课程： 4.12 利用矩阵和技术解决线性线性系统

章节大纲

选择节常规

折叠展开
常规

全部折叠展开全部
- 选择活动新闻通告
  
  新闻通告讨论区
选择节利用矩阵和技术解决线性系统

折叠展开
利用矩阵和技术解决线性系统
Customers of an ice cream shop are asking for more sizes, so the shop decides to test out medium and large cones. During its test, the shop sells 24 medium chocolate cones and 32 large chocolate cones. It also sells 40 medium vanilla cones and 60 X-large vanilla cones. The shop's chocolate sales for the day totaled $292 and its vanilla sales totaled $525. How much did the shop charge for a medium cone versus a large cone?
::一家冰淇淋店的客户要求更多尺寸,因此商店决定测试中大锥体。在测试期间,商店出售24个中中巧克力锥体和32个大巧克力锥体。它也出售40个中香草锥体和60个X大香草锥体。当日,商店的巧克力销售总额达292美元,香草销售总额达525美元。一个中大锥体和一个大锥体的商店收费是多少?

Solving Linear Systems using Matrices
::使用矩阵解决线性系统

In this concept we will be using the calculator to find the inverse matrix and do matrix multiplication . We have already entered matrices into the calculator and multiplied matrices together on the calculator. To find the inverse of a matrix using the calculator we can call up the matrix by name and use the $\begin{align*}x^{-1}\end{align*}$ key on the calculator to get the inverse of the matrix.
::在这个概念中,我们将使用计算器查找反向矩阵并进行矩阵乘法。我们已经将矩阵输入计算器, 并在计算器上将矩阵一起乘以。要找到使用计算器的矩阵的反向, 我们可以用名称调用矩阵, 并在计算器上使用 x-1 键来获取矩阵的反向。

Now that we know how to get an inverse using the calculator, let’s look at solving the system. When we translate our system into a matrix equation we have the expression $\begin{align*}AX=B\end{align*}$ . From the last concept, we know that $\begin{align*}X=A^{-1}B\end{align*}$ . We can enter the coefficient matrix into matrix $\begin{align*}A\end{align*}$ and the constants into matrix $\begin{align*}B\end{align*}$ and have the calculator find $\begin{align*}A^{-1}B\end{align*}$ for us.
::现在我们知道如何用计算器获得反向的计算器,让我们看看如何解决系统。当我们把我们的系统转换成矩阵方程式时,我们有了AX=B的表达式。从最后一个概念中,我们知道X=A-1B。我们可以将系数矩阵输入矩阵A,常数输入矩阵B,让计算器为我们找到A-1B。

Let's use the calculator to solve the following problems.
::让我们用计算器来解决以下问题。

Find the inverse of the matrix:
::查找矩阵的反向 :

$\begin{aligned} [\begin{matrix} - 3 & - 5 \\ 1 & 2 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} -3 & -5 \\ 1 & 2 \end{bmatrix}\end{align*}$

First, enter the matrix into the calculator in matrix $\begin{align*}[A]\end{align*}$ . Now, $\begin{align*}2^{nd}\end{align*}$ QUIT to return to the home screen. From here, go back into the MATRIX menu and select $\begin{align*}[A]\end{align*}$ under NAMES and press ENTER. You should see $\begin{align*}[A]\end{align*}$ in the home screen. Now press the $\begin{align*}x^{-1}\end{align*}$ button to get $\begin{align*}[A]^{-1}\end{align*}$ and press ENTER. The result is the inverse matrix
::首先,在矩阵 [A] 中将矩阵输入计算器。现在, 第二 QUIT 将返回主屏幕。从这里, 返回 MATRIX 菜单, 在 NAMES 下选择 [A] 并按 ENTER 键。您应该在主屏幕中看到 [A] 。现在按 x-1 按钮获取 [A] - 1 并按 ENTER 键。结果是反向矩阵。

$\begin{aligned} [\begin{matrix} - 2 & - 5 \\ 1 & 3 \end{matrix}] . \end{aligned}$
$\begin{align*}\begin{bmatrix} -2 & -5 \\ 1 & 3 \end{bmatrix}.\end{align*}$

