线性代数示例

$7 x + 23 y + 3 z = 17$ , $3 x + 10 y + 4 z = - 2$ , $x + 3 y - 5 z = 6$

解题步骤 1

以矩阵形式书写方程组。

$[\begin{array}{c} 7 & 23 & 3 & 17 \\ 3 & 10 & 4 & - 2 \\ 1 & 3 & - 5 & 6 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{7}{7} & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 3 & 10 & 4 & - 2 \\ 1 & 3 & - 5 & 6 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 3 & 10 & 4 & - 2 \\ 1 & 3 & - 5 & 6 \end{array}]$

解题步骤 2.2

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 3 - 3 \cdot 1 & 10 - 3 (\frac{23}{7}) & 4 - 3 (\frac{3}{7}) & - 2 - 3 (\frac{17}{7}) \\ 1 & 3 & - 5 & 6 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 & \frac{1}{7} & \frac{19}{7} & - \frac{65}{7} \\ 1 & 3 & - 5 & 6 \end{array}]$

解题步骤 2.3

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 2.3.1

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 & \frac{1}{7} & \frac{19}{7} & - \frac{65}{7} \\ 1 - 1 & 3 - \frac{23}{7} & - 5 - \frac{3}{7} & 6 - \frac{17}{7} \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 & \frac{1}{7} & \frac{19}{7} & - \frac{65}{7} \\ 0 & - \frac{2}{7} & - \frac{38}{7} & \frac{25}{7} \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $7$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.4.1

Multiply each element of $R_{2}$ by $7$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 7 \cdot 0 & 7 (\frac{1}{7}) & 7 (\frac{19}{7}) & 7 (- \frac{65}{7}) \\ 0 & - \frac{2}{7} & - \frac{38}{7} & \frac{25}{7} \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 & 1 & 19 & - 65 \\ 0 & - \frac{2}{7} & - \frac{38}{7} & \frac{25}{7} \end{array}]$

解题步骤 2.5

Perform the row operation $R_{3} = R_{3} + \frac{2}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 2.5.1

Perform the row operation $R_{3} = R_{3} + \frac{2}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 & 1 & 19 & - 65 \\ 0 + \frac{2}{7} \cdot 0 & - \frac{2}{7} + \frac{2}{7} \cdot 1 & - \frac{38}{7} + \frac{2}{7} \cdot 19 & \frac{25}{7} + \frac{2}{7} \cdot - 65 \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 & 1 & 19 & - 65 \\ 0 & 0 & 0 & - 15 \end{array}]$

解题步骤 2.6

Multiply each element of $R_{3}$ by $- \frac{1}{15}$ to make the entry at $3, 4$ a $1$ .

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解题步骤 2.6.1

Multiply each element of $R_{3}$ by $- \frac{1}{15}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 & 1 & 19 & - 65 \\ - \frac{1}{15} \cdot 0 & - \frac{1}{15} \cdot 0 & - \frac{1}{15} \cdot 0 & - \frac{1}{15} \cdot - 15 \end{array}]$

解题步骤 2.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 & 1 & 19 & - 65 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.7

Perform the row operation $R_{2} = R_{2} + 65 R_{3}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{2} = R_{2} + 65 R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 + 65 \cdot 0 & 1 + 65 \cdot 0 & 19 + 65 \cdot 0 & - 65 + 65 \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & \frac{17}{7} \\ 0 & 1 & 19 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.8

Perform the row operation $R_{1} = R_{1} - \frac{17}{7} R_{3}$ to make the entry at $1, 4$ a $0$ .

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解题步骤 2.8.1

Perform the row operation $R_{1} = R_{1} - \frac{17}{7} R_{3}$ to make the entry at $1, 4$ a $0$ .

$[\begin{array}{c} 1 - \frac{17}{7} \cdot 0 & \frac{23}{7} - \frac{17}{7} \cdot 0 & \frac{3}{7} - \frac{17}{7} \cdot 0 & \frac{17}{7} - \frac{17}{7} \cdot 1 \\ 0 & 1 & 19 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{23}{7} & \frac{3}{7} & 0 \\ 0 & 1 & 19 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.9

Perform the row operation $R_{1} = R_{1} - \frac{23}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.9.1

Perform the row operation $R_{1} = R_{1} - \frac{23}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{23}{7} \cdot 0 & \frac{23}{7} - \frac{23}{7} \cdot 1 & \frac{3}{7} - \frac{23}{7} \cdot 19 & 0 - \frac{23}{7} \cdot 0 \\ 0 & 1 & 19 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & - 62 & 0 \\ 0 & 1 & 19 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 3

使用结果矩阵定义方程组的最终解。

$x - 62 z = 0$

$y + 19 z = 0$

$0 = 1$

解题步骤 4

在等式两边都加上 $62 z$ 。

$x = 62 z$

$y + 19 z = 0$

$0 = 1$

解题步骤 5

从等式两边同时减去 $19 z$ 。

$y = - 19 z$

$x = 62 z$

$0 = 1$

解题步骤 6

因为 $0 \neq 1$ ，所以没有解。

无解

线性代数 示例

线性代数示例