10.8 决定因素

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  决定因素

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  Introduction
  ::导言

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  Suppose your math professo r provided you with the coordinates of three points in the coordinate plane ,  $\begin{array}{r}A=(2,1),B=(5,6),\text{ and }C=(9,-1).\end{array}$  Your professor then  asked you to determine if the points are collinear (lie on the same line), and, if not, to find the area of the triangle formed by the points. How would you answer your teacher's questions?
  ::假设你的数学教授给了你坐标平面上三个点的坐标, A=( 2) 1, B=( 5) 和 C=( 9) - ( 1) 。 然后,你的教授要求你确定这些点是否是圆线( 在同一条线上) , 如果不是的话, 找到点所形成的三角形区域。 您将如何回答老师的问题 ?

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  Determinant
  ::决定因素

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  A determinant is a number computed from the entries in a square matrix . The determinant of a matrix  is defined only when a matrix is a square matrix.  It has many properties and interpretations that you will explore in linear algebra. This concept is focused on the procedure of calculating determinants. Once you know how to calculate the determinant of a $\begin{array}{r}2\times 2\end{array}$  matrix, you will be able to calculate the determinant of a $\begin{array}{r}3\times 3\end{array}$ matrix. Once you know how to calculate the determinant of a $\begin{array}{r}3\times 3\end{array}$  matrix, you can calculate the determinant of a $\begin{array}{r}4\times 4,\end{array}$  and so on. 
  ::决定因素是从一个平方矩阵中的条目中计算出来的数字。只有当一个矩阵是一个平方矩阵时,矩阵的决定因素才会被定义。它有许多属性和解释,将在线性代数中加以探讨。这个概念侧重于计算决定因素的程序。一旦你知道如何计算一个 2x2 矩阵的决定因素,你就可以计算一个 3x3 矩阵的决定因素。一旦知道如何计算一个 3x3 矩阵的决定因素,你可以计算一个 4x4 的决定因素,等等。

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  The determinant of matrix  $\begin{array}{r}A\end{array}$ is written as $\begin{array}{r}|A|\end{array}$  or det A.  For a $\begin{array}{r}2\times 2\end{array}$  matrix $\begin{array}{r}A\end{array}$ , the value is calculated as
  ::矩阵 A 的决定因素以 A 或 det A 写成。 对于 2x2 矩阵 A , 数值按

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  $$\begin{array}{rl}A& =\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\\ \text{det}A& =|A|=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc.\end{array}$$

  
  ::A=[abcd]did AAAabcdad-bc。

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  Notice how the elements on each diagonal are multiplied and then subtracted. 
  ::注意每个对角的元素如何乘以,然后减去。

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  A logical question about determinants is, where does the procedure come from? Why are determinants defined the way they are? Determinants for $\begin{array}{r}2\times 2\end{array}$  matrices are defined the way they are because of the general solution to a system of 2 variables and 2 equations:
  ::关于决定因素的一个逻辑问题是,程序来自何方?为什么决定因素是按其方式界定的?2x2矩阵的决定因素是按对两个变量和两个方程式系统的一般解决办法界定的:

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

  
  ::ax+by=ecx+dy=f。

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  To eliminate the $\begin{array}{r}x\end{array}$ , scale the first equation by  $\begin{array}{r}c\end{array}$ and the second equation by a.
  ::要删除 x, 将第一个方程式缩放为 c, 第二个方程式缩放为 a 。

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  $$\begin{array}{rl}acx+bcy& =ec\\ acx+ady& =af\end{array}$$

  
  ::acx+bcy=ecacx+ady=af 缩写

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  Subtract the second equation by the first, and solve for $\begin{array}{r}y\end{array}$ . 
  ::将第二个方程减去第一个方程,然后解决y。

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  $$\begin{array}{rl}ady-bcy& =af-ec\\ y(ad-bc)& =af-ec\\ y& =\frac{af-ec}{ad-bc}\end{array}$$

  
  ::a-bcy=af-ecy(ad-bc)=af-ecy=af-ecad-bc

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  When you solve for $\begin{array}{r}x,\end{array}$  you also get $\begin{array}{r}ad-bc\end{array}$  in the denominator of the general solution. This pattern led people to start using this strategy in solving systems of equations. The determinant is defined in this way so it will always be the denominator of the general solution of either variable . 
  ::当您解决 x 时,您也会在通用解决方案的分母中获得 ad-bc 。这种模式引导人们开始使用此策略解决方程式系统。此方式定义了决定因素,因此它始终是任一变量的一般解决方案的分母。

