4.8 Cramer规则

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  Cramer 规则

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  At your school book fair, paperbacks cost one price and hardcovers cost another. You buy 3 paperbacks and 2 hardcovers. Your total comes to $54. Your best friend buys 2 paperbacks and 4 hardcovers. His total comes to $76. How could you use a matrix to find the price of each type of book?
  ::在你的学校书展上,纸背成本为1美元,硬封面成本为1美元。您购买了3张纸背和2张硬封面。总计为54美元。您最好的朋友购买了2张纸背和4张硬封面。他的总数为76美元。您如何使用一个矩阵来找到每本书的价格?

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  Cramer's Rule
  ::Cramer 规则

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  We have already learned how to solve systems of linear equations using graphing, substitution and linear combinations. In this section , we will explore one way to use matrices and to solve linear systems.
  ::我们已经学会了如何用图形化、替代和线性组合来解决线性方程系统。在本节,我们将探索一种使用矩阵和解决线性系统的方法。

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  Cramer’s Rule in Two Variables
  ::两种变数的Cramer规则

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  Given the system:

  $$\begin{align*}ax+by &= e\\ cx+dy &= f,\end{align*}$$ we can set up the matrix $\begin{align*}A\end{align*}$ and solve for $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ as shown below:
  ::鉴于这个系统:ax+by=ecx+dy=f,我们可以设置矩阵A,为 x 和 y 确定如下:

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  $$\begin{align*}A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}, \ x=\frac{\begin{vmatrix}{\color{red}e} & d\\ {\color{red}f} & b\end{vmatrix}}{det \ A}\end{align*}$$ and $$\begin{align*}y = \frac{\begin{vmatrix}a & {\color{red}e}\\c & {\color{red}f}\end{vmatrix}}{det \ A},\end{align*}$$ provided $\begin{align*}|A| \ne 0\end{align*}$ .
  ::A = [abcd], xedfbdet A 和 yececdet A, 提供 A0 。

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  Note: When $\begin{align*}|A| = 0\end{align*}$ , there is no unique solution. We have to investigate further to determine if there are infinite solutions or no solutions to the system. Notice that there is a pattern here. The coefficients of the variable we are trying to find are replaced with the constants.
  ::注:当 {A_0}0 时,没有独特的解决方案。我们必须进一步调查,以确定系统是否有无限的解决方案或没有解决方案。请注意这里有一个模式。我们试图找到的变量的系数被常数所取代。

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  Cramer’s Rule in Three Variables
  ::三个变数的Cramer规则

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  Given the system:

  $$\begin{align*}ax+by+cz &= j\\ dx+ey+fz &= k\\ gx+hy+iz &= l,\end{align*}$$ we can set up the matrix $\begin{align*}A\end{align*}$ and solve for $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ as shown below:
  ::鉴于这个系统:x++by+cz=jdx+ey+fz=kgx+hy+iz=l,我们可以设置矩阵A,解决x和y的如下:

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  $$\begin{align*}A = \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}\end{align*}$$
  ::A = [abcdefghi]

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  $$\begin{align*}x = \frac{\begin{vmatrix}{\color{red}j} & b & c\\{\color{red}k} & e & f\\{\color{red}l} & h & i\end{vmatrix}}{det \ A}\end{align*}$$ , $$\begin{align*}y = \frac{\begin{vmatrix}a & {\color{red}j} & c\\d & {\color{red}k} & f\\g & {\color{red}l} & i\end{vmatrix}}{det \ A}\end{align*}$$ and $$\begin{align*}z = \frac{\begin{vmatrix}a & b & {\color{red}j}\\d & e & {\color{red}k}\\g & h & {\color{red}l}\end{vmatrix}}{det \ A}\end{align*}$$
  ::A、yajcdkflidet A和zabjdekhldet A

