有限数学示例

$[\begin{array}{c} 6 & 3 & 8 & 1 & 0 & 0 & 0 & 784 \\ 1 & 1 & 2 & 0 & 1 & 0 & 0 & 140 \\ 4 & 5 & 0 & 0 & 0 & 1 & 0 & 504 \\ - 60 & - 50 & - 100 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1

求行简化阶梯形矩阵。

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解题步骤 1.1

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

$[\begin{array}{c} \frac{6}{6} & \frac{3}{6} & \frac{8}{6} & \frac{1}{6} & \frac{0}{6} & \frac{0}{6} & \frac{0}{6} & \frac{784}{6} \\ 1 & 1 & 2 & 0 & 1 & 0 & 0 & 140 \\ 4 & 5 & 0 & 0 & 0 & 1 & 0 & 504 \\ - 60 & - 50 & - 100 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 1 & 1 & 2 & 0 & 1 & 0 & 0 & 140 \\ 4 & 5 & 0 & 0 & 0 & 1 & 0 & 504 \\ - 60 & - 50 & - 100 & 0 & 0 & 0 & 1 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 1 - 1 & 1 - \frac{1}{2} & 2 - \frac{4}{3} & 0 - \frac{1}{6} & 1 - 0 & 0 - 0 & 0 - 0 & 140 - \frac{392}{3} \\ 4 & 5 & 0 & 0 & 0 & 1 & 0 & 504 \\ - 60 & - 50 & - 100 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & \frac{1}{2} & \frac{2}{3} & - \frac{1}{6} & 1 & 0 & 0 & \frac{28}{3} \\ 4 & 5 & 0 & 0 & 0 & 1 & 0 & 504 \\ - 60 & - 50 & - 100 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.3

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

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解题步骤 1.3.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & \frac{1}{2} & \frac{2}{3} & - \frac{1}{6} & 1 & 0 & 0 & \frac{28}{3} \\ 4 - 4 \cdot 1 & 5 - 4 (\frac{1}{2}) & 0 - 4 (\frac{4}{3}) & 0 - 4 (\frac{1}{6}) & 0 - 4 \cdot 0 & 1 - 4 \cdot 0 & 0 - 4 \cdot 0 & 504 - 4 (\frac{392}{3}) \\ - 60 & - 50 & - 100 & 0 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.3.2

化简 $R_{3}$ 。

解题步骤 1.4

Perform the row operation $R_{4} = R_{4} + 60 R_{1}$ to make the entry at $4, 1$ a $0$ .

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解题步骤 1.4.1

Perform the row operation $R_{4} = R_{4} + 60 R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & \frac{1}{2} & \frac{2}{3} & - \frac{1}{6} & 1 & 0 & 0 & \frac{28}{3} \\ 0 & 3 & - \frac{16}{3} & - \frac{2}{3} & 0 & 1 & 0 & - \frac{56}{3} \\ - 60 + 60 \cdot 1 & - 50 + 60 (\frac{1}{2}) & - 100 + 60 (\frac{4}{3}) & 0 + 60 (\frac{1}{6}) & 0 + 60 \cdot 0 & 0 + 60 \cdot 0 & 1 + 60 \cdot 0 & 0 + 60 (\frac{392}{3}) \end{array}]$

解题步骤 1.4.2

化简 $R_{4}$ 。

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 $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 2 \cdot 0 & 2 (\frac{1}{2}) & 2 (\frac{2}{3}) & 2 (- \frac{1}{6}) & 2 \cdot 1 & 2 \cdot 0 & 2 \cdot 0 & 2 (\frac{28}{3}) \\ 0 & 3 & - \frac{16}{3} & - \frac{2}{3} & 0 & 1 & 0 & - \frac{56}{3} \\ 0 & - 20 & - 20 & 10 & 0 & 0 & 1 & 7840 \end{array}]$

解题步骤 1.5.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & - \frac{1}{3} & 2 & 0 & 0 & \frac{56}{3} \\ 0 & 3 & - \frac{16}{3} & - \frac{2}{3} & 0 & 1 & 0 & - \frac{56}{3} \\ 0 & - 20 & - 20 & 10 & 0 & 0 & 1 & 7840 \end{array}]$

解题步骤 1.6

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

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解题步骤 1.6.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & - \frac{1}{3} & 2 & 0 & 0 & \frac{56}{3} \\ 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & - \frac{16}{3} - 3 (\frac{4}{3}) & - \frac{2}{3} - 3 (- \frac{1}{3}) & 0 - 3 \cdot 2 & 1 - 3 \cdot 0 & 0 - 3 \cdot 0 & - \frac{56}{3} - 3 (\frac{56}{3}) \\ 0 & - 20 & - 20 & 10 & 0 & 0 & 1 & 7840 \end{array}]$

解题步骤 1.6.2

化简 $R_{3}$ 。

解题步骤 1.7

Perform the row operation $R_{4} = R_{4} + 20 R_{2}$ to make the entry at $4, 2$ a $0$ .

