线性代数示例

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 2 & - 1 & 2 & 0 & 1 & 0 & 0 & 30 \\ - 3 & 1 & 3 & 0 & 0 & 1 & 0 & 40 \\ - 3 & - 2 & 4 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1

求行简化阶梯形矩阵。

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解题步骤 1.1

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

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解题步骤 1.1.1

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

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 2 - 2 \cdot 1 & - 1 - 2 \cdot - 1 & 2 - 2 \cdot 1 & 0 - 2 \cdot 1 & 1 - 2 \cdot 0 & 0 - 2 \cdot 0 & 0 - 2 \cdot 0 & 30 - 2 \cdot 10 \\ - 3 & 1 & 3 & 0 & 0 & 1 & 0 & 40 \\ - 3 & - 2 & 4 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ - 3 & 1 & 3 & 0 & 0 & 1 & 0 & 40 \\ - 3 & - 2 & 4 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.2

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

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解题步骤 1.2.1

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

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ - 3 + 3 \cdot 1 & 1 + 3 \cdot - 1 & 3 + 3 \cdot 1 & 0 + 3 \cdot 1 & 0 + 3 \cdot 0 & 1 + 3 \cdot 0 & 0 + 3 \cdot 0 & 40 + 3 \cdot 10 \\ - 3 & - 2 & 4 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & - 2 & 6 & 3 & 0 & 1 & 0 & 70 \\ - 3 & - 2 & 4 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.3

Perform the row operation $R_{4} = R_{4} + 3 R_{1}$ to make the entry at $4, 1$ a $0$ .

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解题步骤 1.3.1

Perform the row operation $R_{4} = R_{4} + 3 R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & - 2 & 6 & 3 & 0 & 1 & 0 & 70 \\ - 3 + 3 \cdot 1 & - 2 + 3 \cdot - 1 & 4 + 3 \cdot 1 & 0 + 3 \cdot 1 & 0 + 3 \cdot 0 & 0 + 3 \cdot 0 & 1 + 3 \cdot 0 & 0 + 3 \cdot 10 \end{array}]$

解题步骤 1.3.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & - 2 & 6 & 3 & 0 & 1 & 0 & 70 \\ 0 & - 5 & 7 & 3 & 0 & 0 & 1 & 30 \end{array}]$

解题步骤 1.4

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

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解题步骤 1.4.1

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

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 6 + 2 \cdot 0 & 3 + 2 \cdot - 2 & 0 + 2 \cdot 1 & 1 + 2 \cdot 0 & 0 + 2 \cdot 0 & 70 + 2 \cdot 10 \\ 0 & - 5 & 7 & 3 & 0 & 0 & 1 & 30 \end{array}]$

解题步骤 1.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & 0 & 6 & - 1 & 2 & 1 & 0 & 90 \\ 0 & - 5 & 7 & 3 & 0 & 0 & 1 & 30 \end{array}]$

解题步骤 1.5

Perform the row operation $R_{4} = R_{4} + 5 R_{2}$ to make the entry at $4, 2$ a $0$ .

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解题步骤 1.5.1

Perform the row operation $R_{4} = R_{4} + 5 R_{2}$ to make the entry at $4, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & 0 & 6 & - 1 & 2 & 1 & 0 & 90 \\ 0 + 5 \cdot 0 & - 5 + 5 \cdot 1 & 7 + 5 \cdot 0 & 3 + 5 \cdot - 2 & 0 + 5 \cdot 1 & 0 + 5 \cdot 0 & 1 + 5 \cdot 0 & 30 + 5 \cdot 10 \end{array}]$

解题步骤 1.5.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & 0 & 6 & - 1 & 2 & 1 & 0 & 90 \\ 0 & 0 & 7 & - 7 & 5 & 0 & 1 & 80 \end{array}]$

解题步骤 1.6

Multiply each element of $R_{3}$ by $\frac{1}{6}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 1.6.1

