8.8 木板规则

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  木板规则

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  A system of equations can be represented and solved in general using matrices and .  This method can be significantly more efficient than eliminating variables in equations.   What does it mean for a solution method to be more efficient?  Is Cramer’s Rule the most efficient means of solving a system of equations?  
  ::使用矩阵和 。 这种方法比消除方程式中的变量效率高得多。 解决方案方法效率高意味着什么? Cramer 规则是否是解决方程式系统的最有效方法?

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  Using Cramer's Rule
  ::使用Cramer规则

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  The determinant is defined in a seemingly arbitrary way; however, when you look at the general solution for a $\begin{array}{r}2\times 2\end{array}$  matrix, the reasoning why it is defined this way is apparent. 
  ::决定因素的界定似乎是任意的;然而,当你审视2×2矩阵的总体解决办法时,为什么以这种方式界定它的理由显而易见。

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  $$\begin{array}{rl}ax+by& =e\\ cx+dy& =f\end{array}$$

  
  ::ax+by=ecx+dy=f =x+by=ecx+dy=f =xx=ecx+by=ecx+dy=f = ax+by=ecx+dy=f

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  When you solve the system above for $\begin{array}{r}y\end{array}$  and $\begin{array}{r}x\end{array}$ , you get the following:
  ::当为 y 和 x 解析上面的系统时,可获得以下信息:

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  $$\begin{array}{rl}y& =\frac{-af+ce}{-ad+bc}\\ x& =\frac{-bf+de}{ad-bc}\end{array}$$

  
  ::yaf+ce-ce-ad+bcx_bf+死-bc

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  Note that the system can be represented by the matrix and the solutions can be written as ratios of two determinants.  The determinant in the denominator is of the coefficient matrix. Cramer's Rule states that for two equations, the numerator of the  $\begin{array}{r}x\end{array}$ solution is the determinant of the new matrix whose columns are made up of the  $\begin{array}{r}y\end{array}$ coefficients and the solution coefficients.  The numerator of the  $\begin{array}{r}y\end{array}$ solution is the determinant of the new matrix made up of the  $\begin{array}{r}x\end{array}$ coefficients and the solution coefficients.
  ::注意系统可以用矩阵表示,解决办法可以写成两个决定因素的比率,分母的决定因素是系数矩阵。Cramer规则规定,对于两个方程式,X解决办法的分子是新矩阵的决定因素,其柱子由y系数和溶解系数组成。y解决办法的分子是x系数和溶解系数组成的新矩阵的决定因素。

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  $$\begin{array}{rl}\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =\left[\begin{array}{c}e\\ f\end{array}\right]\\ x& =\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}\\ y& =\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}\end{array}$$

  
  ::[abcd][xy]=[ef]x[ef]

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  This is a fantastic improvement over solving or elimination.  Cramer’s Rule also works with larger order matrices.  For a system of 3 variables and 3 equations the reasoning is identical.
  ::这是与解决或消除相比的一个惊人的改进。 Cramer规则(Cramer Rules)也与更大的顺序矩阵合作。 对于由3个变量和3个方程式组成的系统来说,推理是完全相同的。

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  $$\begin{array}{rl}ax+by+cz& =j\\ dx+ey+fz& =k\\ gx+hy+iz& =l\end{array}$$

  
  ::ax++by+cz=jdx+ey+fz=kkx+hy+iz=l

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  The system can be represented as a matrix.
  ::该系统可以作为矩阵表来代表。

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  $\begin{array}{r}\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}j\\ k\\ l\end{array}\right]\end{array}$
  ::[abcdefghi] [xyz]=[jkl] [jkl] [abcdefghi] ] [xyz]=[jkl] [jkl]

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  The three solutions can be represented as a ratio of determinants.
  ::三种解决办法可以作为决定因素的比例来代表。

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  $$\begin{array}{rl}x& =\frac{\left|\begin{array}{ccc}j& b& c\\ k& e& f\\ l& h& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}\\ y& =\frac{\left|\begin{array}{ccc}a& j& c\\ d& k& f\\ g& l& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}\\ z& =\frac{\left|\begin{array}{ccc}a& b& j\\ d& e& k\\ g& h& l\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}\end{array}$$

