10.7 强化儿科

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  Introduction
  ::导言

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  The reason why the rules for row reducing matrices are the same as the rules for eliminating coefficients when solving a system of equations is because you are essentially doing the same thing in each case. When you write and rewrite the equation every time, you end up writing down lots of extra information. Matrices take care of this information by embedding it in the location of each entry. How would you use matrices to write the system of equations below?
  ::为何行排减缩矩阵的规则与解决方程式体系时消除系数的规则相同? 原因是您在每种情况下基本上都做同样的事情。 每次写和重写方程式时,最后都会写出大量额外信息。 矩阵通过将信息嵌入每个条目的位置来照顾这些信息。 您如何使用矩阵来写入下面的方程式系统?

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  $$\begin{array}{rl}5x+y& =6\\ x+y& =10\end{array}$$

  
  ::5x+y=6x+y=10 5x+y=6x+y=10

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  Augmented Matrix
  ::增强矩阵

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  To represent a system as a matrix equation , first write all the equations in standard form so that the coefficients of the variables line up in columns. Then copy down just the coefficients in a matrix array. Next, copy the variables in a variable matrix and the constants into a constant matrix.
  ::要将一个系统作为矩阵方程式代表, 请首先以标准格式将所有方程式写入标准格式, 这样就可以将变量的系数排成列列。 然后复制一个矩阵数组中的系数。 然后, 在变量矩阵和常数中将变量复制成一个常数矩阵 。

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  $$\begin{array}{rl}x+y+z& =9\\ x+2y+3z& =22\\ 2x+3y+4z& =31\end{array}$$

  
  ::x+y+z=9x+2y+3z=222x+3y+4z=31

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  $$\begin{array}{r}\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 2& 3& 4\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}9\\ 22\\ 31\end{array}\right]\end{array}$$

  
  ::[111123234] [xyz]=[92231]

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  The reason why this works is because of the way matrix multiplication is defined.
  ::之所以这样做,是因为矩阵乘法的定义。

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  $$\begin{array}{r}\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 2& 3& 4\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1x+1y+1z\\ 1x+2y+3z\\ 2z+3y+4z\end{array}\right]=\left[\begin{array}{c}9\\ 22\\ 31\end{array}\right]\end{array}$$

  
  ::[111123234] [xyz] =[1x+1y+1y+1z1x+2y+3z2z+3y+4z] =[92231]

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  Notice how putting brackets around the two matrices on the right does very little to hide the fact that this is just a regular system of three equations and three variables. 
  ::注意在右侧两个矩阵上加括号会如何掩盖这样一个事实,即这只是由三个方程和三个变量组成的常规系统。

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  Once you have your system represented as a matrix, you can solve it using an augmented matrix. An augmented matrix is two matrices that are joined together and operated on as if they were a single matrix. In the case of solving a system, you need to augment the coefficient matrix and the constant matrix. The vertical line indicates the separation between the coefficient matrix and the constant matrix.
  ::一旦您将您的系统作为矩阵来代表, 您可以使用一个扩大的矩阵来解析它。 增强的矩阵是两个矩阵, 它们合并在一起, 并像一个矩阵一样运行。 在解决一个系统时, 您需要增加系数矩阵和恒定矩阵。 垂直线表示系数矩阵和恒定矩阵之间的分离 。

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  $$\begin{array}{r}\left[\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 2& 3& 4\end{array}|\begin{array}{c}9\\ 22\\ 31\end{array}\end{array}\right]\\ R_1\cdot -1+R_2,R_1\cdot -2+R_3\to & \left[\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 1& 2\end{array}|\begin{array}{c}9\\ 13\\ 13\end{array}\end{array}\right]\\ R_2\cdot -1+R_3\to & \left[\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 0& 0\end{array}|\begin{array}{c}9\\ 13\\ 0\end{array}\end{array}\right]\end{array}$$

  
  ::[112323492231]R11+R2,R12+R3}[111012012913]R21+R3}[1110120009130]

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  Reduce the matrix to reduced row echelon form and you will find the solution to the system, if one exists.
  ::将矩阵减为排梯格式,如果系统存在,您将找到系统解决方案。

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  The following video demonstrates  how to transform an augmented matrix to row echelon form to solve a system of equations:
  ::以下视频展示了如何将扩大的矩阵转换成排梯式,以解决方程式系统:

   

   

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  Examples
  ::实例

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  Example 1
  ::例1

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  Solve the following system using an augmented matrix: 
  ::使用扩大的矩阵表解决以下系统:

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  $$\begin{array}{rl}x+y+z& =6\\ x-y-z& =-4\\ x+2y+3z& =14\end{array}$$

