Section: 强化儿科 | 8.6 强化儿科

Section outline

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 following system of equations?
::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。每次您在写入大量额外信息时都写和重写方程式。矩阵通过将信息嵌入每个条目的位置来照顾这些信息。您将如何使用矩阵来写入以下方程式系统 ?

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

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

Solving Systems of Equations with Augmented Matrices
::增加分数等同溶解系统

In order 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 coefficient matrix array . Next copy the variables in a variable matrix and the constants into a constant matrix.
::为了作为矩阵方程式代表一个系统,请首先以标准格式将所有方程式写入标准格式,以便将变量的系数排成列列。然后将系数矩阵数组中的系数复制下来。然后将变量在变量矩阵中的变量和常数复制成一个常数矩阵。

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

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

\begin{aligned} [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{matrix}] \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 9 \\ 22 \\ 31 \end{matrix}] \end{aligned}

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

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。

\begin{aligned} [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{matrix}] \cdot [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 1 x + 1 y + 1 z \\ 1 x + 2 y + 3 z \\ 2 z + 3 y + 4 z \end{matrix}] = [\begin{matrix} 9 \\ 22 \\ 31 \end{matrix}] \end{aligned}

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

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 3 equations and 3 variables.
::注意在右侧两个矩阵上加括号对掩盖以下事实没有多大作用:这只是一个由3个方程和3个变量组成的常规系统。

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。

$\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{matrix} | \begin{matrix} 9 \\ 22 \\ 31 \end{matrix} \end{matrix}] \end{aligned}$

To solve, reduce the matrix to reduced row echelon form .
::要解析, 将矩阵缩小为排梯格式。

\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{matrix} | \begin{matrix} 9 \\ 22 \\ 31 \end{matrix} \end{matrix}] \\ R_{1} \cdot - 1 + R_{2} \to [\begin{matrix} \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 2 & 3 & 4 \end{matrix} | \begin{matrix} 9 \\ 13 \\ 31 \end{matrix} \end{matrix}] \\ R_{1} \cdot - 2 + R_{3} \to [\begin{matrix} \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \end{matrix} | \begin{matrix} 9 \\ 13 \\ 13 \end{matrix} \end{matrix}] \\ R_{2} \cdot - 1 + R_{3} \to [\begin{matrix} \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix} | \begin{matrix} 9 \\ 13 \\ 0 \end{matrix} \end{matrix}] \end{aligned}

::[112323492231]R11+R2}[11101223491331]R12+R3}[111012012913]R21+R3}[1110120009130]

Because the last row is all 0's, this system is dependent . Therefore , there are an infinite number of solutions.
::因为最后一行是全部 0, 这个系统是依附于。因此, 有无限数量的解决方案。

Examples
::实例

Example 1
::例1

Earlier, you were asked how to write a system of equations as a matrix equation. If you were to write the system as a matrix equation, you could write:
::早些时候,有人问您如何将方程式系统写成矩阵方程式。如果您要将系统写成矩阵方程式,您可以写:

\begin{aligned} 5 x + y & = 6 \\ x + y & = 10 \\ [\begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix}] \cdot [\begin{matrix} x \\ y \end{matrix}] & = [\begin{matrix} 6 \\ 10 \end{matrix}] \end{aligned}

::5x+y=6x+y=10[5111]+[xy]=[610]