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  provided $\begin{align*}|A| \ne 0\end{align*}$ . Once again, if the $\begin{align*}|A| = 0\end{align*}$ , then there is no unique solution to the system. Once again there is a pattern here. The coefficients of the variable for which we are trying to solve are replaced with the constants.
  ::提供 *A0。再次,如果 A0 , 那么系统就不会有唯一的解决方案。 这里再次出现一个模式。 我们试图解答的变量的系数会被常数取代 。

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  Let's use Cramer's Rule to solve the following systems of equations.
  ::让我们使用Cramer规则 解决以下的方程式系统

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  1.     
     $$\begin{align*}3x-7y &=13\\ -5x+9y &=-19\end{align*}$$
     ::3x-7y=13-5x+9y19

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  Matrix $\begin{align*}A\end{align*}$ is the matrix made up of the coefficients of $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ :
  ::矩阵A是由x和y的系数组成的矩阵:

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  $$\begin{align*}A = \begin{bmatrix} 3 & -7\\ -5 & 9\end{bmatrix}\end{align*}$$
  ::A=[3-7-59]

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  Now we can find the $\begin{align*}det \ A = (3\cdot9)-(-7\cdot-5) = 27-35=-8\end{align*}$
  ::现在我们可以找到A=(39)-(-75)=27-358

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  So, using the formulas above:
  ::因此,使用上面的公式:

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  $$\begin{align*}x = \frac{\begin{vmatrix}{\color{red}13} & -7\\{\color{red}-19} & 9\end{vmatrix}}{-8}=\frac{(13\cdot9)-(-7\cdot-19)}{-8}=\frac{-16}{-8}=2\end{align*}$$
  ::x13-7-1998=(139)-(-719)-8_16-8=2

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  $$\begin{align*}y = \frac{\begin{vmatrix}3 & {\color{red}13}\\-5 & {\color{red}-19}\end{vmatrix}}{-8}=\frac{(3\cdot-19)-(13\cdot-5)}{-8}=\frac{-57-(-65)}{-8}=\frac{8}{-8}=-1\end{align*}$$
  ::y... 313 - 5 - 19... 8= (319) - (135) - 857 - (- 65) - 8=8-8 - 81

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  Therefore , the solution is (2, -1).
  ::因此,解决办法是(2,-1)。

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  2.      
     $$\begin{align*}6x+3y &=-12\\ 2x+y &=20\end{align*}$$
     ::6x+3y 122x+y=20

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  Matrix $\begin{align*}A\end{align*}$ is the matrix made up of the coefficients of $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ :
  ::矩阵A是由x和y的系数组成的矩阵:

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  $$\begin{align*}A = \begin{bmatrix} 6 & 3\\ 2 & 1\end{bmatrix}\end{align*}$$
  ::A=[6321]

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  Now we can find the $\begin{align*}det \ A = (6\cdot1)-(3\cdot2) = 6-6=0\end{align*}$
  ::现在我们可以找到A=(6)1-(3)2=6-6=0的分母 A=(6)1-(3)2=(6)-6=0

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  Because the determinant is zero, there is no unique solution and we cannot solve the system further using Cramer's Rule. Looking at the system, we see that the left-hand side of the first equation is a multiple of the second equation, by 3. The right sides are not multiples of each other, therefore there is no solution.
  ::因为决定因素是零,所以没有独特的解决方案,我们无法用Cramer规则进一步解决系统问题。 看看这个系统,我们看到第一个等式的左侧是第二个等式的倍数,乘以3。 右侧不是彼此的倍数,因此没有解决办法。

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  3.     
     $$\begin{align*}2x+2y-z &=-7\\ 5x+y-2z &=-3\\ x-3y+2z &=21\end{align*}$$
     ::2x+2y-z%75x+y-2z%3x-3y+2z=21

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  Matrix $\begin{align*}A\end{align*}$ is the matrix made up of the coefficients of $\begin{align*}x, y\end{align*}$ and $\begin{align*}z\end{align*}$ :
  ::矩阵A是由x、y和z系数组成的矩阵:

  $$\begin{align*}\begin{bmatrix} 2 & 2 & -1\\ 5 & 1 & -2\\ 1 & -3 & 2\end{bmatrix}\end{align*}$$

