Name: Cramer's rule, explained geometrically | Essence of linear algebra, chapter 12
Uploaded: 2021-02-16T09:26:11.000Z
Duration: 12 min 12 s
Description: 3blue1brown 線性代數精髓系列第 12 章 - Cramer 公式的幾何意義学べる英語：

In a previous video, I've talked about linear systems of equations, and I sort of brushed

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aside the discussion of actually computing solutions to these systems.

And while it's true that number-crunching is something we typically leave to the computers,

digging into some of these computational methods is a good litmus test for whether or not you

actually understand what's going on, since this is really where the rubber meets the

Here I want to describe the geometry behind a certain method for computing solutions to

The relevant background needed here is an understanding of determinants, dot products,

and of linear systems of equations, so be sure to watch the relevant videos on those

I should say up front that Cramer's rule is not the best way for computing solutions

Gaussian elimination, for example, will always be faster.

Think of this as a sort of cultural excursion; it's a helpful exercise in deepening your

knowledge of the theory of these systems.

Wrapping your mind around this concept will help consolidate ideas from linear algebra,

like the determinant and linear systems, by seeing how they relate to each other.

Also, from a purely artistic standpoint, the ultimate result is just really pretty to think

about, much more so that Gaussian elimination.

Alright, so the setup here will be some linear system of equations, say with two unknowns,

In principle, everything we're talking about will work systems with a larger number of

unknowns, and the same number of equations.

But for simplicity, a smaller example is nicer to hold in our heads.

So as I talked about in a previous video, you can think of this setup geometrically

as a certain known matrix transforming an unknown vector, [x; y], where you know what

the output is going to be, in this case [-4; -2].

Remember, the columns of this matrix tell you how the matrix acts as a transform, each

one telling you where the basis vectors of the input space land.

So this is a sort of puzzle, what input [x; y], is going to give you this

Remember, the  type of answer you get here can depend on

whether or not the transformation squishes all of space into a lower dimension.

In that case, either none of the inputs land on our given output or there are a whole bunch

But for this video we'll limit our view to the case of a non-zero determinant, meaning

the output of this transformation still spans the full n-dimensional space it started in;

every input lands on one and only one output and every output has one and only one input.

One way to think about our puzzle is that we know the given output vector is some linear

combination of the columns of the matrix; x*(the vector where i-hat lands) + y*(the

vector where j-hat lands), but we wish to compute what exactly x and y are.

As a first pass, let me show an idea that is wrong, but in the right direction.

The x-coordinate of this mystery input vector is what you get by taking its dot product

Likewise, the y-coordinate is what you get by dotting it with the second basis vector,

So maybe you hope that after the transformation, the dot products with the transformed version

of the mystery vector with the transformed versions of the basis vectors will also be

That'd be fantastic because we know the transformed versions of each of these vectors.

There's just one problem with this: it's not at all true!

For most linear transformations, the dot product before and after the transformation will be

For example, you could have two vectors generally pointing in the same direction, with a positive

dot product, which get pulled away from each other during the transformation, in such a

way that they then have a negative dot product.

Likewise, if things start off perpendicular, with dot product zero, like the two basis

vectors, there's no guarantee that they will stay perpendicular after the transformation,

In the example we were looking at, dot products certainly aren't preserved.

They tend to get bigger since most vectors are getting stretched.

In fact, transformations which do preserve dot products are special enough to have their

These are the ones which leave all the basis vectors perpendicular to each other with unit

You often think of these as rotation matrices.

The correspond to rigid motion, with no stretching, squishing or morphing.

Solving a linear system with an orthonormal matrix is very easy: Since dot products are

preserved, taking the dot product between the output vector and all the columns of your

matrix will be the same as taking the dot products between the input vector and all

the basis vectors, which is the same as finding the coordinates of the input vector.

So, in that very special case, x would be the dot product of the first column with the

output vector, and y would be the dot product of the second column with the output vector.

Now, even though this idea breaks down for most linear systems, it points us in the direction

of something to look for: Is there an alternate geometric understanding for the coordinates

of our input vector which remains unchanged after the transformation?

If your mind has been mulling over determinants, you might think of this clever idea: Take

the parallelogram defined by the first basis vector, i-hat, and the mystery input vector

The area of this parallelogram is its base, 1, times the height perpendicular to that

base, which is the y-coordinate of our input vector.

So, the area of this parallelogram is sort of a screwy roundabout way to describe the

vector's y-coordinate; it's a wacky way to talk about coordinates, but run with me.

Actually, to be more accurate, you should think of the signed area of this parallelogram,

in the sense described by the determinant video.

That way, a vector with negative y-coordinate would correspond to a negative area for this

Symmetrically, if you  look at the parallelogram spanned by the vector

and the second basis vector, j-hat, its area will be the x-coordinate of the vector.

Again, it's a strange way to represent the x-coordinate, but you'll see what it buys

Here's what this would look like in three-dimensions: Ordinarily the way you might think of one

of a vector's coordinate, say its z-coordinate, would be to take its dot product with the

But instead, consider the parallelepiped it creates with the other two basis vectors,