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  The determinant of a $\begin{array}{r}3\times 3\end{array}$  matrix is more involved.
  ::3x3矩阵的决定因素涉及较多。

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  $$\begin{array}{r}B=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\end{array}$$

  
  ::B = [abcdefghi]

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  Usually you will start by looking at the top row, although any row or column will work. Then use the checkerboard pattern for signs (shown below) and create smaller $\begin{array}{r}2\times 2\end{array}$  matrices.
  ::通常您会首先查看上行, 尽管任何行或列都会工作。 然后使用检查板模式来显示符号( 显示在下面) , 并创建较小的 2x2 矩阵 。

  $$\begin{array}{r}\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right]\end{array}$$

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  The smaller  $\begin{array}{r}2\times 2\end{array}$ matrices are the entries that remain when the row and column of the coefficient you are working with are ignored.
  ::较小的 2x2 矩阵是当您正在使用的系数的行和列被忽略时留下的条目 。

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  $$\begin{array}{r}\text{det}B=|B|=+a\cdot \left|\begin{array}{cc}e& f\\ h& i\end{array}\right|-b\cdot \left|\begin{array}{cc}d& f\\ g& i\end{array}\right|+c\cdot \left|\begin{array}{cc}d& e\\ g& h\end{array}\right|\end{array}$$

  
  ::

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  Next, take the determinant of the smaller $\begin{array}{r}2\times 2\end{array}$  matrices, and you get a long string of computations.
  ::接下来,选择小于 2x2 矩阵的决定因素, 然后你得到一长串计算。

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  $$\begin{array}{rl}\text{det }B& =+a(ei-fh)-b(di-fg)+c(dh-eg)\\ & =aei-afh-bdi+bfg+cdh-ceg\\ & =aei+bfg+cdh-ceg-afh-bdi\end{array}$$

  
  ::=aei-afh+c(dh-eg)=aei-afh-bdi+bfg+cdh-ceg=aei+bfg+cdh-cdh-ceg-afh-bdi

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  Most people do not remember this sequence. A French mathematician named Pierre Frédéric Sarrus demonstrated a great device to memorize the computation of the determinant for $\begin{array}{r}3\times 3\end{array}$  matrices. The first step is simply to copy the first two columns to the right of the matrix. Then draw three diagonal lines going down and to the right.
  ::大多数人不记得这个序列。 一位叫皮埃尔·弗雷德里克·萨尔鲁斯的法国数学家展示了一个伟大的装置,可以记住计算3x3 矩阵的决定因素。 第一步只是将前两列复制到矩阵右侧。 然后绘制三条对角线向下和向右。

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  $$\begin{array}{r}B=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\end{array}$$

  
  ::B = [abcdefghi]

  

  
  

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  Recall the formula derived above:
  ::回顾上述公式:

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  $$\begin{array}{r}\text{det}B=aei+bfg+cdh-ceg-afh-bdi\end{array}$$

  
  ::dat B=aei+bfg+cdh-ceg-afh-bdi =aei+bfg+cdh-ceg-afh-bdi =aei+bfg+cdh-ceg-efh-bdi

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  These three diagonals run through the variables  aei, bfg, and cdh . These are precisely the first three positive products in the formula.
  ::这三个对角线通过变量aei, bfg, 和 cdh。 这些正是公式中前三个正值产品 。

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  Next, draw three diagonals going up and to the right. These three diagonals run through the variables  ceg, afh, and bdi. These are precisely the last three  negative products in the formula. It is important to note that  Sarrus's Rule does not work for the determinants of matrices that are not of order $\begin{array}{r}3\times 3\end{array}$ .
  ::接下来, 绘制三个对角线向上和向右。 这三个对角线通过变量 geg、 afh 和 bdi 。 这些正是公式中最后三个负值产品。 必须指出, Sarrus 的规则不适用于非顺序 3x3 的矩阵决定因素 。

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  The following video demonstrates how to evaluate $\begin{array}{r}2\times 2\end{array}$  and $\begin{array}{r}3\times 3\end{array}$  determinants:  
  ::以下视频展示了如何评价2×2和3×3决定因素:

   

   

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  Examples
  ::实例

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  Example 1
  ::例1

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  Find $\begin{array}{r}\text{det}A\end{array}$  for the matrix:
  

  $$\begin{array}{r}A=\left[\begin{array}{cc}3& 2\\ 1& 5\end{array}\right]\end{array}$$

