节: 行操作和行梯表 | 10.6 直排操作和直排梯层表格

章节大纲

Introduction
::导言

Applying row operations to reduce a matrix is a procedural skill that takes lots of writing, rewriting, and careful arithmetic. The payoff for being able to transform a matrix into a simplified form will become clear later. For now, what does the simplified form mean for a matrix?
::应用行操作来减少矩阵是一个程序技能,需要大量书写、重写和仔细算术。能够将矩阵转换为简化格式的回报稍后会变得清晰。现在,简化格式对矩阵意味着什么?

Row Operations
::行操作

The process of solving a system of linear equations by elimination is similar to the process of solving matrices. Notice the connection between a system of linear expressions and its coefficient matrix.

$\begin{aligned} A x & + B y + C z \\ D x & + E y + F z \to [\begin{matrix} A & B & C \\ D & E & F \\ G & H & J \end{matrix}] \\ G x & + H y + J z \end{aligned}$
$\begin{align*}Ax & +By+Cz \\ Dx&+Ey+Fz \rightarrow \begin{bmatrix} A & B & C\\ D & E & F\\ G & H & J \end{bmatrix} \\ Gx&+Hy+Jz\end{align*}$
::通过消除解决线性方程系统的过程与解决矩阵的过程相似。注意线性表达式系统与其系数矩阵之间的联系。Ax+By+CzDx+Ey+Fz[ABCDEFGHJ]Gx+Hy+Jz

Thus, the operations that are permitted when solving a system of linear equations by elimination are the same as those permitted to act on matrices. Also, the goal for solving a matrix is the same as the goal of solving a system: solve for the unknown variables.
::因此,在通过删除解决线性方程式系统时允许的操作与允许在矩阵上采取行动的操作相同,此外,解决矩阵的目标与解决系统的目标相同:解决未知变量。

There are only three operations that are permitted to act on matrices:
::只有三项行动允许按矩阵表采取行动:

Add a multiple of one row to another row.
::将多个行加到另一行。

Scale a row by multiplying through by a nonzero constant .
::通过乘以非零常数来缩放一行。

Swap two rows.
::交换两排

Using these three operations, your job is to simplify matrices into row echelon form . Row echelon form must meet three requirements:
::使用这三种操作,您的工作是将矩阵简化为排梯表格。行梯层表格必须符合以下三项要求:

The leading coefficient of each row must be a one.
::每个行的主要系数必须是一。

All entries in a column below a leading one must be zero.
::前列一栏下的所有条目必须是零。

All rows that just contain zeros are at the bottom of the matrix .
::所有只包含零的行都在矩阵的底部。

Here are some examples of matrices in row echelon form:
::以下是以行梯层形式列出的一些矩阵实例:

$\begin{aligned} [\begin{matrix} 1 & 14 \\ 0 & 1 \end{matrix}], [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 4 \end{matrix}], [\begin{matrix} 1 & 2 & 3 & 5 & 6 \\ 0 & 0 & 1 & 4 & 7 \\ 0 & 0 & 0 & 1 & - 2 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 1 & 14\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 4 \end{bmatrix}, \begin{bmatrix} 1 & 2 & 3 & 5 & 6\\ 0 & 0 & 1 & 4 & 7\\ 0 & 0 & 0 & 1 & -2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\end{align*}$

Reduced row echelon form also has one extra stipulation compared with row echelon form:
::减少的排梯表与排梯表相比,还有一项额外规定:

Every leading coefficient of one must be the only nonzero element in that column.
::每一主要系数必须是该栏中唯一的非零系数。

Here are some examples of matrices in reduced row echelon form:
::以下是一些缩排梯层表格的矩阵实例:

$\begin{aligned} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}], [\begin{matrix} 1 & 0 & 3 \\ 0 & 1 & 4 \end{matrix}], [\begin{matrix} 1 & 2 & 0 & 0 & 6 \\ 0 & 0 & 1 & 0 & 7 \\ 0 & 0 & 0 & 1 & - 2 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 & 3\\ 0 & 1 & 4 \end{bmatrix}, \begin{bmatrix} 1 & 2 & 0 & 0 & 6\\ 0 & 0 & 1 & 0 & 7\\ 0 & 0 & 0 & 1 & -2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\end{align*}$

