so that just stays alone.
Let's go ahead and replace E2 with E2 minus 2E3.
So I'm trying to get a 0 right here.
So equations 1 and 2 didn't change, but equation 2 changed, so I've done that computation.
Now, we're almost there.
I now have 1's down the diagonal, but I would really like to have 0's right here.
Well, let's go ahead and let E1 equal E1 plus two E2 minus E3.
So you can see what I'm doing.
I'm using this entry to cancel this, and I'm going to use this entry to cancel that.
So after I do that, equations 2 and 3 don't change, so I'll just recopy those down.
But equation 1 will now be equation 1 plus 2 times equation 2 minus equation 3.
But notice, there's all 0's here, so nothing happens in that first entry.
This will not be a 0 because I'm going to subtract off or add 2 times 1.
So negative 2 plus 2 is 0.
Same thing happens here.
This one and this one, they'll cancel, so I have a 0.
And then, this turns into a negative 13.
So at this point, now, I have basically solved this equation.
We look here, and this equation right here says that 0x1 plus 0x2 plus x3 equals 4.
Well, that's the same thing as x3 equals 4.
This equation says that 0x1 plus 1x2 plus x3 equals a negative 6, which is the same
thing as saying x2 equals a negative 6.
And this equation right here says, x1 plus 0x2 plus 0x3 equals a negative 13, which is
the same thing as saying, x1 equals a negative 13.
So we can actually just look at this and write down the solution to our equation.
So that is the solution of this system of linear equations.
We use the augmented matrix approach, and then we performed row operations to get this
answer.
This is the first video in a sequence of three videos.
In the next two videos, we'll look at a very similar system of equations, but something
that's different is going to happen.
This system of equations had a unique single solution.
In the following videos, we'll look at cases where we might not have any solutions, or
we might have an infinite number of solutions.
So those are coming up next.