代数示例

3 x + 1 = 0 y

$3 x + 1 = 0 y$ ,

y + 2 x > - 9

$y + 2 x > - 9$

解题步骤 1

引入松弛变量

u

$u$ 和

v

$v$ 以便使用方程替换不等式。

y + 2 x - Z = - 9

$y + 2 x - Z = - 9$

3 x + 1 = 0

$3 x + 1 = 0$

解题步骤 2

从等式两边同时减去

1

$1$ 。

y + 2 x - Z = - 9, 3 x = - 1

$y + 2 x - Z = - 9, 3 x = - 1$

解题步骤 3

以矩阵形式书写方程组。

[\begin{matrix} 2 & 1 & 0 & - 9 3 & 0 & 0 & - 1 \end{matrix}]

$[\begin{array}{c} 2 & 1 & 0 & - 9 \\ 3 & 0 & 0 & - 1 \end{array}]$

解题步骤 4

求行简化阶梯形矩阵。

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解题步骤 4.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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解题步骤 4.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{matrix} \frac{2}{2} & \frac{1}{2} & \frac{0}{2} & \frac{- 9}{2} 3 & 0 & 0 & - 1 \end{matrix}]

$[\begin{array}{c} \frac{2}{2} & \frac{1}{2} & \frac{0}{2} & \frac{- 9}{2} \\ 3 & 0 & 0 & - 1 \end{array}]$

解题步骤 4.1.2

化简

R_{1}

$R_{1}$ 。

[\begin{matrix} 1 & \frac{1}{2} & 0 & - \frac{9}{2} 3 & 0 & 0 & - 1 \end{matrix}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 3 & 0 & 0 & - 1 \end{array}]$

[\begin{matrix} 1 & \frac{1}{2} & 0 & - \frac{9}{2} 3 & 0 & 0 & - 1 \end{matrix}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 3 & 0 & 0 & - 1 \end{array}]$

解题步骤 4.2

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

$R_{2} = R_{2} - 3 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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解题步骤 4.2.1

Perform the row operation

R_{2} = R_{2} - 3 R_{1}

$R_{2} = R_{2} - 3 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & \frac{1}{2} & 0 & - \frac{9}{2} 3 - 3 \cdot 1 & 0 - 3 (\frac{1}{2}) & 0 - 3 \cdot 0 & - 1 - 3 (- \frac{9}{2}) \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 3 - 3 \cdot 1 & 0 - 3 (\frac{1}{2}) & 0 - 3 \cdot 0 & - 1 - 3 (- \frac{9}{2}) \end{array}]$

解题步骤 4.2.2

化简

R_{2}

$R_{2}$ 。

[\begin{matrix} 1 & \frac{1}{2} & 0 & - \frac{9}{2} 0 & - \frac{3}{2} & 0 & \frac{25}{2} \end{matrix}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & - \frac{3}{2} & 0 & \frac{25}{2} \end{array}]$

[\begin{matrix} 1 & \frac{1}{2} & 0 & - \frac{9}{2} 0 & - \frac{3}{2} & 0 & \frac{25}{2} \end{matrix}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & - \frac{3}{2} & 0 & \frac{25}{2} \end{array}]$

解题步骤 4.3

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{2}{3}

$- \frac{2}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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解题步骤 4.3.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{2}{3}

$- \frac{2}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & \frac{1}{2} & 0 & - \frac{9}{2} - \frac{2}{3} \cdot 0 & - \frac{2}{3} (- \frac{3}{2}) & - \frac{2}{3} \cdot 0 & - \frac{2}{3} \cdot \frac{25}{2} \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ - \frac{2}{3} \cdot 0 & - \frac{2}{3} (- \frac{3}{2}) & - \frac{2}{3} \cdot 0 & - \frac{2}{3} \cdot \frac{25}{2} \end{array}]$

解题步骤 4.3.2

化简

R_{2}

$R_{2}$ 。

[\begin{matrix} 1 & \frac{1}{2} & 0 & - \frac{9}{2} 0 & 1 & 0 & - \frac{25}{3} \end{matrix}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

[\begin{matrix} 1 & \frac{1}{2} & 0 & - \frac{9}{2} 0 & 1 & 0 & - \frac{25}{3} \end{matrix}]

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & - \frac{9}{2} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

解题步骤 4.4

Perform the row operation

R_{1} = R_{1} - \frac{1}{2} R_{2}

$R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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解题步骤 4.4.1

Perform the row operation

R_{1} = R_{1} - \frac{1}{2} R_{2}

$R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot 0 & - \frac{9}{2} - \frac{1}{2} (- \frac{25}{3}) 0 & 1 & 0 & - \frac{25}{3} \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot 0 & - \frac{9}{2} - \frac{1}{2} (- \frac{25}{3}) \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

解题步骤 4.4.2

化简

R_{1}

$R_{1}$ 。

[\begin{matrix} 1 & 0 & 0 & - \frac{1}{3} 0 & 1 & 0 & - \frac{25}{3} \end{matrix}]

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

[\begin{matrix} 1 & 0 & 0 & - \frac{1}{3} 0 & 1 & 0 & - \frac{25}{3} \end{matrix}]

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

[\begin{matrix} 1 & 0 & 0 & - \frac{1}{3} 0 & 1 & 0 & - \frac{25}{3} \end{matrix}]

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{3} \\ 0 & 1 & 0 & - \frac{25}{3} \end{array}]$

解题步骤 5

使用结果矩阵定义方程组的最终解。

x = 0

$x = 0$

y = 0

$y = 0$

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