Putting a matrix into reduced row echelon form is a result of performing Gauss-Jordan elimination . The process illustrated in this concept is named after those mathematicians.
::将矩阵制成减排梯式是消除高斯-约旦的结果,这一概念所说明的过程是以这些数学家的名字命名的。

The following video discusses for the purpose of solving systems of equations, and row echelon and reduced row echelon forms of matrices:
::下列视频讨论,目的是解决方程系统、单排梯层和减排梯层矩阵表:

Examples
::实例

Example 1
::例1

Put the following matrix into reduced row echelon form:
::将下列矩阵划入缩排的梯层表:

$\begin{aligned} [\begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 3 & 7\\ 2 & 5 \end{bmatrix}\end{align*}$

Solution:
::解决方案 :

In each step of the solution, only one of the three row operations will be used. Specific shorthand will be introduced—namely, $\begin{align*}R_1\end{align*}$ refers to the first row, and $\begin{align*}R_2\end{align*}$ refers to the 2nd row.
::在解决方案的每个步骤中,将只使用三行操作中的一个。将引入具体的缩略语——即R1指第一行,R2指第二行。

$\begin{aligned} [\begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix}] \\ 3 \cdot R_{2} & \to [\begin{matrix} 3 & 7 \\ 6 & 15 \end{matrix}] \\ - 2 \cdot R_{1} + R_{2} & \to [\begin{matrix} 3 & 7 \\ 0 & 1 \end{matrix}] \\ - 7 R_{2} + R_{1} & \to [\begin{matrix} 3 & 0 \\ 0 & 1 \end{matrix}] \\ R_{1} \cdot \frac{1}{3} & \to [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \end{aligned}$
$\begin{align*}&\begin{bmatrix} 3 & 7\\ 2 & 5 \end{bmatrix} \\ 3 \cdot R_2 &\rightarrow \begin{bmatrix} 3 & 7\\ 6 & 15 \end{bmatrix} \\ -2 \cdot R_1 + R_2 & \rightarrow \begin{bmatrix} 3 & 7\\ 0 & 1 \end{bmatrix} \\ -7R_2+R_1 & \rightarrow \begin{bmatrix} 3 & 0\\ 0 & 1 \end{bmatrix} \\ R_1 \cdot \frac{1}{3} & \rightarrow \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\end{align*}$
::[37615]-2R1+R2}[3701]-7R2+R1}[3001]-7R2+R1][3001]-R1}[1001]

Row reducing a $\begin{align*}2 \times 2\end{align*}$ matrix to become the identity matrix is an exercise that illustrates the fact that the rows were linearly independent .
::将一个 2x2 矩阵缩小为身份矩阵的行说明各行是线性独立的。

Example 2
::例2

Put the following matrix into reduced row echelon form:
::将下列矩阵划入缩排的梯层表:

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

Solution:
::解决方案 :

$\begin{aligned} [\begin{matrix} 2 & 4 & 0 \\ 0 & 3 & 1 \\ 1 & 2 & 4 \end{matrix}] \\ R_{1} \cdot - \frac{1}{2} + R_{3} & \to [\begin{matrix} 2 & 4 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 4 \end{matrix}] \\ R_{3} \div 4 & \to [\begin{matrix} 2 & 4 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix}] \\ R_{1} \div 2, R_{2} \div 3 & \to [\begin{matrix} 1 & 2 & 0 \\ 0 & 1 & \frac{1}{3} \\ 0 & 0 & 1 \end{matrix}] \end{aligned}$
$\begin{align*}&\begin{bmatrix} 2 & 4 & 0\\ 0 & 3 & 1\\ 1 & 2 & 4 \end{bmatrix} \\ R_1 \cdot -\frac{1}{2} +R_3 & \rightarrow \begin{bmatrix} 2 & 4 & 0\\ 0 & 3 & 1\\ 0 & 0 & 4 \end{bmatrix} \\ R_3 \div 4 & \rightarrow \begin{bmatrix} 2 & 4 & 0\\ 0 & 3 & 1\\ 0 & 0 & 1 \end{bmatrix} \\ R_1 \div 2, \; R_2 \div 3 & \rightarrow \begin{bmatrix} 1 & 2 & 0\\ 0 & 1 & \frac{1}{3}\\ 0 & 0 & 1 \end{bmatrix}\end{align*}$
::[240031144]R112+R3}[240031004]R3}4}[240031001]R1}2,R2}3}3}3}4}[1(200113001]