  
  ::为矩阵查找 det A: A=[3215]

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  Solution:
  

  $$\begin{array}{r}\left|\begin{array}{cc}3& 2\\ 1& 5\end{array}\right|=3\cdot 5-2\cdot 1=15-2=13\end{array}$$

  
  ::解答:%321535-21=15-2=13

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  Example 2
  ::例2

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  Find $\begin{array}{r}\text{det}B\end{array}$  for the matrix:
  

  $$\begin{array}{r}B=\left[\begin{array}{ccc}3& 2& 1\\ 5& 0& 2\\ 2& 1& 5\end{array}\right]\end{array}$$

  
  ::B=[321502215]

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  Solution:
  ::解决方案 :

  $$\begin{array}{rl}\left|\begin{array}{ccc}3& 2& 1\\ 5& 0& 2\\ 2& 1& 5\end{array}\right|& =3\left|\begin{array}{cc}0& 2\\ 1& 5\end{array}\right|-2\left|\begin{array}{cc}5& 2\\ 2& 5\end{array}\right|+1\left|\begin{array}{cc}5& 0\\ 2& 1\end{array}\right|\\ & =3(0\cdot 5-2\cdot 1)-2(5\cdot 5-2\cdot 2)+1(5\cdot 1-2\cdot 0)\\ & =-6-42+5\\ & =-43\end{array}$$

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  Example 3
  ::例3

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  Find the determinant of $\begin{array}{r}B\end{array}$  from Example 2 using Sarrus's Rule. 
  ::根据Sarrus的规则,从例2中找出B的决定因素。

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  Solution:
  ::解决方案 :

  $$\begin{array}{r}\begin{array}{ccccc}3& 2& 1& 3& 2\\ 5& 0& 2& 5& 0\\ 2& 1& 5& 2& 1\end{array}\end{array}$$

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  $$\begin{array}{r}\text{det}B=0+8+5-0-6-50=-43\end{array}$$

  
  ::B=0+8+5-5-0-6-50{_________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  As you can see, Sarrus's Rule is efficient, and much of the calculations can be done mentally. Additionally, zero values make much of the multiplication easier.
  ::如你所见,Sarrus的规则是有效的, 大部分的计算可以在精神上进行。此外,零值让乘法更容易很多。

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  Example 4
  ::例4

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  Recall the problem from the Introduction. A re  the points $\begin{array}{r}A=(2,1),B=(5,6),\text{ and }C=(9,-1)\end{array}$  collinear (lie on the same line)? If not, what is the area of the triangle formed by the points?
  ::回顾导言中的问题。 A = 2, 1, B = (5,6) 和 C = (9, - 1) 圆线( 在同一行上) 吗 ? 如果不是, 点构成的三角区域是什么 ?

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  Solution:
  ::解决方案 :

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  Finding the determinant   of the matrix formed by the points could help you answer both questions. 
  ::找到由各点构成的矩阵的决定因素可有助于回答这两个问题。

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  Three points are collinear if and only if the determinant of the matrix with   $\begin{array}{r}x\end{array}$ -coordinates in the first column,  $\begin{array}{r}y\end{array}$ -coordinates in the second column, and ones in the third column is equal to zero.
  ::如果且只有在第一栏内带有x坐标的矩阵的决定因素为第二栏,Y坐标为第二栏,而第三栏内坐标等于零时,三点为圆线。

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  $$\begin{array}{rl}A& =\left[\begin{array}{ccc}2& 1& 1\\ 5& 6& 1\\ 9& -1& 1\end{array}\right]\\ |A|& =\left|\begin{array}{ccc}2& 1& 1\\ 5& 6& 1\\ 9& -1& 1\end{array}\right|\\ & =2\left|\begin{array}{cc}6& 1\\ -1& 1\end{array}\right|-1\left|\begin{array}{cc}5& 1\\ 9& 1\end{array}\right|+1\left|\begin{array}{cc}5& 6\\ 9& -1\end{array}\right|\\ & =2(6\cdot 1-1\cdot -1)-1(5\cdot 1-1\cdot 9)+1(5\cdot -1-6\cdot 9)\\ & =14+4-59\\ & =-41\end{array}$$

   
  ::A=[2115619-11] A2115619-11261-1111519111569-12(61-11)-1-1-1-1(51-19)+1(51-69)=14+4-5941

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  Since the  $\begin{array}{r}\text{det }A\ne 0\end{array}$ , the given points are not collinear.  
  ::自第AQQ0号指令以来,给定的点不是线性线性点 。

