课程： 1.5 高维矩阵 | The Way To Learn

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选择节高维矩阵

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高维矩阵
In this lecture we will be continuing the last lecture and discussing how to do some of the same operations on higher dimensional matrices such as 3x3 and 4x4 matrices.
::在这次演讲中,我们将继续最后一次讲座,并讨论如何对3x3和4x4等高维矩阵进行一些同样的操作。

The same properties hold as to how to do matrix addition and matrix multiplication , but as the dimensions get higher it is a bit harder to do.
::如何增加矩阵和矩阵乘法的属性相同,但随着尺寸越高,就越难做到。

Generalizing from the last lecture we have that each entry from each row $\begin{aligned} C_{i, j} \end{aligned}$ where $\begin{aligned} C = A \cdot B \end{aligned}$ is the $\begin{aligned} i \end{aligned}$ -th row of A dot product with the j-th column of B.
::从上次演讲中归纳出,每行Ci,j的每个条目,C=AB是A点产品的i行,B列为j-th。

This means that given
::这意味着给

$\begin{aligned} A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}] \\ B = [\begin{matrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{matrix}] \\ C = A \cdot B \\ C = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}] \cdot [\begin{matrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{matrix}] \\ C_{1, 1} = [\begin{matrix} a_{11} \\ a_{12} \\ a_{13} \end{matrix}] \cdot [\begin{matrix} b_{11} \\ b_{21} \\ b_{31} \end{matrix}] = [\begin{matrix} a_{11} b_{11} + a_{12} b_{21} + a_{13} b_{31} \end{matrix}] \\ \dots \\ C = [\begin{matrix} a_{11} b_{11} + a_{12} b_{21} + a_{13} b_{31} & a_{11} b_{12} + a_{12} b_{22} + a_{13} b_{32} & a_{11} b_{13} + a_{12} b_{23} + a_{13} b_{33} \\ a_{21} b_{11} + a_{22} b_{21} + a_{23} b_{31} & a_{21} b_{12} + a_{22} b_{22} + a_{23} b_{32} & a_{21} b_{13} + a_{22} b_{23} + a_{23} b_{33} \\ a_{31} b_{11} + a_{32} b_{21} + a_{33} b_{31} & a_{31} b_{12} + a_{32} b_{22} + a_{33} b_{32} & a_{31} b_{13} + a_{32} b_{23} + a_{33} b_{33} \end{matrix}] \end{aligned}$
::=C=C[a11b1121+a321+a1333]C[a11b11+a321+a33]C=ABBC=[a111212a13a13a21a22a22a23a23a23a3333]C1,1=[a11bbb12bb13bb21b22b22b23b23b32b32b3333]C1,1=[a11b21b21b31][b1121+a13b21bbb22b22bbbb23b33b33b3=a323221b3+23b33b3b

I included the ... just because the math is relatively simple and tedious to calculate these entries and I did not want to have you read me do that for all 9 entries in this 3x3 matrix.
::我加入了... 仅仅因为数学比较简单和无聊来计算这些条目我不想让你读我这个3x3矩阵中的所有9个条目

Let's do some multiplication problems now.
::让我们现在做一些乘法问题。

1.

$\begin{aligned} A = [\begin{matrix} 1 & 2 & - 3 \\ 4 & - 5 & 2 \\ 1 & - 3 & 5 \end{matrix}] \\ B = [\begin{matrix} - 1 & 2 & 3 \\ - 4 & 1 & 2 \\ 5 & 3 & - 7 \end{matrix}] \\ A \cdot B \\ = \\ [\begin{matrix} 1 & 2 & - 3 \\ 4 & - 5 & 2 \\ 1 & - 3 & 5 \end{matrix}] \cdot [\begin{matrix} - 1 & 2 & 3 \\ - 4 & 1 & 2 \\ 5 & 3 & - 7 \end{matrix}] \\ = \\ Applying the formula we had from earlier \\ [\begin{matrix} 6 & 13 & - 14 \\ 26 & 9 & - 12 \\ 36 & 14 & - 38 \end{matrix}] \end{aligned}$
::A=[12-34-521-35]B=[-123-4253-7]AB=[12-34-521-35]__[-123-41-253-7]=适用我们早先的公式[613-14269-123614-38]