Example 2
::例2

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

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

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

\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 1 & 1 \\ 1 & - 1 & - 1 \\ 1 & 2 & 3 \end{matrix} | \begin{matrix} 6 \\ - 4 \\ 14 \end{matrix} \end{matrix}] \\ R_{1} \cdot - 1 + R_{2} \to [\begin{matrix} \begin{matrix} 1 & 1 & 1 \\ 0 & - 2 & - 2 \\ 1 & 2 & 3 \end{matrix} | \begin{matrix} 6 \\ - 10 \\ 8 \end{matrix} \end{matrix}] \\ R_{1} \cdot - 1 + R_{3} \to [\begin{matrix} \begin{matrix} 1 & 1 & 1 \\ 0 & - 2 & - 2 \\ 0 & 1 & 2 \end{matrix} | \begin{matrix} 6 \\ - 10 \\ 8 \end{matrix} \end{matrix}] \\ R_{3} \cdot - 1 + R_{1} \to [\begin{matrix} \begin{matrix} 1 & 0 & - 1 \\ 0 & - 2 & - 2 \\ 0 & 1 & 2 \end{matrix} | \begin{matrix} - 2 \\ - 10 \\ 8 \end{matrix} \end{matrix}] \\ R_{3} \cdot 3 + R_{2} \to [\begin{matrix} \begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & 4 \\ 0 & 1 & 2 \end{matrix} | \begin{matrix} - 2 \\ 14 \\ 8 \end{matrix} \end{matrix}] \\ R_{2} \cdot - 1 + R_{3} \to [\begin{matrix} \begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & 4 \\ 0 & 0 & - 2 \end{matrix} | \begin{matrix} - 2 \\ 14 \\ - 6 \end{matrix} \end{matrix}] \\ R_{3} \div - 2 \to [\begin{matrix} \begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} - 2 \\ 14 \\ 3 \end{matrix} \end{matrix}] \\ R_{3} + R_{1} \to [\begin{matrix} \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} 1 \\ 14 \\ 3 \end{matrix} \end{matrix}] \\ R_{3} \cdot - 4 + R_{2} \to [\begin{matrix} \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \end{matrix}] \end{aligned}

::[1111-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-2-1-1-1-1-1-1-414-6-108]R1-1-2-2-2-2-12-2012-6-108]R3-1-+R1-1-1-R1-1-[10-2--2012-2-108]R3-3-+R2__[10-10-10-140-12-2148]R2-1+R3/13__[10-1400-2-1-1-1-1-14-1-1-1-1-1-14-1-14-114-6]R3-2__[10-114-001-2143]R3+R1____[10014001}1143]R3-4+R2____[10010010010001____-123]

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。

\begin{aligned} 1 x + 0 y + 0 z & = 1 \\ 0 x + 1 y + 0 z & = 2 \\ 0 x + 0 y + 1 z & = 3 \end{aligned}

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

Which means $\begin{aligned} x = 1, y = 2, z = 3 \end{aligned}$ .
::这意味着 x=1,y=2,z=3。

Example 3
::例3

Solve the following system using augmented Matrices.
::使用扩增的母体,解决以下系统。

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

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

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。

\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{matrix} | \begin{matrix} 11 \\ 9 \\ 7 \\ 5 \end{matrix} \end{matrix}] \\ S w i t c h R_{2}, R_{3}, a n d R_{4} \to [\begin{matrix} \begin{matrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \end{matrix} | \begin{matrix} 11 \\ 7 \\ 5 \\ 9 \end{matrix} \end{matrix}] \\ R_{1} \cdot - 1 + R_{4} \to [\begin{matrix} \begin{matrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & - 1 \end{matrix} | \begin{matrix} 11 \\ 7 \\ 5 \\ - 2 \end{matrix} \end{matrix}] \\ R_{4} \cdot - 1 \to [\begin{matrix} \begin{matrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{matrix} | \begin{matrix} 11 \\ 7 \\ 5 \\ 2 \end{matrix} \end{matrix}] \\ R_{4} \cdot - 1 + R_{1} \to [\begin{matrix} \begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{matrix} | \begin{matrix} 9 \\ 7 \\ 5 \\ 2 \end{matrix} \end{matrix}] \\ R_{4} \cdot - 1 + R_{3} \to [\begin{matrix} \begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} | \begin{matrix} 9 \\ 7 \\ 3 \\ 2 \end{matrix} \end{matrix}] \\ R_{3} \cdot - 1 + R_{2} \to [\begin{matrix} \begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} | \begin{matrix} 9 \\ 4 \\ 3 \\ 2 \end{matrix} \end{matrix}] \\ R_{2} \cdot - 1 + R_{1} \to [\begin{matrix} \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} | \begin{matrix} 5 \\ 4 \\ 3 \\ 2 \end{matrix} \end{matrix}] \end{aligned}

