节: 逆矩阵 | 10.10逆矩阵

章节大纲

Introduction
::导言

Cryptography, the study of encoding and decoding important information, is used today to protect data like credit card numbers, computer passwords, and electronic transactions. Before modern times, cryptography was primarily used as a means of encrypting messages that needed to remain secret.
::密码学是重要信息的编码和解码研究,如今被用于保护信用卡号码、计算机密码和电子交易等数据。在现代时代,加密学主要被用作加密需要保密的信息的手段。

The invention of computers has made breaking codes incredibly easy, so encryption and cryptography have become increasingly complex. One method of more advanced cryptography is the use of a matrix for encoding a message. By assigning each letter of the alphabet a value between 1 and 26, the message to be passed is translated to a series of numbers, with the zero being used as a space between words. The series of numbers is converted into a matrix with any extra spaces filled in with zeros . The matrix is then multiplied by an agreed-upon encoding matrix, and the coded matrix is sent. To decrypt the original message, the recipient needs to find the inverse of the encoded matrix and work backwards to decipher the message.
::计算机的发明使破解代码变得非常容易,因此加密和加密变得日益复杂。更先进的加密方法之一是使用矩阵来编码一个电文。如果给字母中的每个字母指定一个值在1到26之间,要传递的信息将被转换成一系列数字,零则用作字词之间的空格。数字序列被转换成一个矩阵,任何额外的空格都以零填满。然后将矩阵乘以一个商定的编码矩阵,并发送编码矩阵。要解密原始电文,收件人需要找到编码矩阵的反面,然后向后对电文进行解密。

Two numbers are multiplicative inverses if their product is 1, which is the multiplicative identity for the set of real numbers. Every number besides the number 0 has a multiplicative inverse . For matrices, two matrices are inverses of each other if they multiply to be the identity matrix .
::两个数字如果其产品是 1, 即真实数字组的多倍化特性, 则是多倍反函数。除数字 0 外, 每个数字都有多倍反函数。对于矩阵, 如果两个矩阵乘以身份矩阵, 则彼此反函数。

What kinds of matrices do not have inverses?
::哪些类型的矩阵没有反言?

Inverting Matrices
::倒转矩阵

Consider a matrix $\begin{aligned} A \end{aligned}$ that has inverse $\begin{aligned} A^{- 1} \end{aligned}$ . How do you find matrix $\begin{aligned} A^{- 1} \end{aligned}$ if you just have matrix $\begin{aligned} A \end{aligned}$ ?
::考虑一个具有A-1反差的矩阵A。如果只有矩阵A,你如何找到A-1?

$\begin{aligned} A = [\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix}], A^{- 1} = ? \end{aligned}$

We can find $\begin{aligned} A^{- 1} \end{aligned}$ by creating an of matrix $\begin{aligned} A \end{aligned}$ and the identity matrix seen below.
::通过建立矩阵A和身份矩阵,我们可以看到A-1。

$\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \end{matrix}] \end{aligned}$

The next step is to perform row reduction. The row reduction of this matrix can be found in Example 1 below. The right part of the augmented matrix is the inverse matrix $\begin{aligned} A^{- 1} \end{aligned}$ .
::下一步是执行行排减。此矩阵的排减可在下面的例1中找到。扩充矩阵的右部分是逆矩阵A-1。

$\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & 0 \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix} \end{matrix}] \\ A^{- 1} = [\begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & 0 \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix}] \end{aligned}$

Fractions are often unavoidable when computing inverses.
::当计算反向时,分数往往是不可避免的。

One reason why inverses are so powerful is that they allow you to solve systems of equations with the same logic as you would solve a single linear equation . Consider the following system based on the coefficients of matrix $\begin{aligned} A \end{aligned}$ from above:
::反向如此强大的一个原因就是它们允许你用与您相同的逻辑来解决单一线性方程式的方程式系统。根据以上矩阵A的系数来考虑以下系统:

$\begin{aligned} x + 2 y + 3 z & = 96 \\ x + 0 y + z & = 36 \\ 0 x + 2 y - z & = - 12 \end{aligned}$

By writing this system as a matrix equation , you get:
::将这个系统写成矩阵方程式后,您可以得到:

$\begin{aligned} [\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix}] \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ A \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \end{aligned}$

If this were a normal linear equation where you had a constant times the variable equals a constant, you would multiply both sides by the multiplicative inverse of the coefficient . Do the same in this case.
::如果这是正常的线性方程式, 变量的常数乘以常数, 变量等于常数, 则以乘法乘以乘以系数的倍数反差。在此情况下, 也这样做。