2. Let's try another example:
::2. 让我们再举一个例子:

$\begin{aligned} A = [\begin{matrix} 1 & - 1 & 2 \\ 4 & 3 & - 1 \\ 1 & 5 & - 2 \end{matrix}] \\ B = [\begin{matrix} 4 & 2 & - 3 \\ 1 & - 2 & 6 \\ 2 & 1 & - 3 \end{matrix}] \\ A \cdot B \\ = \\ [\begin{matrix} 1 & - 1 & 2 \\ 4 & 3 & - 1 \\ 1 & 5 & - 2 \end{matrix}] \cdot [\begin{matrix} 4 & 2 & - 3 \\ 1 & - 2 & 6 \\ 2 & 1 & - 3 \end{matrix}] \\ = \\ [\begin{matrix} 7 & 6 & 15 \\ 17 & 1 & 9 \\ 5 & - 10 & 33 \end{matrix}] \end{aligned}$
::A=[1-1243-115-2,B=[42-31-2621-33]AB=[1-1-1243-115-2][42-1243-21-2621-3]=[76-1517195-1033]

3. Finally, let's try one more example:
::3. 最后,我们再举一个例子:

$\begin{aligned} A = [\begin{matrix} 2 & 3 & - 5 \\ 4 & - 1 & 2 \\ 7 & 8 & - 4 \end{matrix}] \\ B = [\begin{matrix} - 8 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 36 & 24 \end{matrix}] \\ A \cdot B \\ = \\ [\begin{matrix} 2 & 3 & - 5 \\ 4 & - 1 & 2 \\ 7 & 8 & - 4 \end{matrix}] \cdot [\begin{matrix} - 8 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 36 & 24 \end{matrix}] \\ = [\begin{matrix} - 42 & - 157 & - 113 \\ - 21 & 83 & 55 \\ - 60 & - 76 & - 74 \end{matrix}] \end{aligned}$
::A=[23-54-1278-4]B=[-842-33517376264]AB=[23-54-54-1278-44][-842-335-1773624][-42-157-113-113-218355-60-76-774]]

Now let's look at taking the inverse of a 3x3 matrix and derive the inverse matrix formula . We will learn this in a much simpler way once we use , but let's try and apply the skills that we have now.
::现在让我们来看看3x3矩阵的反向, 并得出反向矩阵公式。一旦我们使用, 我们将以简单得多的方式了解这一点, 但是让我们尝试运用我们现在拥有的技能。

I won't go through the full steps of derivation, but I will take you through it.
::我不会走整个步骤的衍生, 但我会带你通过它。

Given a matrix $\begin{aligned} A \end{aligned}$ , it's inverse is the matrix $\begin{aligned} B \end{aligned}$ such that $\begin{aligned} A \cdot B = I \end{aligned}$ where $\begin{aligned} I \end{aligned}$ is the identity matrix .
::从矩阵A来看, 矩阵B是相反的, 因此AB=I代表我的身份矩阵。

So let's say $\begin{aligned} A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}] \\ B = [\begin{matrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{matrix}] \\ A \cdot B = [\begin{matrix} a_{11} b_{11} + a_{12} b_{21} + a_{13} b_{31} & a_{11} b_{12} + a_{12} b_{22} + a_{13} b_{32} & a_{11} b_{13} + a_{12} b_{23} + a_{13} b_{33} \\ a_{21} b_{11} + a_{22} b_{21} + a_{23} b_{31} & a_{21} b_{12} + a_{22} b_{22} + a_{23} b_{32} & a_{21} b_{13} + a_{22} b_{23} + a_{23} b_{33} \\ a_{31} b_{11} + a_{32} b_{21} + a_{33} b_{31} & a_{31} b_{12} + a_{32} b_{22} + a_{33} b_{32} & a_{31} b_{13} + a_{32} b_{23} + a_{33} b_{33} \end{matrix}] \end{aligned}$
::所以让我们说A=[a11a12a13a13a21a22a23a31a33]B=[b11b12bb21b22b2222b23b23b23b23b2323b33]A=B=[a11b11+a12b21+a12b21+a13b21b+a12bb22+a13b32a1113+a12b23b23+a13b21b21b21b21b23+a2222b22+a2323b13+a23b33b33ab31b11+a32b21+a33b31b12+a32b22+a33b3222+a33b3222+a33b32b23b23+a33]