$\begin{align*}^*\end{align*}$ Note that the calculator can find the inverse of any square matrix .
::* 注意计算器能找到任何平方矩阵的反面。

Solve the system:
::解决系统:

$\begin{aligned} 10 x + 6 y & = 11 \\ - 6 x + 9 y & = - 15 \end{aligned}$
$\begin{align*}10x+6y &= 11\\ -6x+9y &= -15\end{align*}$
::10x+6y=11-6x+9y15

Let
$\begin{aligned} A = [\begin{matrix} 10 & 6 \\ - 6 & 9 \end{matrix}] \end{aligned}$
$\begin{align*} A = \begin{bmatrix} 10 & 6\\ -6 & 9 \end{bmatrix}\end{align*}$ and
$\begin{aligned} B = [\begin{matrix} 11 \\ - 15 \end{matrix}] . \end{aligned}$
$\begin{align*} B = \begin{bmatrix} 11\\ -15 \end{bmatrix}.\end{align*}$ Enter these matrices into the calculator. Now, from the home screen, we can call up matrix $\begin{align*}A\end{align*}$ (from NAMES in the matrix menu) and use the $\begin{align*}x^{-1}\end{align*}$ button to get $\begin{align*}[A]^{-1}\end{align*}$ on the home screen. Finally we can call up matrix $\begin{align*}B\end{align*}$ from the matrix menu to get $\begin{align*}[A]^{-1}[B]\end{align*}$ and press ENTER. The result is
::让 A = [106- 69] 和 B = [11- 15] 。请将这些矩阵输入计算器中。现在, 我们可以从主屏幕上调用矩阵 A (在矩阵菜单中的 NAMES ) , 并使用 x-1 按钮在主屏幕上调用 [A] - 1 。最后, 我们可以从矩阵菜单中调用矩阵 B 以获取 [A] - 1 [B] 并按 ENTER 。结果是。

$\begin{aligned} [\begin{matrix} \frac{3}{2} \\ \frac{- 2}{3} \end{matrix}] . \end{aligned}$
$\begin{align*}\begin{bmatrix} \frac{3}{2}\\ \frac{-2}{3} \end{bmatrix}.\end{align*}$ The solution to the system is $\begin{align*}\left(\frac{3}{2},- \frac{2}{3} \right)\end{align*}$ .
::[32-23]. 系统的解决办法是(32,-23)。

Solve the system:
::解决系统:

$\begin{aligned} 3 x + 11 y & = 2 \\ - 9 x - 33 y & = - 6 \end{aligned}$
$\begin{align*}3x+11y &= 2\\ -9x-33y &= -6\end{align*}$
::3x+11y=2-9x-33y=6

Let
$\begin{aligned} A = [\begin{matrix} 3 & 11 \\ - 9 & - 33 \end{matrix}] \end{aligned}$
$\begin{align*} A = \begin{bmatrix} 3 & 11 \\ -9 & -33 \end{bmatrix}\end{align*}$ and
$\begin{aligned} B = [\begin{matrix} 2 \\ - 6 \end{matrix}] . \end{aligned}$
$\begin{align*} B = \begin{bmatrix} 2 \\ -6 \end{bmatrix}.\end{align*}$ Enter these matrices into the calculator. Now, from the home screen, we can call up matrix $\begin{align*}A\end{align*}$ (from NAMES in the matrix menu) and use the $\begin{align*}x^{-1}\end{align*}$ button to get $\begin{align*}[A]^{-1}\end{align*}$ on the home screen. Finally we can call up matrix $\begin{align*}B\end{align*}$ from the matrix menu to get $\begin{align*}[A]^{-1}[B]\end{align*}$ and press ENTER. The result is ERR: SINGULAR MATRIX. Recall that a singular matrix is a matrix for which the inverse does not exist. This message tells us that there is no unique solution to this system. We must now use an alternative method to determine whether there are infinite solutions or no solution to the system. Using linear combination as shown below we can see that there are infinite solutions to this system.
::允许 A = [311- 9- 33] 和 B = [2-6] 。请将这些矩阵输入计算器。现在, 从主屏幕上, 我们可以调用矩阵 A (在矩阵菜单中的 NAMES ) , 并使用 x-1 按钮在主屏幕上取 [A] - 1 。最后, 我们可以调用矩阵菜单中的矩阵 B 来调用 [A] - 1 [B] 和按 EnTER 。结果是 ERR: SINGULAR MATRIX 。提醒注意单项矩阵是一个不存在反向的矩阵。此信息告诉我们, 这个系统没有独特的解决方案。我们现在必须使用替代方法来确定系统是否有无限的解决方案或没有解决方案。使用下文所示的线性组合, 我们可以看到这个系统有无限的解决方案。