Note that in the preceding step, two operations were used. This is acceptable when the operations do not interfere or interact with each other.
::请注意,前一步使用了两个操作,当操作不相互干扰或互动时可以接受。

$\begin{aligned} [\begin{matrix} 1 & 2 & 0 \\ 0 & 1 & \frac{1}{3} \\ 0 & 0 & 1 \end{matrix}] \\ R_{3} \cdot - \frac{1}{3} + R_{2} & \to [\begin{matrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \\ R_{2} \cdot - 2 + R_{1} & \to [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \end{aligned}$
$\begin{align*}& \begin{bmatrix} 1 & 2 & 0\\ 0 & 1 & \frac{1}{3}\\ 0 & 0 & 1 \end{bmatrix} \\ R_3 \cdot -\frac{1}{3} + R_2 & \rightarrow \begin{bmatrix} 1 & 2 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \\ R_2 \cdot -2 + R_1 & \rightarrow \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\end{align*}$
::[20010001]R313+R2}[20010001]R2+R1}[100000001]

Again, row reducing a $\begin{align*}3 \times 3\end{align*}$ matrix to become the identity matrix is just an exercise that illustrates the fact that the rows were linearly independent.
::同样,将3x3矩阵减为身份矩阵的行数只是表明各行线性独立这一事实的一种做法。

Example 3
::例3

In a single $\begin{align*}3 \times 3\end{align*}$ matrix, describe the general approach of Gauss-Jordan elimination. In other words, which locations would you try to focus on first?
::在单一的3x3矩阵中,描述消除高斯-约旦的一般方法。换句话说,你会首先关注哪个地点?

Solution:
::解决方案 :

One approach is to try to get a one in the A position. Then get a zero in position B and position C by multiplying by a multiple of row 1. Then try to get a zero in position D.
::一种方法是尝试在 A 位置上找到一个。然后在 B 位置和 C 位置上找到一个零, 乘以第 1 行的多个。然后尝试在 D 位置上找到一个零。

$\begin{aligned} [\begin{matrix} A & I & G \\ B & H & F \\ C & D & E \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} A & I & G\\ B & H & F\\ C & D & E \end{bmatrix}\end{align*}$
::[英GBHFCDE] [英GBHFCDE]

Every matrix may have a different strategy, and as long as you use the three row operations, you will be on the right track. One thing to be very careful of is to try to avoid fractions within your matrix. Scale the row to eliminate the fraction .
::每个矩阵都可能有不同的策略, 只要你使用三行操作, 你就会走上正确的轨道。需要非常小心的一件事就是尽量避免在矩阵中出现分数。缩放行以删除分数。

Example 4
::例4

Recall the problem from the Introduction. There are two forms of a matrix that are most simplified. The most important is reduced row echelon form that follows the four stipulations from the guidance section. An example of a matrix in reduced row echelon form is:
::回顾导言中的问题。有两种形式最简化的矩阵。最重要的是按照指导部分的四项规定降低行阶表。减排梯表的一个例子是:

$\begin{aligned} [\begin{matrix} 1 & 0 & 0 & 2 & 43 \\ 0 & 1 & 0 & 2 & 3 \\ 0 & 0 & 1 & 98 & 5 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 1 & 0 & 0 & 2 & 43\\ 0 & 1 & 0 & 2 & 3\\ 0 & 0 & 1 & 98 & 5 \end{bmatrix}\end{align*}$

Example 5
::例5

Reduce the following matrix to reduced row echelon form:
::将下列矩阵减为减缩的行梯层表:

$\begin{aligned} [\begin{matrix} 0 & 4 & 5 \\ 2 & 6 & 8 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 0 & 4 & 5\\ 2 & 6 & 8 \end{bmatrix}\end{align*}$

Solution:
::解决方案 :