  
  ::x+y+z=6x-y-z4x+2y+3z=14

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  Solution:
  ::解决方案 :

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  $$\begin{array}{rl}& \left[\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 1& -1& -1\\ 1& 2& 3\end{array}|\begin{array}{c}6\\ -4\\ 14\end{array}\end{array}\right]\\ R_1\cdot -1+R_2,R_1\cdot -1+R_3\to & \left[\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 0& -2& -2\\ 0& 1& 2\end{array}|\begin{array}{c}6\\ -10\\ 8\end{array}\end{array}\right]\\ R_3\cdot -1+R_1,R_3\cdot 3+R_2\to & \left[\begin{array}{c}\begin{array}{ccc}1& 0& -1\\ 0& 1& 4\\ 0& 1& 2\end{array}|\begin{array}{c}-2\\ 14\\ 8\end{array}\end{array}\right]\\ R_2\cdot -1+R_3\to & \left[\begin{array}{c}\begin{array}{ccc}1& 0& -1\\ 0& 1& 4\\ 0& 0& -2\end{array}|\begin{array}{c}-2\\ 14\\ -6\end{array}\end{array}\right]\\ R_3÷-2\to & \left[\begin{array}{c}\begin{array}{ccc}1& 0& -1\\ 0& 1& 4\\ 0& 0& 1\end{array}|\begin{array}{c}-2\\ 14\\ 3\end{array}\end{array}\right]\\ R_3+R_1,R_3\cdot -4+R_2\to & \left[\begin{array}{c}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{c}1\\ 2\\ 3\end{array}\end{array}\right]\end{array}$$

  
  ::[1111-1-1-1-11-1123}6-414]R1-11+R2、R1-1+R3}[1110-2-2012__6-108]R3}1+R1、R3}3}3+R2}[10-10140122148]R2}1+R3}[10-101400-2__2}214-6]R3}[10-1014000012}2}[10-10140.0012143]R3+R1,R3__4+R2}[1001001}123]

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  Every matrix can be interpreted as its own linear system. The final augmented matrix can be interpreted as:
  ::每一矩阵可被解释为其本身的线性系统。

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  $$\begin{array}{rl}1x+0y+0z& =1\\ 0x+1y+0z& =2\\ 0x+0y+1z& =3\end{array}$$

  
  ::1x+0y+0z=10x+1y+0z=20x+0y+1z=3

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  Thus,  $\begin{array}{r}x=1,y=2,\text{ and }z=3.\end{array}$
  ::因此,x=1,y=2,z=3。

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  Example 2
  ::例2

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  Attempt to solve the system from the first section. 
  ::试图从第一部分解析系统 。

  $$\begin{array}{r}\left[\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 2& 3& 4\end{array}|\begin{array}{c}9\\ 22\\ 31\end{array}\end{array}\right]\end{array}$$

   

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  Solution:
  ::解决方案 :

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  $$\begin{array}{rl}& \left[\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 2& 3& 4\end{array}|\begin{array}{c}9\\ 22\\ 31\end{array}\end{array}\right]\\ R_1\cdot -1+R_2,R_1\cdot -2+R_3\to & \left[\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 1& 2\end{array}|\begin{array}{c}9\\ 13\\ 13\end{array}\end{array}\right]\\ R_2\cdot -1+R_3\to & \left[\begin{array}{c}\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 0& 0\end{array}|\begin{array}{c}9\\ 13\\ 0\end{array}\end{array}\right]\end{array}$$

  
  ::[112323492231]R11+R2,R12+R3}[111012012913]R21+R3}[1110120009130]

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  This system is dependent , so there are infinitely many solutions.
  ::这个系统是依赖性的,所以有无数的解决方案。

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  Example 3
  ::例3

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  Solve the following system using augmented matrices:
  ::使用扩充矩阵解决以下系统:

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  $$\begin{array}{rl}w+x+z& =11\\ w+x& =9\\ x+y& =7\\ y+z& =5\end{array}$$

  
  ::w+x+z=11w+x=9x+y=7y+z=5

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  Solution:
  ::解决方案 :

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  While substitution would work in this problem, the idea is to demonstrate how augmented matrices will work even with larger matrices. 
  ::虽然替代在这一问题上是可行的,但设想是要表明,即使使用较大的矩阵,扩大的矩阵将如何运作。