$\begin{aligned} A^{- 1} \cdot A \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] & = A^{- 1} \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] & = A^{- 1} \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \end{aligned}$

All that is left is for you to perform the matrix multiplication to get the solution. See Example 2.
::剩下的是您执行矩阵乘法以获得解决方案。请参见例2。

The following video demonstrates how to determine the inverse of a matrix using augmented matrices.
::以下视频展示了如何利用扩大的矩阵来确定矩阵的反向。

Examples
::实例

Example 1
::例1

Show the steps for finding the inverse matrix $\begin{aligned} A \end{aligned}$ :

::显示查找反矩阵A的步骤 :

$\begin{aligned} A = [\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix}], A^{- 1} = ? \end{aligned}$

Solution:
::解决方案 :

$\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \end{matrix}] \\ R_{1} \cdot - 1 + R_{2} \to & [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 0 & - 2 & - 2 \\ 0 & 2 & - 1 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \end{matrix}] \\ R_{2} + R_{3} \to & [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 0 & - 2 & - 2 \\ 0 & 0 & - 3 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - 1 & 1 & 1 \end{matrix} \end{matrix}] \\ R_{2} \div - 2, R_{3} \div - 3 \to & [\begin{matrix} \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} 1 & 0 & 0 \\ \frac{1}{2} & - \frac{1}{2} & 0 \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix} \end{matrix}] \\ R_{3} \cdot - 3 + R_{1}, R_{3} \cdot - 1 + R_{2} \to & [\begin{matrix} \begin{matrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} 0 & 1 & 1 \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix} \end{matrix}] \\ R_{2} \cdot - 2 + R_{1} \to & [\begin{matrix} \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix} \end{matrix}] \end{aligned}$

The matrix on the right is the inverse matrix $\begin{aligned} A^{- 1} \end{aligned}$ .
::右边的矩阵是逆矩阵A-1。

$\begin{aligned} A^{- 1} = [\begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix}] \end{aligned}$

Example 2
::例2

Solve the following system of equations using inverse matrices:
::使用反向矩阵解决下列方程式系统:

$\begin{aligned} [\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix}] \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \end{aligned}$

Solution:
::解决方案 :

Note: Use the inverse matrix you found in Example 1 since $\begin{aligned} A \end{aligned}$ is the same.
::注:使用例1中发现的反向矩阵,因为A是相同的。

$\begin{aligned} A \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ A^{- 1} \cdot A \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] & = A^{- 1} \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] & = A^{- 1} \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix}] \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} - \frac{1}{3} \cdot 96 + \frac{4}{3} \cdot 36 + \frac{1}{3} \cdot (- 12) \\ \frac{1}{6} \cdot 96 - \frac{1}{6} \cdot 36 + \frac{1}{3} \cdot (- 12) \\ \frac{1}{3} \cdot 96 - \frac{1}{3} \cdot 36 - \frac{1}{3} \cdot (- 12) \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] & = [\begin{matrix} 12 \\ 6 \\ 24 \end{matrix}] \end{aligned}$

Example 3
::例3

Find the inverse of the following matrix:
::查找以下矩阵的反向 :

$\begin{aligned} [\begin{matrix} 1 & 6 \\ 4 & 24 \end{matrix}] \end{aligned}$

Solution:
::解决方案 :

$\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 6 \\ 4 & 24 \end{matrix} | \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \end{matrix}] \\ R_{1} \cdot - 4 + R_{2} \to & [\begin{matrix} \begin{matrix} 1 & 6 \\ 0 & 0 \end{matrix} | \begin{matrix} 1 & 0 \\ - 4 & 1 \end{matrix} \end{matrix}] \end{aligned}$

This matrix is not invertible because its rows are not linearly independent . To test to see if a square matrix is invertible, check whether or not the determinant is zero. If the determinant is zero, then the matrix is not invertible because the rows are not linearly independent.
::此矩阵不是不可倒置的, 因为它的行并非线性独立。要测试一个平方矩阵是否是不可倒置的, 请检查决定数是否为零。如果决定因素为零, 那么矩阵就不会倒置, 因为行并非线性独立。

Example 4
::例4

Recall the question from the Introduction: What kinds of matrices do not have inverses?
::回顾导言中的问题:哪些类型的矩阵没有反差?