$\begin{aligned} 3 (3 x + 11 y = 2) \Rightarrow 9 x + 33 y = 6 \\ - 9 x - 33 y = - 6 \underline{- 9 x - 33 y = - 6} \\ 0 = 0 \end{aligned}$
$\begin{align*}& 3(3x+11y=2) \quad \Rightarrow \quad \ 9x+33y=6\\ & -9x-33y=-6 \qquad \quad \underline{-9x-33y=-6 \; \; }\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 0=0\end{align*}$
::3(3x+11y=2) 9x+33y=6-9x-33y _6-9x-33y_6_9x-33y_6_0=0

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the cost of a medium cone versus a large cone.
::早些时候,你被要求寻找中型锥体相对于大型锥体的成本。

Solve the system using matrices on the calculator:
::使用计算器上的矩阵解决系统 :

$\begin{aligned} 24 x + 32 y & = 292 \\ 40 x + 60 y & = 525 \end{aligned}$
$\begin{align*}24x+32y &= 292\\ 40x+60y &= 525\end{align*}$
::24x+32y=29240x60y=525

The shop charges $4.50 for a medium cone and $5.75 for a large cone.
::商店收取4.50美元的中型锥体和5.75美元的大型锥体。

Example 2
::例2

Find the inverse of
$\begin{aligned} [\begin{matrix} - 2 & 4 & - 5 \\ 3 & - 1 & 4 \\ 7 & 2 & 1 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} -2 & 4 & -5 \\ 3 & -1 & 4 \\ 7 & 2 & 1 \end{bmatrix}\end{align*}$ using your calculator.

$\begin{aligned} {[\begin{matrix} - 2 & 4 & - 5 \\ 3 & - 1 & 4 \\ 7 & 2 & 1 \end{matrix}]}^{- 1} = [\begin{matrix} - \frac{9}{53} & - \frac{14}{53} & \frac{11}{53} \\ \frac{25}{53} & \frac{33}{53} & - \frac{7}{53} \\ \frac{13}{53} & \frac{32}{53} & - \frac{10}{53} \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} -2 & 4 & -5 \\ 3 & -1 & 4 \\ 7 & 2 & 1 \end{bmatrix}^{-1}= \begin{bmatrix} -\frac{9}{53} & -\frac{14}{53} & \frac{11}{53} \\ \frac{25}{53} & \frac{33}{53} & -\frac{7}{53} \\ \frac{13}{53} & \frac{32}{53} & -\frac{10}{53} \end{bmatrix}\end{align*}$

The calculator will convert all the values in the matrix to fractions. Press MATH and ENTER to choose $\begin{align*}1: >FRAC\end{align*}$ .
::计算器将把矩阵中的所有值转换为分数。按 MATH 和 ENTER 选择 1: > FRAC 。

Solve the following systems using matrices on the calculator.
::使用计算器上的矩阵解决以下系统。

Example 3

$\begin{aligned} 2 x + 3 y & = 13 \\ - 6 x - 7 y & = - 21 \end{aligned}$
$\begin{align*}2x+3y &= 13\\ -6x-7y &= -21\end{align*}$
::例3 2x+3y=13-6x-7y21

Let
$\begin{aligned} A = [\begin{matrix} 2 & 3 \\ - 6 & - 7 \end{matrix}] \end{aligned}$
$\begin{align*}A=\begin{bmatrix} 2 & 3 \\ -6 & -7 \end{bmatrix}\end{align*}$ and
$\begin{aligned} B = [\begin{matrix} 13 \\ - 21 \end{matrix}] . \end{aligned}$
$\begin{align*}B=\begin{bmatrix} 13 \\ -21 \end{bmatrix}.\end{align*}$ Using the calculator
$\begin{aligned} A^{- 1} B = [\begin{matrix} - 7 \\ 9 \end{matrix}] . \end{aligned}$
$\begin{align*}A^{-1}B=\begin{bmatrix} -7 \\ 9 \end{bmatrix}.\end{align*}$ The solution is (-7, 9).
::Let A=[23-6-7]和B=[13-21]。使用计算器 A-1B=[-79]。解决方案是(-7,9)

