线性代数示例

$[\begin{array}{c} - 2 & - 5 & 8 & 0 & 17 & 0 \\ 1 & 3 & - 5 & 1 & 5 & 0 \\ 3 & 11 & - 19 & 7 & 1 & 0 \\ 1 & 7 & - 13 & 5 & - 3 & 0 \end{array}]$

解题步骤 1

求行简化阶梯形矩阵。

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解题步骤 1.1

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

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解题步骤 1.1.1

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

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 5 & - \frac{1}{2} \cdot 8 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 17 & - \frac{1}{2} \cdot 0 \\ 1 & 3 & - 5 & 1 & 5 & 0 \\ 3 & 11 & - 19 & 7 & 1 & 0 \\ 1 & 7 & - 13 & 5 & - 3 & 0 \end{array}]$

解题步骤 1.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 1 & 3 & - 5 & 1 & 5 & 0 \\ 3 & 11 & - 19 & 7 & 1 & 0 \\ 1 & 7 & - 13 & 5 & - 3 & 0 \end{array}]$

解题步骤 1.2

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

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解题步骤 1.2.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 1 - 1 & 3 - \frac{5}{2} & - 5 + 4 & 1 - 0 & 5 + \frac{17}{2} & 0 - 0 \\ 3 & 11 & - 19 & 7 & 1 & 0 \\ 1 & 7 & - 13 & 5 & - 3 & 0 \end{array}]$

解题步骤 1.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & \frac{1}{2} & - 1 & 1 & \frac{27}{2} & 0 \\ 3 & 11 & - 19 & 7 & 1 & 0 \\ 1 & 7 & - 13 & 5 & - 3 & 0 \end{array}]$

解题步骤 1.3

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 1.3.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & \frac{1}{2} & - 1 & 1 & \frac{27}{2} & 0 \\ 3 - 3 \cdot 1 & 11 - 3 (\frac{5}{2}) & - 19 - 3 \cdot - 4 & 7 - 3 \cdot 0 & 1 - 3 (- \frac{17}{2}) & 0 - 3 \cdot 0 \\ 1 & 7 & - 13 & 5 & - 3 & 0 \end{array}]$

解题步骤 1.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & \frac{1}{2} & - 1 & 1 & \frac{27}{2} & 0 \\ 0 & \frac{7}{2} & - 7 & 7 & \frac{53}{2} & 0 \\ 1 & 7 & - 13 & 5 & - 3 & 0 \end{array}]$

解题步骤 1.4

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

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解题步骤 1.4.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & \frac{1}{2} & - 1 & 1 & \frac{27}{2} & 0 \\ 0 & \frac{7}{2} & - 7 & 7 & \frac{53}{2} & 0 \\ 1 - 1 & 7 - \frac{5}{2} & - 13 + 4 & 5 - 0 & - 3 + \frac{17}{2} & 0 - 0 \end{array}]$

解题步骤 1.4.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & \frac{1}{2} & - 1 & 1 & \frac{27}{2} & 0 \\ 0 & \frac{7}{2} & - 7 & 7 & \frac{53}{2} & 0 \\ 0 & \frac{9}{2} & - 9 & 5 & \frac{11}{2} & 0 \end{array}]$

解题步骤 1.5

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

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解题步骤 1.5.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 2 \cdot 0 & 2 (\frac{1}{2}) & 2 \cdot - 1 & 2 \cdot 1 & 2 (\frac{27}{2}) & 2 \cdot 0 \\ 0 & \frac{7}{2} & - 7 & 7 & \frac{53}{2} & 0 \\ 0 & \frac{9}{2} & - 9 & 5 & \frac{11}{2} & 0 \end{array}]$

解题步骤 1.5.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 & \frac{7}{2} & - 7 & 7 & \frac{53}{2} & 0 \\ 0 & \frac{9}{2} & - 9 & 5 & \frac{11}{2} & 0 \end{array}]$

解题步骤 1.6

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

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解题步骤 1.6.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 - \frac{7}{2} \cdot 0 & \frac{7}{2} - \frac{7}{2} \cdot 1 & - 7 - \frac{7}{2} \cdot - 2 & 7 - \frac{7}{2} \cdot 2 & \frac{53}{2} - \frac{7}{2} \cdot 27 & 0 - \frac{7}{2} \cdot 0 \\ 0 & \frac{9}{2} & - 9 & 5 & \frac{11}{2} & 0 \end{array}]$

解题步骤 1.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 & 0 & 0 & 0 & - 68 & 0 \\ 0 & \frac{9}{2} & - 9 & 5 & \frac{11}{2} & 0 \end{array}]$

解题步骤 1.7

Perform the row operation $R_{4} = R_{4} - \frac{9}{2} R_{2}$ to make the entry at $4, 2$ a $0$ .