Solution:
::解决方案 :

Non- square matrices do not have inverses. Square matrices that have equal to zero do not have inverses.
::非平方矩阵没有反差。等于零的平方矩阵没有反差。

Example 5
::例5

1) Confirm matrix $\begin{aligned} A \end{aligned}$ and $\begin{aligned} A^{- 1} \end{aligned}$ are inverses by computing $\begin{aligned} A^{- 1} \cdot A \end{aligned}$ and $\begin{aligned} A \cdot A^{- 1} \end{aligned}$ .
::1) 确认矩阵A和A-1是反向的,计算为A-1-A和A-1-A。

$\begin{aligned} A = [\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix}], A^{- 1} = [\begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix}] \end{aligned}$

Solution:
::解决方案 :

$\begin{aligned} A^{- 1} \cdot A & = [\begin{matrix} - \frac{1}{3} & \frac{4}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & - \frac{1}{3} & - \frac{1}{3} \end{matrix}] \cdot [\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 0 & 2 & - 1 \end{matrix}] = \\ a_{11} & = - \frac{1}{3} \cdot 1 + \frac{4}{3} \cdot 1 + \frac{1}{3} \cdot 0 = 1 \\ a_{22} & = \frac{1}{6} \cdot 2 - \frac{1}{6} \cdot 0 + \frac{1}{3} \cdot 2 = 1 \\ a_{33} & = \frac{1}{3} \cdot 3 - \frac{1}{3} \cdot 1 - \frac{1}{3} (- 1) = 1 \end{aligned}$

Note that the rest of the entries turn out to be zero. This is left for you to confirm.
::请注意, 其余的条目结果为零。这留待您确认。

2) Use a calculator to compute $\begin{aligned} A^{- 1} \end{aligned}$ , compute $\begin{aligned} A^{- 1} \cdot A \end{aligned}$ , compute $\begin{aligned} A \cdot A^{- 1} \end{aligned}$ and compute
::2 (2) 使用计算器计算 A - 1, 计算 A - 1, 计算 A - 1, 计算 A - 1 A, 计算 A - A - 1, 计算

$\begin{aligned} A^{- 1} \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \end{aligned}$
Solution:

Start by entering just matrix $\begin{aligned} A \end{aligned}$ into the calculator.
::从输入矩阵A进入计算器开始。

To compute matrix $\begin{aligned} A^{- 1}, \end{aligned}$ use the Inverse button programmed into the calculator. Do not try to raise the matrix to the negative one exponent . This will not work.
::要计算矩阵 A - 1, 请在计算器中使用程序输入的反键。不要试图将矩阵提升到负引号中。这行不通。

Note that the calculator may return decimal versions of the fractions, and will not show the entire matrix on its limited display. You will have to scroll to the right to confirm that $\begin{aligned} A^{- 1} \end{aligned}$ matches what you have already found. Once you have found $\begin{aligned} A^{- 1}, \end{aligned}$ go ahead and store it as matrix $\begin{aligned} B \end{aligned}$ so you do not need to type in the entries.
::请注意计算器可以返回分数的十进制版本, 并且不会在其有限的显示中显示整个矩阵。您需要滚动到右侧以确认 A - 1 匹配您已经找到的内容。一旦找到 A - 1 , 请继续作为矩阵 B 存储, 这样您就不需要在条目中输入。

$\begin{aligned} A^{- 1} \cdot A = B \cdot A \end{aligned}$
::A - 1 - A = B = A = B = A

$\begin{aligned} A \cdot A^{- 1} = A \cdot B \end{aligned}$
::A = A = A = 1 = A = B = B = A = A = A = A = B

$\begin{aligned} A^{- 1} \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] = B \cdot [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] = B \cdot C \end{aligned}$
::A-1-[96-36-12] = B = B = [96-36-12] = B = B = C

You need to create matrix $\begin{aligned} C = [\begin{matrix} 96 \\ 36 \\ - 12 \end{matrix}] \end{aligned}$
::您需要创建矩阵 C = [96 36 - 12]

Using a calculator effectively should improve your understanding of matrices and allow you to check all the work you do by hand.
::有效使用计算器可以提高您对矩阵的理解,并允许您亲手检查您所做的所有工作。

3) The identity matrix happens to be its own inverse. Find another matrix that is its own inverse.
::3) 身份矩阵恰好是其自身的反面。另找一个反面矩阵。

Solution:
::解决方案 :