::[1101000100010001111975] Switch R2、R3、R3和R4}[1101000100011110011759]R11+R4}[1101000100010001000-1000-1}1}1}[110100010001000-11001}11752]R41+R1}[111000100010001000100010001100019752]R1+R3}[11100010001000100010001001}9732]R3____110001+R2}[111000100010001000100010009}

Thus, $\begin{aligned} w = 5, x = 4, y = 3, z = 2 \end{aligned}$
::因此, w=5,x=4,y=3,z=2

Example 4
::例4

Use an augmented matrix to solve the following system.
::使用扩大的矩阵来解决以下系统问题。

\begin{aligned} 3 x + y & = - 15 \\ x + 2 y & = 15 \end{aligned}

::3+y15x+2y=15 3+y15x+2y=15

The row reduction steps are not shown, only the initial and final augmented matrices.
::没有显示行削减步骤,只有初始和最终增派矩阵。

\begin{aligned} [\begin{matrix} \begin{matrix} 3 & 1 \\ 1 & 2 \end{matrix} | \begin{matrix} - 15 \\ 15 \end{matrix} \end{matrix}] \to [\begin{matrix} \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} | \begin{matrix} - 9 \\ 12 \end{matrix} \end{matrix}] \end{aligned}

Example 5
::例5

Use an augmented matrix to solve the following system.
::使用扩大的矩阵来解决以下系统问题。

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

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

The row reduction steps are not shown, only the initial and final augmented matrices.
::没有显示行削减步骤,只有初始和最终增派矩阵。

\begin{aligned} [\begin{matrix} \begin{matrix} - 1 & 1 & - 1 \\ 2 & - 2 & - 3 \\ 3 & - 4 & 3 \end{matrix} | \begin{matrix} 0 \\ 25 \\ 2 \end{matrix} \end{matrix}] \to [\begin{matrix} \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} 3 \\ - 2 \\ - 5 \end{matrix} \end{matrix}] \end{aligned}

Summary

To represent a system as a matrix equation, write all equations in standard form, then create a coefficient matrix, a variable matrix, and a constant matrix.
::代表矩阵等式的系统,以标准格式写入所有方程式,然后创建系数矩阵、变量矩阵和不变矩阵。

An augmented matrix is two matrices that are joined together.
::扩大后的矩阵是两个组合在一起的矩阵。

When solving a system, augment the coefficient matrix and the constant matrix with a vertical line indicating the separation between them.
::在解决系统时,增加系数矩阵和恒定矩阵,加上一条垂直线,表明两者的分离。

To solve the system, reduce the augmented matrix to reduced row echelon form.
::为了解决系统问题,将扩充后的矩阵缩小为减缩的排梯表。

If the last row of the reduced matrix is all 0's, the system is dependent, and there are an infinite number of solutions.
::如果减少的矩阵最后一行是全部 0,则系统是依附的,并且有无限数量的解决方案。

Review
::回顾

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

\begin{aligned} 4 x - 2 y & = - 20 \\ x - 3 y & = - 15 \end{aligned}

::4 - 2y 20x - 3y 15

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

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

\begin{aligned} x + 4 y & = 11 \\ 3 x + 12 y & = 33 \end{aligned}

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

\begin{aligned} - 3 x + y & = - 7 \\ - x + 4 y & = 5 \end{aligned}

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

\begin{aligned} 3 x + y & = 6 \\ - 6 x - 2 y & = 10 \end{aligned}

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

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

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

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

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

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

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

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

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

10.

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

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

11.

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

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

12.

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

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

13.

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

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

14.

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

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

15.

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

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

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

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

Solving Systems of Equations with Augmented Matrices ::增加分数等同溶解系统

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)