Now we want to find all of the numbers in B such that the entries of A*B is equal to the identity matrix and the entries of B are written in terms of A. From here, we get a system of 9 linear equations with 9 variables, which though quite tedious to solve is nevertheless solvable.
::现在我们想要找到 B 中的所有数字, 这样 A* B 的条目等于身份矩阵, 而 B 的条目是用 A 写成的。从这里, 我们得到一个有 9 个线性方程式的系统, 包含 9 个变量, 虽然解决这些变量很无聊, 但还是可以解决。

After solving this out we end up getting that
::解决了这个问题之后我们最终会得到这个

$\begin{aligned} B \\ = \\ A^{- 1} \\ = \\ \frac{1}{a_{11} a_{22} a_{33} - a_{11} a_{32} a_{23} - a_{21} a_{33} a_{12} + a_{21} a_{32} a_{13} + a_{31} a_{23} a_{12} - a_{31} a_{22} a_{13}} \cdot [\begin{matrix} a_{33} a_{22} - a_{32} a_{23} & a_{32} a_{13} - a_{33} a_{12} & a_{23} a_{12} - a_{22} a_{13} \\ a_{31} a_{23} - a_{33} a_{21} & a_{33} a_{11} - a_{31} a_{13} & a_{21} a_{13} - a_{23} a_{11} \\ a_{32} a_{21} - a_{31} a_{22} & a_{31} a_{12} - a_{32} a_{11} & a_{22} a_{11} - a_{21} a_{12} \end{matrix}] \end{aligned}$
::B=A-1=1a11a22a33-a11a32a23-a21a33a12+a21a32a32a13+a31a23a12a-a31a22a13+a31a23a12a13a[a33a22a23a23a13a23a23a13a23aa13a21a11aa31aa13a13a13a13a23aaa21a21a21a21a21a21a21a31a22aa12a-a31a32a21a11a11a21a12]

Now that we've seen this definition outright, let's do one example and I'll leave the rest for you to do yourself.
::既然我们直截了当地看到了这个定义, 让我们举一个例子, 我剩下的留给你自己来做。

1. Calculate $\begin{aligned} ([\begin{matrix} 1 & 3 & 2 \\ - 1 & 4 & 5 \\ - 7 & 6 & 2 \end{matrix}])^{- 1} Applying the formula from earlier, we get that the answer is [\begin{matrix} \frac{2}{7} & \frac{- 6}{77} & \frac{- 1}{11} \\ \frac{3}{7} & \frac{- 16}{77} & \frac{1}{11} \\ \frac{- 2}{7} & \frac{27}{77} & \frac{- 1}{11} \end{matrix}] \end{aligned}$
::1. 计算([132-145-762])-1 应用早先的公式,我们得到的答案是[27-677-11137-1677-1111-272777-111]

Videos:
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Problems
::问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题、问题

Calculate
::计算计算

$\begin{aligned} 1. ([\begin{matrix} 1 & - 2 & 3 \\ 4 & 2 & - 5 \\ - 7 & 6 & 8 \end{matrix}])^{- 1} \\ 2. ([\begin{matrix} 7 & - 1 & 2 \\ - 4 & 5 & 6 \\ - 3 & 1 & 12 \end{matrix}])^{- 1} \\ 3. ([\begin{matrix} 3 & - 4 & 5 \\ - 6 & 8 & - 1 \\ 7 & 2 & - 1 \end{matrix}])^{- 1} \\ 4. ([\begin{matrix} 2 & - 1 & 6 \\ 1 & 4 & - 3 \\ 8 & 2 & - 1 \end{matrix}])^{- 1} \\ 5. ([\begin{matrix} 5 & - 2 & 6 \\ 9 & 19 & 13 \\ 2 & 14 & 8 \end{matrix}])^{- 1} \end{aligned}$

章节大纲

常规

高维矩阵