$\begin{aligned} [\begin{matrix} 0 & 4 & 5 \\ 2 & 6 & 8 \end{matrix}] \\ Switch Rows & \to [\begin{matrix} 2 & 6 & 8 \\ 0 & 4 & 5 \end{matrix}] \\ R_{1} \div 2 & \to [\begin{matrix} 1 & 3 & 4 \\ 0 & 4 & 5 \end{matrix}] \\ R_{2} \cdot \frac{1}{4} & \to [\begin{matrix} 1 & 3 & 4 \\ 0 & 1 & \frac{5}{4} \end{matrix}] \\ R_{2} \cdot - 3 + R_{1} & \to [\begin{matrix} 1 & 0 & \frac{1}{4} \\ 0 & 1 & \frac{5}{4} \end{matrix}] \end{aligned}$
$\begin{align*}& \begin{bmatrix} 0 & 4 & 5\\ 2 & 6 & 8 \end{bmatrix}\\ \text{Switch Rows} & \rightarrow \begin{bmatrix} 2 & 6 & 8\\ 0 & 4 & 5 \end{bmatrix} \\ R_1 \div 2 & \rightarrow \begin{bmatrix} 1 & 3 & 4\\ 0 & 4 & 5 \end{bmatrix} \\ R_2 \cdot \frac{1}{4} &\rightarrow \begin{bmatrix} 1 & 3 & 4\\ 0 & 1 & \frac{5}{4} \end{bmatrix} \\ R_2 \cdot -3 + R_1 &\rightarrow \begin{bmatrix} 1 & 0 & \frac{1}{4} \\ 0 & 1 & \frac{5}{4} \end{bmatrix} \end{align*}$
::[045268] 抽动行[268045]R12}[134045]R2}14}[1340154]R2}3+R1}[10140154]

Example 6
::例6

Reduce the following matrix to row echelon form:
::将下列矩阵减为行梯层表:

$\begin{aligned} [\begin{matrix} 3 & 6 \\ 2 & 4 \\ 5 & 17 \end{matrix}] \end{aligned}$
$\begin{align*}\begin{bmatrix} 3 & 6\\ 2 & 4\\ 5 & 17 \end{bmatrix}\end{align*}$

Solution:
$\begin{aligned} [\begin{matrix} 3 & 6 \\ 2 & 4 \\ 5 & 17 \end{matrix}] \\ R_{1} \div 3, R_{2} \div 2 & \to [\begin{matrix} 1 & 2 \\ 1 & 2 \\ 5 & 17 \end{matrix}] \\ R_{1} \cdot - 1 + R_{2}, R_{1} \cdot - 5 + R_{3} & \to [\begin{matrix} 1 & 2 \\ 0 & 0 \\ 0 & 7 \end{matrix}] \\ Switch R_{2} and R_{3} & \to [\begin{matrix} 1 & 2 \\ 0 & 7 \\ 0 & 0 \end{matrix}] \\ R_{2} \div 7 & \to [\begin{matrix} 1 & 2 \\ 0 & 1 \\ 0 & 0 \end{matrix}] \end{aligned}$
$\begin{align*}& \begin{bmatrix} 3 & 6\\ 2 & 4\\ 5 & 17 \end{bmatrix} \\ R_1 \div 3, R_2 \div 2 & \rightarrow \begin{bmatrix} 1 & 2\\ 1 & 2\\ 5 & 17 \end{bmatrix} \\ R_1 \cdot -1 + R_2, R_1 \cdot -5 + R_3 & \rightarrow \begin{bmatrix} 1 & 2\\ 0 & 0\\ 0 & 7 \end{bmatrix} \\ \text{Switch } R_2 \text{ and } R_3 & \rightarrow \begin{bmatrix} 1 & 2\\ 0 & 7\\ 0 & 0 \end{bmatrix} \\ R_2 \div 7 & \rightarrow \begin{bmatrix} 1 & 2\\ 0 & 1\\ 0 & 0 \end{bmatrix}\end{align*}$
::解答:[3624517]R1+R2、R1}5+R3}[120007]Switch R2和R3}[120700]R2}7}[120100]

Example 7
::例7

Reduce the following matrix to reduced row echelon form:
::将下列矩阵减为减缩的行梯层表:

$\begin{align*}\begin{bmatrix} 3 & 4 & 1 & 0\\ 5 & -1 & 0 & 1 \end{bmatrix}\end{align*}$

Solution:
::解决方案 :