  
  ::jbcde fhi jbcde fhi

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  Remember that evaluating the determinants of  $\begin{array}{r}3\times 3\end{array}$ matrices using Sarrus’s rule is very efficient.
  ::请记住,使用萨鲁斯规则评估3x3矩阵的决定因素非常有效。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked about efficient solutions. You have seen that using traditional row reduction to solve a system of equations can take a while and use up a lot of paper.  Efficiency partly means requiring less time and space.  If this was all that efficiency meant then it would not make sense to solve systems of two equations with two unknowns using matrices because the solution could be found more quickly using substitution.  However, the other part of efficiency is minimizing the number of decisions that have to be made.  A computer is very good at adding, subtracting and multiplying numbers, but not very good at deciding whether eliminating  $\begin{array}{r}x\end{array}$ or eliminating  $\begin{array}{r}y\end{array}$ would be better.  This is why a definite algorithm using matrices and Cramer’s Rule is more efficient. 
  ::早些时候,您被问及高效解决方案。 您已经看到, 使用传统的减排来解决方程式系统需要一段时间, 并使用大量纸张。 效率部分意味着需要更少的时间和空间。 如果所有效率都意味着效率, 那么用矩阵来解决两个方程式的系统, 使用两个未知方程式, 因为使用替代可以更快地找到解决方案。 但是, 效率的另一方面是将必须做出的决定数量减少到最低程度。 计算机在添加、 减去和乘法数字方面非常行, 但对于决定删除 x 或删除 y 效果更好并不十分行。 这就是为什么使用矩阵和 Cramer 规则的确定算法更有效率。

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  Example 2
  ::例2

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  Represent the following system of equations as a matrix equation and solve using Cramer's Rule.
  ::代表下列方程式系统,作为矩阵方程式,使用Cramer规则解决。

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  $$\begin{array}{rl}y-13& =-3x\\ x& =19-4y\end{array}$$

  
  ::y - 13 3xx= 19-4yy

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  First write each equation in standard form .
  ::先以标准格式写下每个方程 。

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  $$\begin{array}{rl}3x+y& =13\\ x+4y& =19\end{array}$$

  
  ::3x+y=13x+4y=19

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  Then write as a coefficient matrix times a variable matrix equal to a solution matrix.
  ::然后以系数矩阵方式写入一个变量矩阵乘以等于解决方案矩阵的变量矩阵。

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  $\begin{array}{r}\left[\begin{array}{cc}3& 1\\ 1& 4\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}13\\ 19\end{array}\right]\end{array}$
  ::[3114]][xy]=[1319]

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  $$\begin{array}{rl}x& =\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}=\frac{\left|\begin{array}{cc}13& 1\\ 19& 4\end{array}\right|}{\left|\begin{array}{cc}3& 1\\ 1& 4\end{array}\right|}=\frac{13\cdot 4-19\cdot 1}{3\cdot 4-1\cdot 1}=\frac{33}{11}=3\\ y& =\frac{\left|\begin{array}{cc}3& 13\\ 1& 19\end{array}\right|}{\left|\begin{array}{cc}3& 1\\ 1& 4\end{array}\right|}=\frac{3\cdot 19-13}{11}=\frac{44}{11}=4\end{array}$$

  
  ::

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  Example 3
  ::例3

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  What is  $\begin{array}{r}y\end{array}$ equal to in the following system?
  ::在以下系统中,Y等于什么?

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  $$\begin{array}{rl}x+2y-z& =0\\ 7x-0y+z& =14\\ 0x+y+z& =10\end{array}$$

  
  ::x+2y-z=07x-0y+z=140x+y+z=10

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  If you attempted to solve this using elimination, it would take over a page of writing and rewriting to solve.  Cramer’s Rule speeds up the solving process. 
  ::如果你试图用除去来解决这个问题,它会占用一页书写和重写来解决。 Cramer规则加快了解决过程。

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  $\begin{array}{r}\left[\begin{array}{ccc}1& 2& -1\\ 7& 0& 1\\ 0& 1& 1\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 14\\ 10\end{array}\right]\end{array}$
  ::[[12-1701011] [xyz] = [01410]

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  $\begin{array}{r}y=\frac{\left|\begin{array}{ccc}a& j& c\\ d& k& f\\ g& l& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}=\frac{\left|\begin{array}{ccc}1& 0& -1\\ 7& 14& 1\\ 0& 10& 1\end{array}\right|}{\left|\begin{array}{ccc}1& 2& -1\\ 7& 0& 1\\ 0& 1& 1\end{array}\right|}=\frac{14+0+(-70)-0-10-0}{0+0+(-7)-0-1-14}=\frac{-66}{-22}=3\end{array}$
  ::- 17141010112 - 170101114+0+(-)-(-)0-0-10-00+0+(-7)-0-1-1466-22=3

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  Example 4
  ::例4

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  Solve the following system using Cramer’s Rule.
  ::使用Cramer规则解决以下系统。

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  $$\begin{array}{rl}5x+12y& =72\\ 18x-12y& =108\end{array}$$

  
  ::5x+12y=7218x-12y=108

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  $$\begin{array}{rl}x& =\frac{\left|\begin{array}{cc}72& 12\\ 108& -12\end{array}\right|}{\left|\begin{array}{cc}5& 12\\ 18& -12\end{array}\right|}=\frac{72\cdot (-12)-12\cdot 108}{5\cdot (-12)-12\cdot 18}=\frac{-2160}{-276}=\frac{180}{23}\\ y& =\frac{\left|\begin{array}{cc}5& 72\\ 18& 108\end{array}\right|}{\left|\begin{array}{cc}5& 12\\ 18& -12\end{array}\right|}=\frac{5\cdot 108-72\cdot 18}{-276}=\frac{-756}{-276}=\frac{63}{23}\end{array}$$

  
  ::X7212108-121251218-1212_12_12_12_12_12_12_12_12_12_12_18_18_2160-276=181823y_57218108_51218-12_5_5_108_78_18-18_276_276_756_276=6323

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  Example 5
  ::例5

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  What is the value of $\begin{array}{r}z\end{array}$  in the following system?
  ::兹在以下系统中的价值是什么?