   

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  The area of the triangle formed by three points is half the absolute value of the determinant of this  matrix. 
  ::由三点组成的三角区域是这个矩阵决定因素的绝对值的一半。

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  $$\begin{array}{rl}\text{Area}& =\frac{1}{2}\cdot |\text{ det }A|\\ & =\frac{1}{2}\cdot |-41|\\ & =\frac{41}{2}\\ & =20.5\end{array}$$

   
  ::区域=12* det A* 12* 41* 412=20.5

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  Thus, the area of the triangle formed by the three given points is 20.5. 
  ::因此,三个给定点构成的三角形区域为20.5。

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  Example 5
  ::例5

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  1) Find the determinant of the following matrix:
  :1) 确定下列矩阵的决定因素:

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  $$\begin{array}{r}C=\left[\begin{array}{cc}-4& 12\\ 1& -3\end{array}\right]\end{array}$$

  
  ::C=[-4121-3]

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  Solution:
  

  $$\begin{array}{r}\text{det}C=\left|\begin{array}{cc}-4& 12\\ 1& -3\end{array}\right|=12-12=0\end{array}$$

  
  ::解答: det C4121-312-12=0

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  2) Find the determinant of the following matrix:
  :2) 确定下列矩阵的决定因素:

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  $\begin{array}{r}D=\left[\begin{array}{ccc}4& 8& 3\\ 0& 1& 7\\ 12& 5& 13\end{array}\right]\end{array}$
  ::D=[48301712513]

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  Solution:
  

  $$\begin{array}{r}\text{det}D=\left|\begin{array}{ccc}4& 8& 3\\ 0& 1& 7\\ 12& 5& 13\end{array}\right|=4\cdot 13+8\cdot 7\cdot 12+0-3\cdot 12-5\cdot 7\cdot 4-0=548\end{array}$$

  
  ::解 答: det D48301712513413+8712+0-312-574-0=548

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  3) Find the determinant of the following $\begin{array}{r}4\times 4\end{array}$  matrix by carefully choosing the row or column to work with:
  ::3) 仔细选择要工作的行或列,找出下列 4x4 矩阵的决定因素:

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  $\begin{array}{r}E=\left[\begin{array}{cccc}4& 5& 0& 2\\ -1& -3& 0& 3\\ 4& 8& 1& 5\\ -3& 2& 0& 9\end{array}\right]\end{array}$
  ::E=[4502-1-3034815-3209]

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  Solution:
  ::解决方案 :

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  Notice that the third column is made up with zeros and a one. Choose this column to make up the coefficients, because then instead of having to evaluate the determinant of four individual $\begin{array}{r}3\times 3\end{array}$  matrices, you need to do only  one.
  ::请注意, 第三列由零和一组成。 选择此列来组成系数, 因为这样你不必评估四个 3x3 矩阵的决定因素, 您只需要做一个 。

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  $$\begin{array}{rl}\left|E\right|& =\left|\begin{array}{cccc}4& 5& 0& 2\\ -1& -3& 0& 3\\ 4& 8& 1& 5\\ -3& 2& 0& 9\end{array}\right|\\ & =0\cdot \left|\begin{array}{ccc}-1& -3& 3\\ 4& 8& 5\\ -3& 2& 9\end{array}\right|-0\cdot \left|\begin{array}{ccc}4& 5& 2\\ 4& 8& 5\\ -3& 2& 9\end{array}\right|+1\cdot \left|\begin{array}{ccc}4& 5& 2\\ -1& -3& 3\\ -3& 2& 9\end{array}\right|-0\cdot \left|\begin{array}{ccc}4& 5& 2\\ -1& -3& 3\\ 4& 8& 5\end{array}\right|\\ & =\left|\begin{array}{ccc}4& 5& 2\\ -1& -3& 3\\ -3& 2& 9\end{array}\right|\\ & =4\cdot (-3)\cdot 9+5\cdot 3\cdot (-3)+2\cdot (-1)\cdot 2-2\cdot (-3)\cdot (-3)-4\cdot 3\cdot 2-5\cdot (-1)\cdot 9\\ & =-154\end{array}$$

  
  ::4502-1-3034815-3209-01-33485-3290452485-3291452-1-333-3290452-1-1-329}452-1-33485452-1-1-329}(3-3)__9+533(3-3)+22__(1)-2__(3)____(3)__(3)__(3)____(3)__(3)__(3)__(3)__(3)__(3)__(3)__(3)__(3)__(3)__(3)__(3)__(3)__(3)__(3)__(3)__________(2)______________(9)__________________________________________________________________________________