$\begin{aligned} [\begin{matrix} 3 & 4 & 1 & 0 \\ 5 & - 1 & 0 & 1 \end{matrix}] \\ R_{1} \cdot 5, R_{2} \cdot 3 & \to [\begin{matrix} 15 & 20 & 5 & 0 \\ 15 & - 3 & 0 & 3 \end{matrix}] \\ R_{2} - R_{1} & \to [\begin{matrix} 15 & 20 & 5 & 0 \\ 0 & - 23 & - 5 & 3 \end{matrix}] \\ R_{1} \div 15, R_{2} \div - 23 & \to [\begin{matrix} 1 & \frac{4}{3} & \frac{1}{3} & 0 \\ 0 & 1 & \frac{5}{23} & - \frac{3}{23} \end{matrix}] \\ R_{2} \cdot - \frac{4}{3} + R_{1} & \to [\begin{matrix} 1 & 0 & \frac{1}{23} & \frac{4}{23} \\ 0 & 1 & \frac{5}{23} & - \frac{3}{23} \end{matrix}] \end{aligned}$
$\begin{align*}& \begin{bmatrix} 3 & 4 & 1 & 0\\ 5 & -1 & 0 & 1 \end{bmatrix} \\ R_1 \cdot 5, R_2 \cdot 3 & \rightarrow \begin{bmatrix} 15 & 20 & 5 & 0 \\ 15 & -3 & 0 & 3 \end{bmatrix} \\ R_2 - R_1 &\rightarrow \begin{bmatrix} 15 & 20 & 5 & 0\\ 0 & -23 & -5 & 3 \end{bmatrix} \\ R_1 \div 15, R_2 \div -23 & \rightarrow \begin{bmatrix} 1 & \frac{4}{3} & \frac{1}{3} & 0\\ 0 & 1 & \frac{5}{23} & -\frac{3}{23} \end{bmatrix} \\ R_2 \cdot -\frac{4}{3} + R_1 &\rightarrow \begin{bmatrix} 1 & 0 & \frac{1}{23} & \frac{4}{23}\\ 0 & 1 & \frac{5}{23} & -\frac{3}{23} \end{bmatrix}\end{align*}$

::[34105-101]R1}5,R2}3}[15205015-303]R2—R1}[1520500-23-53]R1}[1520500-23-53]R1}15,R2}23}[14313001523-323]R2}2}43+R1}[1012342301523-323]

Summary
::摘要

Row operations a re swapping rows, adding a multiple of one row to another, or scaling a row by multiplying through by a scalar.
::行的操作是互换行、将一个行的倍数加到另一个行,或者用标标乘来缩放一行。

::行的操作是互换行、将一个行的倍数加到另一个行,或者用标标乘来缩放一行。

Row echelon form is a matrix that has a leading one at the start of every nonzero row, zeros below every leading one, and all rows containing only zeros at the bottom of the matrix.

::排梯表是一个矩阵,在每一个非零行的起始处有一个前导矩阵,在每一个前导行下方零位,所有行在矩阵底部仅包含零位。

Reduced row echelon form is the same as row echelon form with one additional stipulation: that every other entry in a column with a leading one must be zero.
::减少的排梯表与排梯表相同,并附加一项规定:列内每条前列条目必须为零。

There are only three operations that are permitted to act on matrices:

Add a multiple of one row to another row.
::将多个行加到另一行。

Scale a row by multiplying through by a nonzero constant.
::通过乘以非零常数来缩放一行。

Swap two rows.
::交换两排

::只有三种操作允许在矩阵上操作:在另一行增加一行的倍数。通过乘以非零常数来缩放一行。交换两行。

Review
::回顾

1. Give an example of a matrix in row echelon form.
::1. 举行梯层表矩阵为例。

2. Give an example of a matrix in reduced row echelon form.
::2. 举一个缩排梯层表矩阵的例子。

3. What are the three row operations you are allowed to perform when reducing a matrix?
::3. 在减少矩阵时,允许执行的三行操作是什么?

4. If a square matrix reduces to the identity matrix, what does that mean about the rows of the original matrix?
::4. 如果一个平方矩阵缩减为身份矩阵,这对原始矩阵的行意味着什么?