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  Now we can find the determinant of matrix $\begin{align*}A\end{align*}$ :
  ::现在我们可以找到矩阵A的决定因素:

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  $$\begin{align*}det \ A &=\begin{vmatrix}2 & 2 & -1\\ 5 & 1 & -2\\ 1 & -3 & 2\end{vmatrix}\begin{matrix} 2 & 2\\ 5 & 1\\ 1 & -3\end{matrix}\\ &= [(2)(1)(2)+(2)(-2)(1)+(-1)(5)(-3)]-[(1)(1)(-1)+(-3)(-2)(2)+(2)(5)(2)]\\ &= [4-4+15]-[-1+12+20]\\ &=15-31\\ &=-16\end{align*}$$
  ::=[(1)(2)+(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(1)+(-1)(5)(-3)]-[(1)(1)-1-1]+(-3)(-2)(2)+(2)(2)(2)+(2)(5)(2)]=[4-4+15]-[1+12+20]=15-31__16]

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  Now using the formulas above, we can find $\begin{align*}x, y\end{align*}$ and $\begin{align*}z\end{align*}$ as follows:
  ::使用上面的公式,我们可以看到 X,y 和 z 如下:

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  $$\begin{align*}x = \frac{\begin{vmatrix}{\color{red}-7} & 2 & -1\\{\color{red}-3} & 1 & -2\\{\color{red}21} & -3 & 2\end{vmatrix}}{-16}=\frac{-32}{-16}=2 \qquad y = \frac{\begin{vmatrix}2 & {\color{red}-7} & -1\\5 & {\color{red}-3} & -2\\1 & {\color{red}21} & 2\end{vmatrix}}{-16}=\frac{48}{-16}=-3 \qquad z = \frac{\begin{vmatrix}2 & 2 & {\color{red}-7}\\5 & 1 & {\color{red}-3}\\1 & -3 & {\color{red}21}\end{vmatrix}}{-16}=\frac{-80}{-16}=5\end{align*}$$
  ::-1 -31 -221 -32 1632 -16=2y2 -7 - 15 - 3-21216=48 - 163z22 - 751 - 31 - 321 - 1616_80-16=5

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  So the solution is (2, -3, 5).
  ::因此,解决办法是2,3,5。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked how you could use a matrix to find the price of each type of book. 
  ::早些时候,有人问您如何使用矩阵 来找到每种书的价格。

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  The system of linear equation represented by this situation is:
  ::以这种情况为代表的线性等式系统是:

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  $$\begin{align*}3x + 2y = 54\\ 2x + 4y = 76\end{align*}$$
  ::3x+2y=542x+4y=76

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  We can now set up a matrix and apply Cramer's rule to find the price of each type of book.
  ::我们现在可以设置一个矩阵 并运用Cramer的规则 找出每种书的价格。

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  $$\begin{align*}A = \begin{bmatrix} 3 & 2\\ 2 & 4\end{bmatrix}\end{align*}$$
  ::A=[3224]

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  Now we can find the $\begin{align*}det \ A = (3\cdot4)-(2\cdot2) = 12-4=8\end{align*}$
  ::现在我们可以找到A=(34)-(22)=12-4=8

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  So, using the formulas above:
  ::因此,使用上面的公式:

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  $$\begin{align*}x = \frac{\begin{vmatrix}{\color{red}54} & 2\\{\color{red}76} & 4\end{vmatrix}}{8}=\frac{(54\cdot4)-(76\cdot2)}{8}=\frac{216-152}{8}=8\end{align*}$$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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  $$\begin{align*}y = \frac{\begin{vmatrix}3 & {\color{red}54}\\2 & {\color{red}76}\end{vmatrix}}{8}=\frac{(3\cdot76)-(2\cdot54)}{8}=\frac{(228-108)}{8}=15\end{align*}$$
  ::y3542768=(376)-(254)8=(228-108)8=15