Mathematician Friedrich Helmert came up with a very clever matrix that happens to be its own inverse. Here are the $\begin{aligned} 2 \times 2 \end{aligned}$ and the $\begin{aligned} 3 \times 3 \end{aligned}$ versions:
::数学家Friedrich Helmert想出了一个非常聪明的矩阵, 恰好是它自己的反面。以下是 2 x 2 和 3 × 3 版本 :

$\begin{aligned} [\begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \end{matrix}], [\begin{matrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & - \frac{2}{\sqrt{6}} \end{matrix}] \end{aligned}$

Summary
::摘要

Multiplicative inverses are two numbers or matrices whose product is one or the identity matrix.
::倍增效应是产品为一个或身份矩阵的两个数字或矩阵。

Non-square matrices do not have inverses.
::非平方矩阵没有逆差。

Square matrices that have determinants equal to zero do not have inverses.
::具有等于零的决定因素的方位矩阵没有逆差。

Review
::回顾

Find the inverse of each of the matrices below, if possible. Make sure to do some by hand and some with your calculator.
::如果可能的话, 找出下面每个矩阵的反方向。请确保用手做一些, 用计算器做一些。

1. $\begin{aligned} [\begin{matrix} 4 & 5 \\ 2 & 3 \end{matrix}] \end{aligned}$

2. $\begin{aligned} [\begin{matrix} - 3 & 6 \\ 2 & 5 \end{matrix}] \end{aligned}$

3. $\begin{aligned} [\begin{matrix} - 1 & 2 \\ 2 & 0 \end{matrix}] \end{aligned}$

4. $\begin{aligned} [\begin{matrix} 1 & 6 \\ 0 & 1 \end{matrix}] \end{aligned}$

5. $\begin{aligned} [\begin{matrix} 6 & 5 \\ 2 & - 2 \end{matrix}] \end{aligned}$

6. $\begin{aligned} [\begin{matrix} 4 & 2 \\ 6 & 3 \end{matrix}] \end{aligned}$

7. $\begin{aligned} [\begin{matrix} - 1 & 3 & - 4 \\ 4 & 2 & 1 \\ 1 & 2 & 5 \end{matrix}] \end{aligned}$

8. $\begin{aligned} [\begin{matrix} 4 & 5 & 8 \\ 9 & 0 & 1 \\ 0 & 3 & - 2 \end{matrix}] \end{aligned}$

9. $\begin{aligned} [\begin{matrix} 0 & 7 & - 1 \\ 2 & - 3 & 1 \\ 6 & 8 & 0 \end{matrix}] \end{aligned}$

10. $\begin{aligned} [\begin{matrix} 4 & 2 & - 3 \\ 2 & 4 & 5 \\ 1 & 8 & 0 \end{matrix}] \end{aligned}$

11. $\begin{aligned} [\begin{matrix} - 2 & - 6 & - 12 \\ - 1 & - 5 & - 2 \\ 2 & 3 & 4 \end{matrix}] \end{aligned}$

12. $\begin{aligned} [\begin{matrix} - 2 & 6 & 3 \\ 2 & 4 & 0 \\ - 8 & 2 & 1 \end{matrix}] \end{aligned}$

13. Show that Helmert's $\begin{aligned} 2 \times 2 \end{aligned}$ matrix is its own inverse: $\begin{aligned} [\begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \end{matrix}] \end{aligned}$
::13. 显示赫尔默特的 2 × 2 矩阵是其本身的反方向: [1、2 2 2 2 - 1 2

14. Show that Helmert's $\begin{aligned} 3 \times 3 \end{aligned}$ matrix is its own inverse: $\begin{aligned} [\begin{matrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & - \frac{2}{\sqrt{6}} \end{matrix}] \end{aligned}$ .
::14. 显示赫尔默特的 3 × 3 3 矩阵是其本身的反方向: [1 3 1 3 1 3 1 2 - 1 2 0 1 6 6 - 2 6] 。

15. Non-square matrices sometimes have left inverses, where $\begin{aligned} A^{- 1} \cdot A = I \end{aligned}$ , or right inverses, where $\begin{aligned} A \cdot A^{- 1} = I \end{aligned}$ . Why can't non-square matrices have regular inverses?
::15. 非平方矩阵有时会左转,A-1-A=I或右反转,A-A-1=I。为什么非平方矩阵不能经常反转?

章节大纲

Introduction ::导言

Inverting Matrices ::倒转矩阵

Examples ::实例

Summary ::摘要

Review ::回顾

Introduction
::导言

Inverting Matrices
::倒转矩阵

Examples
::实例

Summary
::摘要

Review
::回顾