Use the following matrix for 5-6:
::5-6 使用以下矩阵:

$\begin{align*}A=\begin{bmatrix} -3 & -4 & -12\\ 4 & 4 & 12\\ -11 & -12 & -35 \end{bmatrix}\end{align*}$
::A=[-3-4-4-124412-11-12-35]

5. Reduce matrix $\begin{align*}A\end{align*}$ to row echelon form.
::5. 将表格A到排层表格减少。

6. Reduce matrix $\begin{align*}A\end{align*}$ to reduced row echelon form. Are the rows of matrix $\begin{align*}A\end{align*}$ linearly independent?
::6. 将矩阵A减为排级梯层表:矩阵A各行是否线性独立?

Use the following matrix for 7-8:
::7-8 使用以下矩阵:

$\begin{align*}B=\begin{bmatrix} 3 & -4 & 8\\ 9 & 0 & 1\\ 0 & 1 & -2 \end{bmatrix}\end{align*}$
::B=[3-4890101-2]

7. Reduce matrix $\begin{align*}B\end{align*}$ to row echelon form.
::7. 将表B减为排层表。

8. Reduce matrix $\begin{align*}B\end{align*}$ to reduced row echelon form. Are the rows of matrix $\begin{align*}B\end{align*}$ linearly independent?
::8. 将矩阵B减为排级梯层表:矩阵B的行线性独立吗?

Use the following matrix for 9-10:
::9-10使用以下矩阵:

$\begin{align*}C=\begin{bmatrix} 0 & 0 & -1 & -1\\ 3 & 6 & -3 & 1\\ 6 & 12 & -7 & 0 \end{bmatrix}\end{align*}$
::C=[00-1-136-31612-70]

9. Reduce matrix $\begin{align*}C\end{align*}$ to row echelon form.
::9. 将矩阵C减为排行梯层表。

10. Reduce matrix $\begin{align*}C\end{align*}$ to reduced row echelon form. Are the rows of matrix $\begin{align*}C\end{align*}$ linearly independent?
::10. 将矩阵C减为排排梯层表:矩阵C的行线性独立吗?

Use the following matrix for 11-12:
::11-12 使用以下矩阵:

$\begin{align*}D=\begin{bmatrix} 1 & 1\\ 3 & 4\\ 2 & 3 \end{bmatrix}\end{align*}$
::D=[113423]

11. Reduce matrix $\begin{align*}D\end{align*}$ to row echelon form.
::11. 将表格D减为排排表。

12. Reduce matrix $\begin{align*}D\end{align*}$ to reduced row echelon form. Are the rows of matrix $\begin{align*}D\end{align*}$ linearly independent?
::12. 将矩阵D减为行梯层表:矩阵D的行线性独立吗?

Use the following matrix for 13-14:
::13-14 使用下列矩阵:

$\begin{align*}E=\begin{bmatrix} -5 & -6 & -12\\ -1 & -1 & -2\\ 2 & 2 & 4 \end{bmatrix}\end{align*}$
::E=[-5-6-6 - 12 - 1 - 1 - 1 - 2224]

13. Reduce matrix $\begin{align*}E\end{align*}$ to row echelon form.
::13. 将表E减为表E。

14. Reduce matrix $\begin{align*}E\end{align*}$ to reduced row echelon form. Are the rows of matrix $\begin{align*}E\end{align*}$ linearly independent?
::14. 将矩阵E减为排级梯层表:矩阵E的行线性独立吗?

Use the following matrix for 15-16:
::15-16使用以下矩阵:

$\begin{align*}F=\begin{bmatrix} -23 & 6 & 3\\ 2 & -\frac{1}{2} & 0\\ -8 & 2 & 1 \end{bmatrix}\end{align*}$
::F=[-23632-120-821]

15. Reduce matrix $\begin{align*}F\end{align*}$ to row echelon form.
::15. 将表格F减为排排表。

16. Reduce matrix $\begin{align*}F\end{align*}$ to reduced row echelon form. Are the rows of matrix $\begin{align*}F\end{align*}$ linearly independent?
::16. 将矩阵F减为排排梯层表:矩阵F各行是否线性独立?

Review (Answers)
::回顾(答复)

Please see the Appendix.
::请参看附录。

章节大纲

Introduction ::导言

Row Operations ::行操作

Examples ::实例

Summary ::摘要

Review ::回顾

Review (Answers) ::回顾(答复)

Introduction
::导言

Row Operations
::行操作

Examples
::实例

Summary
::摘要

Review
::回顾

Review (Answers)
::回顾(答复)