Example 4
::例4

$\begin{aligned} 3 x - 5 y + z & = - 24 \\ - 4 x + 3 y - 2 z & = 9 \\ 7 x + y - 3 z & = - 30 \end{aligned}$
$\begin{align*}3x-5y+z &= -24\\ -4x+3y-2z &= 9\\ 7x+y-3z &= -30\end{align*}$
::3x-5y+z24-4x+3y-2z=97x+y-3z=30

Let
$\begin{aligned} A = [\begin{matrix} 3 & - 5 & 1 \\ - 4 & 3 & - 2 \\ 7 & 1 & - 3 \end{matrix}] \end{aligned}$
$\begin{align*}A=\begin{bmatrix} 3 & -5 & 1 \\ -4 & 3 & -2 \\ 7 & 1 & -3 \end{bmatrix}\end{align*}$ and
$\begin{aligned} B = [\begin{matrix} - 24 \\ 9 \\ - 30 \end{matrix}] . \end{aligned}$
$\begin{align*}B=\begin{bmatrix} -24 \\ 9 \\ -30 \end{bmatrix}.\end{align*}$ Using the calculator
$\begin{aligned} A^{- 1} B = [\begin{matrix} - 2 \\ 5 \\ 7 \end{matrix}] . \end{aligned}$
$\begin{align*}A^{-1}B=\begin{bmatrix} -2 \\ 5 \\ 7 \end{bmatrix}.\end{align*}$ The solution is (-2, 5, 7).
::Let A=[3-51-43-271-3]和B=[-249-30]。使用计算器A-1B=[-257]。解决办法是(-2、5、7)。

Review
::回顾

Use a calculator to find the inverse of each matrix below.
::使用计算器查找以下每个矩阵的反向。

$\begin{aligned} [\begin{matrix} 2 & - 3 \\ 8 & 2 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 2 & -3 \\ 8 & 2 \end{bmatrix}\end{align*}$

$\begin{aligned} [\begin{matrix} - 6 & 8 \\ - 3 & 2 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} -6 & 8 \\ -3 & 2 \end{bmatrix}\end{align*}$

$\begin{aligned} [\begin{matrix} 5 & 2 & - 1 \\ 3 & 5 & 0 \\ - 4 & 1 & 6 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 5 & 2 & -1 \\ 3 & 5 & 0 \\ -4 & 1 & 6 \end{bmatrix}\end{align*}$

Solve the systems using matrices and a calculator.
::使用矩阵和计算器解决系统。

$\begin{aligned} 2 x + y & = - 1 \\ - 3 x - 2 y & = - 3 \end{aligned}$
$\begin{align*}2x+y &= -1\\ -3x-2y &= -3\end{align*}$
::2+y1-3x-2y3

$\begin{aligned} 2 x + 10 y & = 6 \\ x + 11 y & = - 3 \end{aligned}$
$\begin{align*}2x+10y &= 6\\ x+11y &= -3\end{align*}$
::2x+10y=6x+11y=3

$\begin{aligned} - 3 x + y & = 17 \\ 2 x - 3 y & = - 23 \end{aligned}$
$\begin{align*}-3x+y &= 17\\ 2x-3y &= -23\end{align*}$
::-3x+y=172x-3y23

$\begin{aligned} 10 x - 5 y & = 9 \\ 2.5 y & = 5 x + 4 \end{aligned}$
$\begin{align*}10x-5y &= 9\\ 2.5y &= 5x +4\end{align*}$
::10x-5y=92.5y=5x+4

$\begin{aligned} 4 x - 3 y & = - 12 \\ 7 x & = 2 y + 5 \end{aligned}$
$\begin{align*}4x-3y &= -12\\ 7x &= 2y+5\end{align*}$
::4x-3y 127x=2y+5