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  Summary
  ::摘要

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  • The determinant of a matrix is a number calculated from the entries in a matrix. The procedure is derived from solving linear systems. 
    ::矩阵的决定因素是从矩阵条目中计算的数字,该程序来自解决线性系统。
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  • The determinant of matrix  A  can be expressed  as  det A  or | A |.
    ::矩阵A的决定因素可以表示为“A”或“A”表示。
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  • Sarrus's Rule is a memorization technique that enables you to compute the determinant of $\begin{array}{r}3\times 3\end{array}$  matrices efficiently. 
    ::Sarrus的规则是一种记忆技术 它可以高效地计算 3x3 矩阵的决定因素。

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  Review
  ::回顾

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  Determine  the determinants of each of the following matrices:
  ::确定下列各矩阵的决定因素:

  1. $\begin{array}{r}\left[\begin{array}{cc}4& 5\\ 2& 3\end{array}\right]\end{array}$

  2. $\begin{array}{r}\left[\begin{array}{cc}-3& 6\\ 2& 5\end{array}\right]\end{array}$

  3.  $\begin{array}{r}\left[\begin{array}{cc}-1& 2\\ 2& 0\end{array}\right]\end{array}$

  4.  $\begin{array}{r}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

  5.  $\begin{array}{r}\left[\begin{array}{cc}6& 5\\ 2& -2\end{array}\right]\end{array}$

  6.  $\begin{array}{r}\left[\begin{array}{cc}1& 2\\ 6& 3\end{array}\right]\end{array}$

  7.  $\begin{array}{r}\left[\begin{array}{ccc}-1& 3& -4\\ 4& 2& 1\\ 1& 2& 5\end{array}\right]\end{array}$

  8.  $\begin{array}{r}\left[\begin{array}{ccc}4& 5& 8\\ 9& 0& 1\\ 0& 3& -2\end{array}\right]\end{array}$

  9.  $\begin{array}{r}\left[\begin{array}{ccc}0& 7& -1\\ 2& -3& 1\\ 6& 8& 0\end{array}\right]\end{array}$

  10.  $\begin{array}{r}\left[\begin{array}{ccc}4& 2& -3\\ 2& 4& 5\\ 1& 8& 0\end{array}\right]\end{array}$

  11.  $\begin{array}{r}\left[\begin{array}{ccc}-2& -6& -12\\ -1& -5& -2\\ 2& 3& 4\end{array}\right]\end{array}$

  12.  $\begin{array}{r}\left[\begin{array}{ccc}-2& 6& 3\\ 2& 4& 0\\ -8& 2& 1\end{array}\right]\end{array}$

  13.  $\begin{array}{r}\left[\begin{array}{cccc}2& 6& 4& 6\\ 0& 1& 0& 1\\ 2& 4& 2& 0\\ -6& 2& 3& 1\end{array}\right]\end{array}$

  14. $\begin{array}{r}\left[\begin{array}{cccc}5& 0& 0& 1\\ 2& 1& 8& 3\\ 9& 3& 2& 6\\ -4& 2& 5& 1\end{array}\right]\end{array}$

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  15. Can you find the determinant for any matrix? Explain.
  ::15. 你能找到任何矩阵的决定因素吗?

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  16. The following matrix has a determinant of zero:  $\begin{array}{r}\left[\begin{array}{cc}6& 4\\ 3& 2\end{array}\right]\end{array}$ . If the determinant of a matrix is zero, what does that say about the rows of the matrix?
  ::16. 下列矩阵有一个零的决定因素:[6432],如果矩阵的决定因素是零,那么矩阵的行数又如何?

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  Determine if the given points below are collinear. If not, then determine the area of each triangle with vertices given below.
  ::确定下面给定的点是否为圆线性点。 如果没有, 请用下面给出的顶点来确定每个三角的面积 。

  播放段落

  17. (2, -1), (-5, 2), and (0, 6)
  ::17.(2,-1,(5,5,2)和(0,6)

  播放段落

  18. (-8, 12), (10, 5), and (1, -4)
  ::18(-8, 12), (10, 5), 和 (1, 4)

  播放段落

  19. (-7, 2), (8, 0), and (3, -4)
  ::19. (-7,2), (8,0)和(3,4)

  播放段落

  Review (Answers )
  ::回顾(答复)

  播放段落

  Please see the Appendix.
  ::请参看附录。