有限数学示例

$[\begin{array}{c} 5 & 6 & 8 & 1 & 1 & 0 & 0 & 20 \\ 3 & 7 & 6 & 2 & 0 & 1 & 0 & 10 \\ 1 & 8 & 8 & 2 & 0 & 0 & 0 & 21 \\ 0 & - 3 & - 1 & 1 & 0 & 0 & 1 & 18 \end{array}]$

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 $\frac{1}{5}$ 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 $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & \frac{0}{5} & \frac{0}{5} & \frac{20}{5} \\ 3 & 7 & 6 & 2 & 0 & 1 & 0 & 10 \\ 1 & 8 & 8 & 2 & 0 & 0 & 0 & 21 \\ 0 & - 3 & - 1 & 1 & 0 & 0 & 1 & 18 \end{array}]$

解题步骤 1.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 3 & 7 & 6 & 2 & 0 & 1 & 0 & 10 \\ 1 & 8 & 8 & 2 & 0 & 0 & 0 & 21 \\ 0 & - 3 & - 1 & 1 & 0 & 0 & 1 & 18 \end{array}]$

解题步骤 1.2

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 1.2.1

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

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 3 - 3 \cdot 1 & 7 - 3 (\frac{6}{5}) & 6 - 3 (\frac{8}{5}) & 2 - 3 (\frac{1}{5}) & 0 - 3 (\frac{1}{5}) & 1 - 3 \cdot 0 & 0 - 3 \cdot 0 & 10 - 3 \cdot 4 \\ 1 & 8 & 8 & 2 & 0 & 0 & 0 & 21 \\ 0 & - 3 & - 1 & 1 & 0 & 0 & 1 & 18 \end{array}]$

解题步骤 1.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & \frac{17}{5} & \frac{6}{5} & \frac{7}{5} & - \frac{3}{5} & 1 & 0 & - 2 \\ 1 & 8 & 8 & 2 & 0 & 0 & 0 & 21 \\ 0 & - 3 & - 1 & 1 & 0 & 0 & 1 & 18 \end{array}]$

解题步骤 1.3

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 1.3.1

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

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & \frac{17}{5} & \frac{6}{5} & \frac{7}{5} & - \frac{3}{5} & 1 & 0 & - 2 \\ 1 - 1 & 8 - \frac{6}{5} & 8 - \frac{8}{5} & 2 - \frac{1}{5} & 0 - \frac{1}{5} & 0 - 0 & 0 - 0 & 21 - 4 \\ 0 & - 3 & - 1 & 1 & 0 & 0 & 1 & 18 \end{array}]$

解题步骤 1.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & \frac{17}{5} & \frac{6}{5} & \frac{7}{5} & - \frac{3}{5} & 1 & 0 & - 2 \\ 0 & \frac{34}{5} & \frac{32}{5} & \frac{9}{5} & - \frac{1}{5} & 0 & 0 & 17 \\ 0 & - 3 & - 1 & 1 & 0 & 0 & 1 & 18 \end{array}]$

解题步骤 1.4

Multiply each element of $R_{2}$ by $\frac{5}{17}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 1.4.1

Multiply each element of $R_{2}$ by $\frac{5}{17}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ \frac{5}{17} \cdot 0 & \frac{5}{17} \cdot \frac{17}{5} & \frac{5}{17} \cdot \frac{6}{5} & \frac{5}{17} \cdot \frac{7}{5} & \frac{5}{17} (- \frac{3}{5}) & \frac{5}{17} \cdot 1 & \frac{5}{17} \cdot 0 & \frac{5}{17} \cdot - 2 \\ 0 & \frac{34}{5} & \frac{32}{5} & \frac{9}{5} & - \frac{1}{5} & 0 & 0 & 17 \\ 0 & - 3 & - 1 & 1 & 0 & 0 & 1 & 18 \end{array}]$

解题步骤 1.4.2

化简 $R_{2}$ 。

解题步骤 1.5

Perform the row operation $R_{3} = R_{3} - \frac{34}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 1.5.1

Perform the row operation $R_{3} = R_{3} - \frac{34}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & 1 & \frac{6}{17} & \frac{7}{17} & - \frac{3}{17} & \frac{5}{17} & 0 & - \frac{10}{17} \\ 0 - \frac{34}{5} \cdot 0 & \frac{34}{5} - \frac{34}{5} \cdot 1 & \frac{32}{5} - \frac{34}{5} \cdot \frac{6}{17} & \frac{9}{5} - \frac{34}{5} \cdot \frac{7}{17} & - \frac{1}{5} - \frac{34}{5} (- \frac{3}{17}) & 0 - \frac{34}{5} \cdot \frac{5}{17} & 0 - \frac{34}{5} \cdot 0 & 17 - \frac{34}{5} (- \frac{10}{17}) \\ 0 & - 3 & - 1 & 1 & 0 & 0 & 1 & 18 \end{array}]$

