We're going to take a look at linear transformations again in this example.
It's similar to the previous video, But?
It's a little bit different.
In a previous video, we had a matrix A, and we were given a vector x, and we just had
to compute the linear transformation by computing the matrix product Ax.
In this problem, we still have a linear transform.
So we saw the linear transform LT, and it's equal to t of x equals A of x where the matrix
A is defined right here.
So is the same matrix we were dealing with.
In the previous video, the problem that we're ask to solve is just a little bit different
though.
We're actually told, hey, here's your vector b.
B is equal to a negative 2, 18, and negative 4.
And we are asked to find some matrix Ax, such that Ax equals b.
So give me a vector x such that after by matrix A operates on it, I get the vector b.
So it's phrase just a little bit differently in this problem.
Mathematically, we know how to do this.
It's just going to be solving a system of equations, so it's not too difficult.
But let's go ahead and see if we can find what vector x when operated on by A gives
us b.
Another question that we want to answer in this particular problem is x unique?
It could be that there are lots of different x's that when a operates on them, we get to
the vector b.
Or could be that there is just one vector x, or can be that there is none.
Who knows?
So let's take a look at this.
How are we going to do this?
Well, we can solve this problem by setting up a system of equations.
We know that we're going to let a operate on this unknown vector x whose entries will
denote x1, x2, and x3.
And that needs to result in the vector b, which is provided to us, a negative 2, 18,
and negative 4.
So this is just a system of equations.
So we're framing the problem a little bit differently.
We're thinking of it as a linear transformation taking a vector x, and turning it into a vector
b.
But mathematically once we construct the problem and set it up, it turns into solving the system
of equations like we've been doing in lots of these videos.
So I'm not going to go through all the details.
Let's just go ahead and construct the augmented matrix, the augmented matrix that we need
to solve is this matrix right here.
So it consists of this matrix, right here.
And then the vector b goes right here.
And this is the augmented matrix that we need to solve.
We know how to solve augmented matrices, we just perform row reductions.
So if we do row reductions, we can manipulate this augmented matrix into the following matrix.
So I've skipped all the row reduction steps, but we can turn it into this matrix right
here if we go through all those details.
Looking at this row reduce matrix, we now can just write down the solution.
This first row says that x1 is equal to 4.
The second row says that x2 is equal to 0.
And the third row says that x3 is equal to a negative 2.
So we've found that the vector x is the vector with entries 4, 0, and negative 2.
When a operates on this vector, it results in the vector b.
So we've found an x, such that Ax equals b.
That's what we were asked to do.
The next question though is, is x unique?
From this system of equations right here in the row reduced form of it, we see that there
is a single unique solution to this system of equations.
There's only one solution, the one that we found.
If we'd done row reductions and ended up with a row of zeros, then we would have had a free
variable.
And then it was turned out that there would be an infinite set of values x that all resulted
in the vector b.
That could have happened.
It didn't happen in this case.
In this case, we do have a unique vector x.
So in this case, we do have a unique vector x.
For other problems, you might end up with a free variable, and there might be many different
values of x that all get transformed into the same vector b.
That's a different case, but that could happen.
All right, that's the end of this video.
We took a look at the linear transformation Ax again.
We thought about it a little bit different way.
We thought about it as finding the vector that got transformed into the vector b when
we were provided the vector b.
And to solve that and to find the vector x, all we had to do was set up a system of equations
into our normal augmented matrix and row reduction operations.