Multiply each element of $R_{3}$ by $\frac{1}{6}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ \frac{0}{6} & \frac{0}{6} & \frac{6}{6} & \frac{- 1}{6} & \frac{2}{6} & \frac{1}{6} & \frac{0}{6} & \frac{90}{6} \\ 0 & 0 & 7 & - 7 & 5 & 0 & 1 & 80 \end{array}]$

解题步骤 1.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & 0 & 1 & - \frac{1}{6} & \frac{1}{3} & \frac{1}{6} & 0 & 15 \\ 0 & 0 & 7 & - 7 & 5 & 0 & 1 & 80 \end{array}]$

解题步骤 1.7

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

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解题步骤 1.7.1

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

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & 0 & 1 & - \frac{1}{6} & \frac{1}{3} & \frac{1}{6} & 0 & 15 \\ 0 - 7 \cdot 0 & 0 - 7 \cdot 0 & 7 - 7 \cdot 1 & - 7 - 7 (- \frac{1}{6}) & 5 - 7 (\frac{1}{3}) & 0 - 7 (\frac{1}{6}) & 1 - 7 \cdot 0 & 80 - 7 \cdot 15 \end{array}]$

解题步骤 1.7.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & 0 & 1 & - \frac{1}{6} & \frac{1}{3} & \frac{1}{6} & 0 & 15 \\ 0 & 0 & 0 & - \frac{35}{6} & \frac{8}{3} & - \frac{7}{6} & 1 & - 25 \end{array}]$

解题步骤 1.8

Multiply each element of $R_{4}$ by $- \frac{6}{35}$ to make the entry at $4, 4$ a $1$ .

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解题步骤 1.8.1

Multiply each element of $R_{4}$ by $- \frac{6}{35}$ to make the entry at $4, 4$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & 0 & 1 & - \frac{1}{6} & \frac{1}{3} & \frac{1}{6} & 0 & 15 \\ - \frac{6}{35} \cdot 0 & - \frac{6}{35} \cdot 0 & - \frac{6}{35} \cdot 0 & - \frac{6}{35} (- \frac{35}{6}) & - \frac{6}{35} \cdot \frac{8}{3} & - \frac{6}{35} (- \frac{7}{6}) & - \frac{6}{35} \cdot 1 & - \frac{6}{35} \cdot - 25 \end{array}]$

解题步骤 1.8.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & 0 & 1 & - \frac{1}{6} & \frac{1}{3} & \frac{1}{6} & 0 & 15 \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.9

Perform the row operation $R_{3} = R_{3} + \frac{1}{6} R_{4}$ to make the entry at $3, 4$ a $0$ .

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解题步骤 1.9.1

Perform the row operation $R_{3} = R_{3} + \frac{1}{6} R_{4}$ to make the entry at $3, 4$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 + \frac{1}{6} \cdot 0 & 0 + \frac{1}{6} \cdot 0 & 1 + \frac{1}{6} \cdot 0 & - \frac{1}{6} + \frac{1}{6} \cdot 1 & \frac{1}{3} + \frac{1}{6} (- \frac{16}{35}) & \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{5} & 0 + \frac{1}{6} (- \frac{6}{35}) & 15 + \frac{1}{6} \cdot \frac{30}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.9.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & - 2 & 1 & 0 & 0 & 10 \\ 0 & 0 & 1 & 0 & \frac{9}{35} & \frac{1}{5} & - \frac{1}{35} & \frac{110}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.10

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

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解题步骤 1.10.1

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

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 + 2 \cdot 0 & 1 + 2 \cdot 0 & 0 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 1 + 2 (- \frac{16}{35}) & 0 + 2 (\frac{1}{5}) & 0 + 2 (- \frac{6}{35}) & 10 + 2 (\frac{30}{7}) \\ 0 & 0 & 1 & 0 & \frac{9}{35} & \frac{1}{5} & - \frac{1}{35} & \frac{110}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.10.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 1 & 0 & 0 & 0 & 10 \\ 0 & 1 & 0 & 0 & \frac{3}{35} & \frac{2}{5} & - \frac{12}{35} & \frac{130}{7} \\ 0 & 0 & 1 & 0 & \frac{9}{35} & \frac{1}{5} & - \frac{1}{35} & \frac{110}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.11