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解题步骤 1.7.1

Perform the row operation $R_{4} = R_{4} - \frac{9}{2} R_{2}$ to make the entry at $4, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 & 0 & 0 & 0 & - 68 & 0 \\ 0 - \frac{9}{2} \cdot 0 & \frac{9}{2} - \frac{9}{2} \cdot 1 & - 9 - \frac{9}{2} \cdot - 2 & 5 - \frac{9}{2} \cdot 2 & \frac{11}{2} - \frac{9}{2} \cdot 27 & 0 - \frac{9}{2} \cdot 0 \end{array}]$

解题步骤 1.7.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 & 0 & 0 & 0 & - 68 & 0 \\ 0 & 0 & 0 & - 4 & - 116 & 0 \end{array}]$

解题步骤 1.8

Swap $R_{4}$ with $R_{3}$ to put a nonzero entry at $3, 4$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 & 0 & 0 & - 4 & - 116 & 0 \\ 0 & 0 & 0 & 0 & - 68 & 0 \end{array}]$

解题步骤 1.9

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

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解题步骤 1.9.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot - 116 & - \frac{1}{4} \cdot 0 \\ 0 & 0 & 0 & 0 & - 68 & 0 \end{array}]$

解题步骤 1.9.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 & 0 & 0 & 1 & 29 & 0 \\ 0 & 0 & 0 & 0 & - 68 & 0 \end{array}]$

解题步骤 1.10

Multiply each element of $R_{4}$ by $- \frac{1}{68}$ to make the entry at $4, 5$ a $1$ .

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解题步骤 1.10.1

Multiply each element of $R_{4}$ by $- \frac{1}{68}$ to make the entry at $4, 5$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 & 0 & 0 & 1 & 29 & 0 \\ - \frac{1}{68} \cdot 0 & - \frac{1}{68} \cdot 0 & - \frac{1}{68} \cdot 0 & - \frac{1}{68} \cdot 0 & - \frac{1}{68} \cdot - 68 & - \frac{1}{68} \cdot 0 \end{array}]$

解题步骤 1.10.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 & 0 & 0 & 1 & 29 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.11

Perform the row operation $R_{3} = R_{3} - 29 R_{4}$ to make the entry at $3, 5$ a $0$ .

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解题步骤 1.11.1

Perform the row operation $R_{3} = R_{3} - 29 R_{4}$ to make the entry at $3, 5$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 - 29 \cdot 0 & 0 - 29 \cdot 0 & 0 - 29 \cdot 0 & 1 - 29 \cdot 0 & 29 - 29 \cdot 1 & 0 - 29 \cdot 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.11.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 27 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.12

Perform the row operation $R_{2} = R_{2} - 27 R_{4}$ to make the entry at $2, 5$ a $0$ .

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解题步骤 1.12.1

Perform the row operation $R_{2} = R_{2} - 27 R_{4}$ to make the entry at $2, 5$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 - 27 \cdot 0 & 1 - 27 \cdot 0 & - 2 - 27 \cdot 0 & 2 - 27 \cdot 0 & 27 - 27 \cdot 1 & 0 - 27 \cdot 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.12.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & - \frac{17}{2} & 0 \\ 0 & 1 & - 2 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.13

Perform the row operation $R_{1} = R_{1} + \frac{17}{2} R_{4}$ to make the entry at $1, 5$ a $0$ .

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解题步骤 1.13.1

Perform the row operation $R_{1} = R_{1} + \frac{17}{2} R_{4}$ to make the entry at $1, 5$ a $0$ .

$[\begin{array}{c} 1 + \frac{17}{2} \cdot 0 & \frac{5}{2} + \frac{17}{2} \cdot 0 & - 4 + \frac{17}{2} \cdot 0 & 0 + \frac{17}{2} \cdot 0 & - \frac{17}{2} + \frac{17}{2} \cdot 1 & 0 + \frac{17}{2} \cdot 0 \\ 0 & 1 & - 2 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.13.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & 0 & 0 \\ 0 & 1 & - 2 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.14

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

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解题步骤 1.14.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & 0 & 0 \\ 0 - 2 \cdot 0 & 1 - 2 \cdot 0 & - 2 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & 0 - 2 \cdot 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.14.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & - 4 & 0 & 0 & 0 \\ 0 & 1 & - 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.15

Perform the row operation $R_{1} = R_{1} - \frac{5}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 1.15.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{5}{2} \cdot 0 & \frac{5}{2} - \frac{5}{2} \cdot 1 & - 4 - \frac{5}{2} \cdot - 2 & 0 - \frac{5}{2} \cdot 0 & 0 - \frac{5}{2} \cdot 0 & 0 - \frac{5}{2} \cdot 0 \\ 0 & 1 & - 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.15.2

化简 $R_{1}$ 。

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

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: $a_{11}, a_{22}, a_{34},$ and $a_{45}$

Pivot Columns: $1, 2, 4,$ and $5$

线性代数 示例

线性代数示例