$\begin{aligned} 2 x + 5 y & = 0 \\ - 4 x + 10 y & = - 12 \end{aligned}$
$\begin{align*}2x+5y &= 0\\ -4x+10y &= -12\end{align*}$
::2x+5y=0-4x+10y=12

$\begin{aligned} 7 x + 2 y + 3 z & = - 12 \\ - 8 x - y + 4 z & = - 6 \\ - 10 x + 3 y + 5 z & = 1 \end{aligned}$
$\begin{align*}7x+2y+3z &= -12\\ -8x-y + 4z &= -6\\ -10x+3y+5z &= 1\end{align*}$
::7+2y+3z12-8x-y+4z6-10x+3y+5z=1

$\begin{aligned} 2 x - 4 y + 11 z & = 18 \\ - 3 x + 5 y - 13 z & = - 25 \\ 5 x + 10 y + 10 z & = 5 \end{aligned}$
$\begin{align*}2x-4y+11z &= 18\\ -3x+5y-13z &= -25\\ 5x+10y+10z &= 5\end{align*}$
::2x-4y+11z=18-3x+5y-13z255x+10y+10z=5

$\begin{aligned} 8 x + y - 4 z & = 4 \\ - 2 x - 3 y + 4 z & = - 7 \\ 4 x + 5 y - 8 z & = 11 \end{aligned}$
$\begin{align*}8x+y-4z &= 4\\ -2x-3y+4z &= -7\\ 4 x+5y-8z &= 11\end{align*}$
::8x+y-4z=4-2x-3y+4z74x+5y-8z=11

For the following word problems, set up a system of linear equations to solve using matrices.
::对于以下字词问题,建立线性方程式系统,用矩阵解决。

A mix of 1 lb almonds and 1.5 lbs cashews sells for $15.00. A mix of 2 lbs almonds and 1 lb cashews sells for $17.00. How much does each nut cost per pound?
::1磅杏仁和1.5磅腰果混合销售15美元,2磅杏仁和1磅腰果混合销售17美元。

Maria, Rebecca and Sally are selling baked goods for their math club. Maria sold 15 cookies, 20 brownies and 12 cupcakes and raised $23.50. Rebecca sold 22 cookies, 10 brownies and 11 cupcakes and raised $19.85. Sally sold 16 cookies, 5 brownies and 8 cupcakes and raised $13.30. How much did they charge for each type of baked good?
::Maria, Rebecca和Sally为他们的数学俱乐部出售面包产品。Maria卖了15个饼干、20个布朗尼蛋糕和12个蛋糕,并筹集了23.50美元。Rebecca卖了22个饼干、10个布朗尼蛋糕和11个蛋糕,并筹集了19.85美元。Sally卖了16个饼干、5个布朗尼蛋糕和8个蛋糕,并筹集了13.30美元。每类烤面包都收了多少钱?

Justin, Mark and George go to the farmer’s market to buy fruit. Justin buys 2 lbs apples, 3 lbs pears and 1 lb of peaches for $13.24. Mark buys 1 lb apples, 2 lbs pears and 3 lbs peaches for $13.34. George buys 3 lbs apples, 1 lb pears and 2 lbs peaches for $14.04. How much does each fruit cost per pound?
::Justin、Mark和George去农民市场买水果。Justin以13.24美元购买2磅苹果、3磅梨和1磅桃子。Mark以13.34美元购买1磅苹果、2磅梨和3磅桃子。George以14.04美元购买3磅苹果、1磅梨和2磅桃子。每个水果每磅要花多少钱?

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。

章节大纲

常规

利用矩阵和技术解决线性系统

Solving Linear Systems using Matrices ::使用矩阵解决线性系统

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 2 x + 3 y = 13 − 6 x − 7 y = − 21 \begin{align*}2x+3y &= 13\\ -6x-7y &= -21\end{align*} ::例3 2x+3y=13-6x-7y21

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Solving Linear Systems using Matrices
::使用矩阵解决线性系统

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3

$\begin{aligned} 2 x + 3 y & = 13 \\ - 6 x - 7 y & = - 21 \end{aligned}$
$\begin{align}2x+3y &= 13\\ -6x-7y &= -21\end{align}$
::例3 2x+3y=13-6x-7y21

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)