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

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解题步骤 1.11.1

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

$[\begin{array}{c} 1 - 0 & - 1 - 0 & 1 - 0 & 1 - 1 & 0 + \frac{16}{35} & 0 - \frac{1}{5} & 0 + \frac{6}{35} & 10 - \frac{30}{7} \\ 0 & 1 & 0 & 0 & \frac{3}{35} & \frac{2}{5} & - \frac{12}{35} & \frac{130}{7} \\ 0 & 0 & 1 & 0 & \frac{9}{35} & \frac{1}{5} & - \frac{1}{35} & \frac{110}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.11.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & 1 & 0 & \frac{16}{35} & - \frac{1}{5} & \frac{6}{35} & \frac{40}{7} \\ 0 & 1 & 0 & 0 & \frac{3}{35} & \frac{2}{5} & - \frac{12}{35} & \frac{130}{7} \\ 0 & 0 & 1 & 0 & \frac{9}{35} & \frac{1}{5} & - \frac{1}{35} & \frac{110}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.12

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

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解题步骤 1.12.1

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

$[\begin{array}{c} 1 - 0 & - 1 - 0 & 1 - 1 & 0 - 0 & \frac{16}{35} - \frac{9}{35} & - \frac{1}{5} - \frac{1}{5} & \frac{6}{35} + \frac{1}{35} & \frac{40}{7} - \frac{110}{7} \\ 0 & 1 & 0 & 0 & \frac{3}{35} & \frac{2}{5} & - \frac{12}{35} & \frac{130}{7} \\ 0 & 0 & 1 & 0 & \frac{9}{35} & \frac{1}{5} & - \frac{1}{35} & \frac{110}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.12.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & 0 & 0 & \frac{1}{5} & - \frac{2}{5} & \frac{1}{5} & - 10 \\ 0 & 1 & 0 & 0 & \frac{3}{35} & \frac{2}{5} & - \frac{12}{35} & \frac{130}{7} \\ 0 & 0 & 1 & 0 & \frac{9}{35} & \frac{1}{5} & - \frac{1}{35} & \frac{110}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.13

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

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解题步骤 1.13.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 0 + 0 & \frac{1}{5} + \frac{3}{35} & - \frac{2}{5} + \frac{2}{5} & \frac{1}{5} - \frac{12}{35} & - 10 + \frac{130}{7} \\ 0 & 1 & 0 & 0 & \frac{3}{35} & \frac{2}{5} & - \frac{12}{35} & \frac{130}{7} \\ 0 & 0 & 1 & 0 & \frac{9}{35} & \frac{1}{5} & - \frac{1}{35} & \frac{110}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 1.13.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{2}{7} & 0 & - \frac{1}{7} & \frac{60}{7} \\ 0 & 1 & 0 & 0 & \frac{3}{35} & \frac{2}{5} & - \frac{12}{35} & \frac{130}{7} \\ 0 & 0 & 1 & 0 & \frac{9}{35} & \frac{1}{5} & - \frac{1}{35} & \frac{110}{7} \\ 0 & 0 & 0 & 1 & - \frac{16}{35} & \frac{1}{5} & - \frac{6}{35} & \frac{30}{7} \end{array}]$

解题步骤 2

The pivot positions are the locations with the leading $1$ in each row. The pivot columns are the columns that have a pivot position.

Pivot Positions: $a_{11}, a_{22}, a_{33},$ and $a_{44}$

Pivot Columns: $1, 2, 3,$ and $4$

线性代数 示例

线性代数示例