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  $$\begin{array}{rl}& \left[\begin{array}{c}\begin{array}{cccc}1& 1& 0& 1\\ 1& 1& 0& 0\\ 0& 1& 1& 0\\ 0& 0& 1& 1\end{array}|\begin{array}{c}11\\ 9\\ 7\\ 5\end{array}\end{array}\right]\\ \text{Switch }{R}_2,R_3\text{ and }{R}_4\to & \left[\begin{array}{c}\begin{array}{cccc}1& 1& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 1\\ 1& 1& 0& 0\end{array}|\begin{array}{c}11\\ 7\\ 5\\ 9\end{array}\end{array}\right]\\ R_1\cdot -1+R_4\to & \left[\begin{array}{c}\begin{array}{cccc}1& 1& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 1\\ 0& 0& 0& -1\end{array}|\begin{array}{c}11\\ 7\\ 5\\ -2\end{array}\end{array}\right]\\ R_4\cdot -1\to & \left[\begin{array}{c}\begin{array}{cccc}1& 1& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 1\\ 0& 0& 0& 1\end{array}|\begin{array}{c}11\\ 7\\ 5\\ 2\end{array}\end{array}\right]\\ R_4\cdot -1+R_1,R_4\cdot -1+R_3\to & \left[\begin{array}{c}\begin{array}{cccc}1& 1& 0& 0\\ 0& 1& 1& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}|\begin{array}{c}9\\ 7\\ 3\\ 2\end{array}\end{array}\right]\\ R_3\cdot -1+R_2\to & \left[\begin{array}{c}\begin{array}{cccc}1& 1& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}|\begin{array}{c}9\\ 4\\ 3\\ 2\end{array}\end{array}\right]\\ R_2\cdot -1+R_1\to & \left[\begin{array}{c}\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}|\begin{array}{c}5\\ 4\\ 3\\ 2\end{array}\end{array}\right]\end{array}$$

  
  ::[11010001000100011111975] Switch R2、R3和R4}[1101000100011110011759]R1}1+R4}[1101000100010001000-1000-1}1}1}[11010001175-2}R4}1}1}[11010001000100010001}11752}R41+R1}[1110001000100010001000100010010010019732]

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  Thus,  $\begin{array}{r}w=5,x=4,y=3,\text{ and }z=2.\end{array}$
  ::因此, w=5,x=4,y=3,z=2。

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  Example 4
  ::例4

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  Recall the problem from the Introduction. H ow would you use matrices to write the system of equations below?
  ::回顾导言中的问题。 您将如何使用矩阵来写出下面的方程式系统 ?

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  $$\begin{array}{rl}5x+y& =6\\ x+y& =10\end{array}$$

   
  ::5x+y=6x+y=10 5x+y=6x+y=10

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  Solution: 
  ::解决方案 :

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  $$\begin{array}{rl}\left[\begin{array}{cc}5& 1\\ 1& 1\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}6\\ 10\end{array}\right]\end{array}$$

  
  ::[5111] [xy] = [610]

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  Example 5
  ::例5

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  Use an augmented matrix to solve the following system:
  ::使用扩大的矩阵解决以下系统:

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  $$\begin{array}{rl}2x+y+z& =16\\ 2y+6z& =0\\ x+y& =10\end{array}$$

  
  ::2x+y+z=162y+6z=0x+y=10

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  Solution:
  ::解决方案 :

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  Only the initial and final augmented matrices are shown. T he row reduction steps are not shown. 
  ::仅显示初始和最后的扩展矩阵。不显示减排步骤。

  $$\begin{array}{r}\left[\begin{array}{c}\begin{array}{ccc}2& 1& 1\\ 0& 2& 6\\ 1& 1& 0\end{array}|\begin{array}{c}16\\ 0\\ 10\end{array}\end{array}\right]\to \left[\begin{array}{c}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{c}7\\ 3\\ -1\end{array}\end{array}\right]\end{array}$$

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  Example 6
  ::例6

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  Use an augmented matrix to solve the following system:
  ::使用扩大的矩阵解决以下系统:

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  $$\begin{array}{rl}3x+y& =-15\\ x+2y& =15\end{array}$$

  Solution:
  ::3+y15x+2y=15 解析器 :

  $$\begin{array}{r}\left[\begin{array}{c}\begin{array}{cc}3& 1\\ 1& 2\end{array}|\begin{array}{c}-15\\ 15\end{array}\end{array}\right]\to \left[\begin{array}{c}\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{c}-9\\ 12\end{array}\end{array}\right]\end{array}$$

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  Example 7
  ::例7

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  Use an augmented matrix to solve the following system:
  ::使用扩大的矩阵解决以下系统:

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  $$\begin{array}{rl}-a+b-c& =0\\ 2a-2b-3c& =25\\ 3a-4b+3c& =2\end{array}$$

  
  ::-a+b-c=02a-2b-3c=253a-4b+3c=2

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  Solution:
  ::解决方案 :

  $$\begin{array}{r}\left[\begin{array}{c}\begin{array}{ccc}-1& 1& -1\\ 2& -2& -3\\ 3& -4& 3\end{array}|\begin{array}{c}0\\ 25\\ 2\end{array}\end{array}\right]\to \left[\begin{array}{c}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{c}3\\ -2\\ -5\end{array}\end{array}\right]\end{array}$$