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  Therefore, paperbacks cost $8 and hardcovers cost $15.
  ::因此,回印费用为8美元,硬覆盖费用为15美元。

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  Example 2
  ::例2

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  Use Cramer's Rule to solve the system below.
  

  $$\begin{align*}2x+5y &=7\\ x+3y &=2\end{align*}$$
  ::使用 Cramer 规则解析以下系统 。 2x+5y=7x+3y=2

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  Find the

  $$\begin{align*}det \ A = \begin{vmatrix} 2 & 5\\ 1 & 3\end{vmatrix} = 1.\end{align*}$$ Now find $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ as follows:
  ::查找 det A 2513\\\ 1. 现在找到 x 和 y 如下 :

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  $$\begin{align*}x= \frac{\begin{vmatrix} 7 & 5\\ 2 & 3 \end{vmatrix}}{1} = \frac{11}{1}=11 \qquad y= \frac{\begin{vmatrix} 2 & 7\\ 1 & 2 \end{vmatrix}}{1} = \frac{-3}{1}=-3, \quad \text{solution:} \ (11,-3)\end{align*}$$
  ::X75231=111=11y27121313, 解说: (11,-3)

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  Example 3
  ::例3

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  Use Cramer's Rule to solve the system below.
  

  $$\begin{align*}4x-y &=6\\ -8x+2y &=10\end{align*}$$
  ::使用 Cramer 规则解析以下系统 。 4x-y=6-8x+2y=10

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  Find the
  ::查找

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  $$\begin{align*}det \ A= \begin{vmatrix} 4 & -1 \\ -8 & 2 \end{vmatrix} = 8-8 = 0.\end{align*}$$
  ::A*4 -1 -82 -8 -8=0。

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  Therefore, there is no unique solution. We must use linear combination or the substitution method to determine whether there are an infinite number of solutions or no solutions. Using linear combinations we can multiply the first equation by 2 and get the following:
  ::因此,没有独特的解决方案。 我们必须使用线性组合或替代方法来确定是否存在无限数量的解决方案或没有解决方案。 使用线性组合我们可以将第一个方程式乘以2, 并获得以下结果:

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  $$\begin{align*}& \quad 8x-2y=12 \\ & \ \underline{-8x+2y=10 \; \;} \ \Rightarrow \quad \text{Therefore, there is no solution}.\\ & \qquad \quad \ \ 0=22\end{align*}$$
  ::8 - 2y= 12 - 8x+2y= 10_ @ @ @ 因此,没有解决方案 。 0=22

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  Example 4
  ::例4

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  Use Cramer's Rule to solve the system below.
  ::使用Cramer规则解决下面的系统

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  $$\begin{align*}x+2y+3z &=8\\ 2x-y+4z &=3\\ -x-4y+3z &=14\end{align*}$$
  ::x+2y+3z=82x-y+4z=3-x-4y+3z=14

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  Find the
  ::查找

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  $$\begin{align*}det \ A=\begin{vmatrix} 1 & 2 & 3\\ 2 & -1 & 4\\ -1 & -4 & 3 \end{vmatrix}=-34.\end{align*}$$
  ::A*1232 - 14 - 1 - 43 - 34。

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  Now find $\begin{align*}x, y\end{align*}$ and $\begin{align*}z\end{align*}$ as follows:
  ::现在找到以下 x、y 和 z :

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  $$\begin{align*}x= \frac{\begin{vmatrix} 8 & 2 & 3\\ 3 & -1 & 4\\ 14 & -4 & 3 \end{vmatrix}}{-34} = \frac{204}{-34}=-6 \qquad y= \frac{\begin{vmatrix} 1 & 8 & 3\\ 2 & 3 & 4 \\ -1 & 14 & 3 \end{vmatrix}}{-34} = \frac{-34}{-34}=1 \qquad z= \frac{\begin{vmatrix} 1 & 2 & 8\\ 2 & -1 & 3\\ -1 & -4 & 14 \end{vmatrix}}{-34}=\frac{-136}{-34}=4\end{align*}$$
  ::823 - 1414 - 4334=204 - 346y183234 - 114334 - 3434 - 34=1z1282 - 13 - 1 - 41434\136 - 34=4