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  $$\begin{array}{rl}3x+2y+z& =7\\ 4x+0y+z& =6\\ 6x-y+0z& =5\end{array}$$

  
  ::3x+2y+z=74x+0y+z=66x-y+0z=5

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  $$\begin{array}{rl}z=\frac{\left|\begin{array}{ccc}3& 2& 7\\ 4& 0& 6\\ 6& -1& 5\end{array}\right|}{\left|\begin{array}{ccc}3& 2& 1\\ 4& 0& 1\\ 6& -1& 0\end{array}\right|}& =\frac{0+2\cdot 6\cdot 6+7\cdot 4\cdot (-1)-0-(-1)\cdot 6\cdot 3-5\cdot 4\cdot 2}{0+2\cdot 1\cdot 6+1\cdot 4\cdot (-1)-0-(-1)\cdot 1\cdot 3-0}\\ & =\frac{22}{11}\\ & =2\end{array}$$

  
  ::3274066-153214016-100+26-6+74(-1)-0-(-1)-63-5420+21}6+144(-1)-7-(-1)-1-(-1)11__3-0=2211=2

  Summary

  播放段落 Cramer's Rule states that each solution of a two equation system can calculated as a ratio of two determinants: 播放段落 The numerator of the x solution is the determinant of a new matrix whose columns are made up of the y coefficients and the solution coefficients. The same is done for the y solution with the x coefficients. ::x 溶解的分子是新矩阵的决定因素,其列由 y 系数和溶解系数组成。对于 x 系数的y 溶解也这样做。 播放段落 The denominator for both solutions is simply the determinant of the coefficient matrix. ::这两种解决办法的分母仅仅是系数矩阵的决定因素。 ::Cramer规则指出,两个方程式系统的每种解决办法都可以按两个决定因素的比率计算:x方程式的分子是新矩阵的决定因素,其柱子由y系数和溶解系数组成。对x系数的y溶液也是一样。两种解决办法的分母只是系数矩阵的决定因素。 播放段落 Cramer's Rule can be applied to larger order matrices, such as systems with 3 variables and 3 equations. ::Cramer规则可适用于较大的订单矩阵,例如具有3个变量和3个方程式的系统。

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  Review
  ::回顾

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  Solve the following systems of equations using Cramer’s Rule.  If one solution does not exist, explain. 
  ::使用Cramer规则解决以下方程式系统。 如果一个解决方案不存在,请解释。

  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

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  6. Use Cramer’s Rule to solve for $\begin{array}{r}x\end{array}$  in the following system:
  ::6. 使用Cramer规则在下列系统中解决x的问题:

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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

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  7. Use Cramer’s Rule to solve for $\begin{array}{r}y\end{array}$  in the following system:
  ::7. 使用Cramer规则在下列系统中解决y:

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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

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  8. Use Cramer’s Rule to solve for $\begin{array}{r}z\end{array}$  in the following system:
  ::8. 使用Cramer规则解决下列系统中的z问题:

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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

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  9. Use Cramer’s Rule to solve for $\begin{array}{r}x\end{array}$  in the following system:
  ::9. 使用Cramer规则在下列系统中解决x的问题:

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

  
  ::2x+y-2z%5-4x-2y+3z=23x+y-z=3

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  10. Use Cramer’s Rule to solve for $\begin{array}{r}y\end{array}$  in the following system:
  ::10. 在下列系统中使用Cramer规则解决y:

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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

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  11. Use Cramer’s Rule to solve for $\begin{array}{r}z\end{array}$  in the following system:
  ::11. 使用Cramer规则解决下列系统中的z:

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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

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  Solve the following systems of equations using Cramer’s Rule.  Practice using your calculator to help with at least one problem.  If one solution does not exist, explain. 
  ::使用 Cramer 规则解决以下方程式系统 。 使用计算器至少帮助解决一个问题。 如果没有一个解决方案,请解释 。

  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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  16.  When using Cramer’s Rule to solve a system of equations you will occasionally find that the determinant of the coefficient matrix is zero.  When this happens, how can you tell whether your system has no solution or infinite solutions?
  ::16. 当使用Cramer规则解决方程式体系时,你偶尔会发现系数矩阵的决定因素是零,当这种情况发生时,你如何判断你的体系是没有解决方案还是没有无限解决方案?

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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 。