解题步骤 1.5.2

化简 $R_{3}$ 。

解题步骤 1.6

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

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解题步骤 1.6.1

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

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & 1 & \frac{6}{17} & \frac{7}{17} & - \frac{3}{17} & \frac{5}{17} & 0 & - \frac{10}{17} \\ 0 & 0 & 4 & - 1 & 1 & - 2 & 0 & 21 \\ 0 + 3 \cdot 0 & - 3 + 3 \cdot 1 & - 1 + 3 (\frac{6}{17}) & 1 + 3 (\frac{7}{17}) & 0 + 3 (- \frac{3}{17}) & 0 + 3 (\frac{5}{17}) & 1 + 3 \cdot 0 & 18 + 3 (- \frac{10}{17}) \end{array}]$

解题步骤 1.6.2

化简 $R_{4}$ 。

解题步骤 1.7

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

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解题步骤 1.7.1

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

解题步骤 1.7.2

化简 $R_{3}$ 。

解题步骤 1.8

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

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解题步骤 1.8.1

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

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & 1 & \frac{6}{17} & \frac{7}{17} & - \frac{3}{17} & \frac{5}{17} & 0 & - \frac{10}{17} \\ 0 & 0 & 1 & - \frac{1}{4} & \frac{1}{4} & - \frac{1}{2} & 0 & \frac{21}{4} \\ 0 - \frac{1}{17} \cdot 0 & 0 - \frac{1}{17} \cdot 0 & \frac{1}{17} - \frac{1}{17} \cdot 1 & \frac{38}{17} - \frac{1}{17} (- \frac{1}{4}) & - \frac{9}{17} - \frac{1}{17} \cdot \frac{1}{4} & \frac{15}{17} - \frac{1}{17} (- \frac{1}{2}) & 1 - \frac{1}{17} \cdot 0 & \frac{276}{17} - \frac{1}{17} \cdot \frac{21}{4} \end{array}]$

解题步骤 1.8.2

化简 $R_{4}$ 。

解题步骤 1.9

Multiply each element of $R_{4}$ by $\frac{4}{9}$ to make the entry at $4, 4$ a $1$ .

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解题步骤 1.9.1

Multiply each element of $R_{4}$ by $\frac{4}{9}$ to make the entry at $4, 4$ a $1$ .

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & 1 & \frac{6}{17} & \frac{7}{17} & - \frac{3}{17} & \frac{5}{17} & 0 & - \frac{10}{17} \\ 0 & 0 & 1 & - \frac{1}{4} & \frac{1}{4} & - \frac{1}{2} & 0 & \frac{21}{4} \\ \frac{4}{9} \cdot 0 & \frac{4}{9} \cdot 0 & \frac{4}{9} \cdot 0 & \frac{4}{9} \cdot \frac{9}{4} & \frac{4}{9} (- \frac{37}{68}) & \frac{4}{9} \cdot \frac{31}{34} & \frac{4}{9} \cdot 1 & \frac{4}{9} \cdot \frac{1083}{68} \end{array}]$

解题步骤 1.9.2

化简 $R_{4}$ 。

解题步骤 1.10

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

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解题步骤 1.10.1

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

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & 1 & \frac{6}{17} & \frac{7}{17} & - \frac{3}{17} & \frac{5}{17} & 0 & - \frac{10}{17} \\ 0 + \frac{1}{4} \cdot 0 & 0 + \frac{1}{4} \cdot 0 & 1 + \frac{1}{4} \cdot 0 & - \frac{1}{4} + \frac{1}{4} \cdot 1 & \frac{1}{4} + \frac{1}{4} (- \frac{37}{153}) & - \frac{1}{2} + \frac{1}{4} \cdot \frac{62}{153} & 0 + \frac{1}{4} \cdot \frac{4}{9} & \frac{21}{4} + \frac{1}{4} \cdot \frac{361}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.10.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & 1 & \frac{6}{17} & \frac{7}{17} & - \frac{3}{17} & \frac{5}{17} & 0 & - \frac{10}{17} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.11

Perform the row operation $R_{2} = R_{2} - \frac{7}{17} R_{4}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 1.11.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{17} R_{4}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 - \frac{7}{17} \cdot 0 & 1 - \frac{7}{17} \cdot 0 & \frac{6}{17} - \frac{7}{17} \cdot 0 & \frac{7}{17} - \frac{7}{17} \cdot 1 & - \frac{3}{17} - \frac{7}{17} (- \frac{37}{153}) & \frac{5}{17} - \frac{7}{17} \cdot \frac{62}{153} & 0 - \frac{7}{17} \cdot \frac{4}{9} & - \frac{10}{17} - \frac{7}{17} \cdot \frac{361}{51} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.11.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 4 \\ 0 & 1 & \frac{6}{17} & 0 & - \frac{200}{2601} & \frac{331}{2601} & - \frac{28}{153} & - \frac{3037}{867} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.12