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  Therefore, the solution is (-6, 1, 4).
  ::因此,解决办法是(6,1,4)

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  Review
  ::回顾

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  Solve the systems below using Cramer’s Rule. If there is no unique solution, use an alternate method to determine whether the system has infinite solutions or no solution.
  ::使用 Cramer 规则解决下面的系统。 如果没有独特的解决方案,请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。

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  1. $$\begin{align*}5x-y &= 22\\ -x+6y &= -16\end{align*}$$
     ::5 - y=22 - x+6y16
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  2. $$\begin{align*}2x+5y &= -1\\ -3x-8y &= 1\end{align*}$$
     ::2x+5y1-3x-8y=1
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  3. $$\begin{align*}4x-3y &= 0\\ -6x+9y &= 3\end{align*}$$
     ::4x-3y=0-6x+9y=3
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  4. $$\begin{align*}4x-9y &= -20\\ -5x+15y &= 25\end{align*}$$
     ::4-9y20-5x+15y=25
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  5. $$\begin{align*}2x-3y &= -4\\ 8x+12y &= -24\end{align*}$$
     ::2 - 3y_ 48x+12y24
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  6. $$\begin{align*}3x-7y &= 12\\ -6x+14y &= -24\end{align*}$$
     ::3x-7y=12-6x+14y24
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  7. $$\begin{align*}x+5y &= 8\\ -2x-10y &= 16\end{align*}$$
     ::x+5y=8-2x-10y=16
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  8. $$\begin{align*}2x-y &= -5\\ -3x+2y &= 13\end{align*}$$
     ::2 - y=5 - 3x+2y=13
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  9. $$\begin{align*}x+y &= -1\\ -3x-2y &= 6\end{align*}$$
     ::xy_% 1 - 3x- 2y=6

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  Solve the systems below using Cramer’s Rule. You may wish to use your calculator to evaluate the determinants.
  ::使用 Cramer 规则解决下面的系统。 您不妨使用计算器来评估决定因素 。

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  10. $$\begin{align*}3x-y+2z &= 11\\ -2x+4y+13z &= -3\\ x+2y+9z &=5\end{align*}$$
      ::3 - y+2z=11 - 2x+4y+13z+13z+3x+2y+9z=5
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  11. $$\begin{align*}3x+2y-7z &= 1\\ -4x-3y+11z &= -2\\ x+4y-z &=7\end{align*}$$
      ::3x+2y-7z=1-4x-3y+11z%2x+4y-7z=7
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  12. $$\begin{align*}6x-9y+z &= -6\\ 4x+3y-2z &= 10\\ -2x+6y+z &=0\end{align*}$$
      ::6-9y+z64x+3y-2z=10-2x+6y+z=0
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  13. $$\begin{align*}x-2y+3z &= 5\\ 4x-y+4z &= 14\\ 5x+2y-4z &=-3\end{align*}$$
      ::x-2y+3z=54x-y+4z=145x+2y-4z%3
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  14. $$\begin{align*}-3x+y+z &= 10\\ 2x+y-2z &= 15\\ -4x-2y+4z &=-20\end{align*}$$
      ::-3x+y+z=102x+y-2z=15-4x-2y+4z+20
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  15. The Smith and Jamison families go to the county fair. The Smiths purchase 6 corndogs and 3 cotton candies for $21.75. The Jamisons purchase 3 corndogs and 4 cotton candies for $15.25. Write a system of linear equations and solve it using Cramer’s Rule to find the price of each food.
      ::史密斯和贾米森家族去县集。史密斯夫妇购买了6只玉米狗和3只棉花糖果,价格为21.75美元。杰米森夫妇购买了3只玉米狗和4只棉花糖果,价格为15.25美元。 写一个线性方程系统,用克拉默规则解决它,以寻找每只食物的价格。

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  Review (Answers)
  ::回顾(答复)

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  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 。