Invertible matrices and transformations | Matrices | Precalculus | Khan Academy

We have two two by two matrices here. In other videos, we talk about how a two by two matrix can represent a transformation of the coordinate plane, of the two-dimensional plane, where this, of course, is the x-axis, and this, of course, is the y-axis.

What we're going to do in this video is visualize these transformations and get a visual understanding for why it's reasonable for A to have an inverse, or why matrix A is invertible, and why it's not reasonable for B to have an inverse matrix, or why matrix B is not invertible.

So just as a reminder, these transformation matrices essentially tell us what to do with our unit vectors. For example, we have the (1, 0) unit vector, and the first column of each of these matrices tells us what the (1, 0) unit vector, the vector that goes one unit in the x direction, will get transformed to under each of these transformations. Then, of course, you have the unit vector in the y direction, the (0, 1) unit vector, and the second columns here tell us how we would transform that.

So let's first think about matrix A. Matrix A transforms the (1, 0) vector into the (2, 1) vector. So the (2, 1) vector is going to look something like that. It transforms the (0, 1) vector into the (2, 3) vector. So the (2, 3) vector is going to look something like this.

One way to think about it: instead of our grid looking like this grid that I already had here, which is just our standard coordinate axes, this is defining a new grid that would look like this. I define this new grid by looking at our multiples of what the (1, 0) vector has been transformed into, and we also look at multiples of what the (0, 1) vector has been transformed into.

For example, if I were to take this point right here, before it is transformed, it has one of each of these vectors. Well, under transformation, under this first transformation, it would be one of each of these vectors. So it's going to be (1, 2) vector plus (1, 2, 3) vector. So this point will be mapped to this point.

By that same logic, this point right over here, which is one unit more in the x direction, well, now it's going to have one unit more in the direction that x, the x unit vector, has been transformed into. This point right over here, by the same logic, will be one unit more in the direction that the y unit vector has been transformed into, so it'll be there.

At this point, by the same logic, will be transformed into that point. This region that I'm showing you in white will get transformed to this region. Now, there's some obvious things going on here. We have a two-dimensional area that has been transformed to another two-dimensional area, and in fact, it looks like it has been scaled up.

In other videos, we have talked about that this scale factor is going to be the absolute value of the determinant of A. It's clear that, frankly, not only is this non-zero, but it's going to be greater than one. It looks like we're scaling up our area. But the very fact that this does not equal zero tells us that we are scaling from a two-dimensional area to another two-dimensional area, and so it's completely reasonable to be able to go back.

You will for sure be able to find a transformation that takes you from this region to that region. So that makes us feel good that A inverse is reasonable.

Now, as a point of contrast, let's think about matrix B right over here. Matrix B transforms the (1, 0) unit vector into the (2, 1) unit vector, so it transforms it into this unit vector, actually very similarly to how A did it. But then let's see what it does to the (0, 1) vector: it transforms this vector into the (4, 2) vector.

Notice the (4, 2) vector is just 2 of the (2, 1) vectors. It's going in the same direction; it just has a different size or a different length or a different magnitude. So in this situation, it's taking things in two-dimensional space and it's turning them into combinations of things that sit on this same direction.

So everything in two dimensions is going to be mapped to something along this line here. This region, if I were to apply transformation B, is going to be a region on this line over here. Anything, if I were to map it using matrix B, is going to get mapped onto this line. So you're going to something with area, and then you're mapping it to something that has no area. So the scaling factor here must be zero.

We know, since the absolute value of the determinant of B is zero, we could also fairly say that the determinant of B is equal to zero. How are you going to have an inverse matrix? How are you going to have something that can scale from 0 area to something that has area?

So we know that B inverse does not exist. There's nothing that can transform us back. There are a couple of things here: this reinforces the idea that if your determinant of a transformation matrix is zero, you're not going to have an inverse; it's not invertible.

The other thing to recognize is seeing the patterns in the matrix itself. Here we saw that the second column is just a multiple of the first column. It's twice the first column: 2 times 2 is 4; 1 times 2 is 2. You can also view it the other way around: the first row is a multiple of the second row. You can play around with the math if you wanted to generalize it and see if that's the case; the determinant will always be 0.

That's because if you view them as transformations, they're all going to map to a line, and you're going to lose all your area. If you view it as a representation for at least the left-hand side of a system of equations, you can think about it as lines that have the same slope. But we talk about that in other videos.