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

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解题步骤 1.12.1

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

$[\begin{array}{c} 1 - \frac{1}{5} \cdot 0 & \frac{6}{5} - \frac{1}{5} \cdot 0 & \frac{8}{5} - \frac{1}{5} \cdot 0 & \frac{1}{5} - \frac{1}{5} \cdot 1 & \frac{1}{5} - \frac{1}{5} (- \frac{37}{153}) & 0 - \frac{1}{5} \cdot \frac{62}{153} & 0 - \frac{1}{5} \cdot \frac{4}{9} & 4 - \frac{1}{5} \cdot \frac{361}{51} \\ 0 & 1 & \frac{6}{17} & 0 & - \frac{200}{2601} & \frac{331}{2601} & - \frac{28}{153} & - \frac{3037}{867} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.12.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & 0 & \frac{38}{153} & - \frac{62}{765} & - \frac{4}{45} & \frac{659}{255} \\ 0 & 1 & \frac{6}{17} & 0 & - \frac{200}{2601} & \frac{331}{2601} & - \frac{28}{153} & - \frac{3037}{867} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.13

Perform the row operation $R_{2} = R_{2} - \frac{6}{17} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 1.13.1

Perform the row operation $R_{2} = R_{2} - \frac{6}{17} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & 0 & \frac{38}{153} & - \frac{62}{765} & - \frac{4}{45} & \frac{659}{255} \\ 0 - \frac{6}{17} \cdot 0 & 1 - \frac{6}{17} \cdot 0 & \frac{6}{17} - \frac{6}{17} \cdot 1 & 0 - \frac{6}{17} \cdot 0 & - \frac{200}{2601} - \frac{6}{17} \cdot \frac{29}{153} & \frac{331}{2601} - \frac{6}{17} (- \frac{61}{153}) & - \frac{28}{153} - \frac{6}{17} \cdot \frac{1}{9} & - \frac{3037}{867} - \frac{6}{17} \cdot \frac{358}{51} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.13.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & \frac{8}{5} & 0 & \frac{38}{153} & - \frac{62}{765} & - \frac{4}{45} & \frac{659}{255} \\ 0 & 1 & 0 & 0 & - \frac{22}{153} & \frac{41}{153} & - \frac{2}{9} & - \frac{305}{51} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.14

Perform the row operation $R_{1} = R_{1} - \frac{8}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 1.14.1

Perform the row operation $R_{1} = R_{1} - \frac{8}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{8}{5} \cdot 0 & \frac{6}{5} - \frac{8}{5} \cdot 0 & \frac{8}{5} - \frac{8}{5} \cdot 1 & 0 - \frac{8}{5} \cdot 0 & \frac{38}{153} - \frac{8}{5} \cdot \frac{29}{153} & - \frac{62}{765} - \frac{8}{5} (- \frac{61}{153}) & - \frac{4}{45} - \frac{8}{5} \cdot \frac{1}{9} & \frac{659}{255} - \frac{8}{5} \cdot \frac{358}{51} \\ 0 & 1 & 0 & 0 & - \frac{22}{153} & \frac{41}{153} & - \frac{2}{9} & - \frac{305}{51} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.14.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{6}{5} & 0 & 0 & - \frac{14}{255} & \frac{142}{255} & - \frac{4}{15} & - \frac{147}{17} \\ 0 & 1 & 0 & 0 & - \frac{22}{153} & \frac{41}{153} & - \frac{2}{9} & - \frac{305}{51} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.15

Perform the row operation $R_{1} = R_{1} - \frac{6}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 1.15.1

Perform the row operation $R_{1} = R_{1} - \frac{6}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{6}{5} \cdot 0 & \frac{6}{5} - \frac{6}{5} \cdot 1 & 0 - \frac{6}{5} \cdot 0 & 0 - \frac{6}{5} \cdot 0 & - \frac{14}{255} - \frac{6}{5} (- \frac{22}{153}) & \frac{142}{255} - \frac{6}{5} \cdot \frac{41}{153} & - \frac{4}{15} - \frac{6}{5} (- \frac{2}{9}) & - \frac{147}{17} - \frac{6}{5} (- \frac{305}{51}) \\ 0 & 1 & 0 & 0 & - \frac{22}{153} & \frac{41}{153} & - \frac{2}{9} & - \frac{305}{51} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

解题步骤 1.15.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 0 & \frac{2}{17} & \frac{4}{17} & 0 & - \frac{25}{17} \\ 0 & 1 & 0 & 0 & - \frac{22}{153} & \frac{41}{153} & - \frac{2}{9} & - \frac{305}{51} \\ 0 & 0 & 1 & 0 & \frac{29}{153} & - \frac{61}{153} & \frac{1}{9} & \frac{358}{51} \\ 0 & 0 & 0 & 1 & - \frac{37}{153} & \frac{62}{153} & \frac{4}{9} & \frac{361}{51} \end{array}]$

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 $4$

有限数学 示例

有限数学示例