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  Summary
  ::摘要

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  • An augmented matrix is a matrix formed when two matrices are joined together and operated on as if they were a single matrix. 
    ::扩大的矩阵是一个矩阵,在将两个矩阵合并在一起并运行时组成一个矩阵,似乎它们是一个单一的矩阵。
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  • In the case of solving a system, you need to augment the coefficient matrix and the constant matrix.
    ::在解决系统时,您需要增加系数矩阵和恒定矩阵。
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  • The vertical line indicates the separation between the coefficient matrix and the constant matrix.
    ::垂直线表示系数矩阵和恒定矩阵之间的分离。
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  • Reduce the matrix to reduced row echelon form and you will find the solution to the system, if one exists.
    ::将矩阵减为排梯格式,如果系统存在,您将找到系统解决方案。

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  Review
  ::回顾

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  Solve the systems of equations below using augmented matrices. If one solution does not exist, explain why not.
  ::使用增强的矩阵解决下面的方程式系统。 如果没有一个解决方案,请解释为什么没有。

  1.

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  $$\begin{array}{rl}4x-2y& =-20\\ x-3y& =-15\end{array}$$

   
  ::4 - 2y 20x - 3y 15

  2. 

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  $$\begin{array}{rl}3x+5y& =33\\ -x-2y& =-13\end{array}$$

  
  ::3x+5y=33 -x-2y13

  3.

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  $$\begin{array}{rl}x+4y& =11\\ 3x+12y& =33\end{array}$$

  
  ::x+4y=113x+12y=33

  4.

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  $$\begin{array}{rl}-3x+y& =-7\\ -x+4y& =5\end{array}$$

  
  ::- 3x+y7-x+4y=5

  5.

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  $$\begin{array}{rl}3x+y& =6\\ -6x-2y& =10\end{array}$$

  
  ::3x+y=6-6x-2y=10 3x+y=6-6x-2y=10

  6.

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  $$\begin{array}{rl}2x-y+z& =4\\ 4x+7y-z& =38\\ -x+3y+2z& =23\end{array}$$

  
  ::2-y+z=44x+7y-z=38-x+3y+2z=23

  7.

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  $$\begin{array}{rl}4x+y-z& =-16\\ -3x+4y+z& =18\\ x+y-3z& =-17\end{array}$$

  
  ::4x+y-z16-3x+4y+z=18x+y-3z17

  8. 

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  $$\begin{array}{rl}3x+2y-3z& =7\\ -x+5y+2z& =29\\ x+2y+z& =15\end{array}$$

  
  ::3x+2y-3z=7-x+5y+2z=29x+2y+z=15

  9. 

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  $$\begin{array}{rl}2x+y-2z& =4\\ -4x-2y+4z& =-8\\ 3x+y-z& =5\end{array}$$

  
  ::2x+y-2z=4-4x-2y+4z83x+y-z=5

  10. 

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  $$\begin{array}{rl}-x+3y+z& =11\\ 3x+y+2z& =27\\ 5x-y-z& =5\end{array}$$

  
  ::-x+3y+z=113x+y+2z=275x-y-z=5

  11.

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  $$\begin{array}{rl}3x+2y+4z& =21\\ -2x+3y+z& =-11\\ x+2y-3z& =-3\end{array}$$

  
  ::3x+2y+4z=21-2x+3y+z11x+2y+3z+3z%3

  12. 

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  $$\begin{array}{rl}-x+2y-6z& =4\\ 8x+5y+3z& =-8\\ 2x-4y+12z& =5\end{array}$$

  
  ::-x+2y-6z=48x+5y+3z82x-4y+12z=5

  13. 

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  $$\begin{array}{rl}3x+5y+8z& =37\\ -6x+3y+z& =42\\ x+3y-2z& =5\end{array}$$

  
  ::3x+5y+8z=37-6x+3y+Z=42x+3y+2z=5

  14.

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  $$\begin{array}{rl}4x+y-6z& =-38\\ 2x+7y+8z& =108\\ -3x+2y-3z& =-15\end{array}$$

  
  ::4x+y-6z382x+7y+8z=108-3x+2y_3z}=15

  15.

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  $$\begin{array}{rl}6x+3y-2z& =-22\\ -4x-2y+4z& =28\\ 3x+3y+2z& =7\end{array}$$

  
  ::6x+3y-2z22-4x-2y+4z=283x+3y+2z=7

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  Review (Answer s)
  ::回顾(答复)

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  Please see the Appendix.
  ::请参看附录。