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解题步骤 1.7.1

Perform the row operation $R_{4} = R_{4} + 20 R_{2}$ to make the entry at $4, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & - \frac{1}{3} & 2 & 0 & 0 & \frac{56}{3} \\ 0 & 0 & - \frac{28}{3} & \frac{1}{3} & - 6 & 1 & 0 & - \frac{224}{3} \\ 0 + 20 \cdot 0 & - 20 + 20 \cdot 1 & - 20 + 20 (\frac{4}{3}) & 10 + 20 (- \frac{1}{3}) & 0 + 20 \cdot 2 & 0 + 20 \cdot 0 & 1 + 20 \cdot 0 & 7840 + 20 (\frac{56}{3}) \end{array}]$

解题步骤 1.7.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & - \frac{1}{3} & 2 & 0 & 0 & \frac{56}{3} \\ 0 & 0 & - \frac{28}{3} & \frac{1}{3} & - 6 & 1 & 0 & - \frac{224}{3} \\ 0 & 0 & \frac{20}{3} & \frac{10}{3} & 40 & 0 & 1 & \frac{24640}{3} \end{array}]$

解题步骤 1.8

Multiply each element of $R_{3}$ by $- \frac{3}{28}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 1.8.1

Multiply each element of $R_{3}$ by $- \frac{3}{28}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & - \frac{1}{3} & 2 & 0 & 0 & \frac{56}{3} \\ - \frac{3}{28} \cdot 0 & - \frac{3}{28} \cdot 0 & - \frac{3}{28} (- \frac{28}{3}) & - \frac{3}{28} \cdot \frac{1}{3} & - \frac{3}{28} \cdot - 6 & - \frac{3}{28} \cdot 1 & - \frac{3}{28} \cdot 0 & - \frac{3}{28} (- \frac{224}{3}) \\ 0 & 0 & \frac{20}{3} & \frac{10}{3} & 40 & 0 & 1 & \frac{24640}{3} \end{array}]$

解题步骤 1.8.2

化简 $R_{3}$ 。

解题步骤 1.9

Perform the row operation $R_{4} = R_{4} - \frac{20}{3} R_{3}$ to make the entry at $4, 3$ a $0$ .

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解题步骤 1.9.1

Perform the row operation $R_{4} = R_{4} - \frac{20}{3} R_{3}$ to make the entry at $4, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & - \frac{1}{3} & 2 & 0 & 0 & \frac{56}{3} \\ 0 & 0 & 1 & - \frac{1}{28} & \frac{9}{14} & - \frac{3}{28} & 0 & 8 \\ 0 - \frac{20}{3} \cdot 0 & 0 - \frac{20}{3} \cdot 0 & \frac{20}{3} - \frac{20}{3} \cdot 1 & \frac{10}{3} - \frac{20}{3} (- \frac{1}{28}) & 40 - \frac{20}{3} \cdot \frac{9}{14} & 0 - \frac{20}{3} (- \frac{3}{28}) & 1 - \frac{20}{3} \cdot 0 & \frac{24640}{3} - \frac{20}{3} \cdot 8 \end{array}]$

解题步骤 1.9.2

化简 $R_{4}$ 。

解题步骤 1.10

Multiply each element of $R_{4}$ by $\frac{7}{25}$ to make the entry at $4, 4$ a $1$ .

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解题步骤 1.10.1

Multiply each element of $R_{4}$ by $\frac{7}{25}$ to make the entry at $4, 4$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & - \frac{1}{3} & 2 & 0 & 0 & \frac{56}{3} \\ 0 & 0 & 1 & - \frac{1}{28} & \frac{9}{14} & - \frac{3}{28} & 0 & 8 \\ \frac{7}{25} \cdot 0 & \frac{7}{25} \cdot 0 & \frac{7}{25} \cdot 0 & \frac{7}{25} \cdot \frac{25}{7} & \frac{7}{25} \cdot \frac{250}{7} & \frac{7}{25} \cdot \frac{5}{7} & \frac{7}{25} \cdot 1 & \frac{7}{25} \cdot 8160 \end{array}]$

解题步骤 1.10.2

化简 $R_{4}$ 。

解题步骤 1.11

Perform the row operation $R_{3} = R_{3} + \frac{1}{28} R_{4}$ to make the entry at $3, 4$ a $0$ .

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解题步骤 1.11.1

Perform the row operation $R_{3} = R_{3} + \frac{1}{28} R_{4}$ to make the entry at $3, 4$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & - \frac{1}{3} & 2 & 0 & 0 & \frac{56}{3} \\ 0 + \frac{1}{28} \cdot 0 & 0 + \frac{1}{28} \cdot 0 & 1 + \frac{1}{28} \cdot 0 & - \frac{1}{28} + \frac{1}{28} \cdot 1 & \frac{9}{14} + \frac{1}{28} \cdot 10 & - \frac{3}{28} + \frac{1}{28} \cdot \frac{1}{5} & 0 + \frac{1}{28} \cdot \frac{7}{25} & 8 + \frac{1}{28} \cdot \frac{11424}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.11.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & - \frac{1}{3} & 2 & 0 & 0 & \frac{56}{3} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.12

Perform the row operation $R_{2} = R_{2} + \frac{1}{3} R_{4}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 1.12.1

Perform the row operation $R_{2} = R_{2} + \frac{1}{3} R_{4}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 + \frac{1}{3} \cdot 0 & 1 + \frac{1}{3} \cdot 0 & \frac{4}{3} + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & 2 + \frac{1}{3} \cdot 10 & 0 + \frac{1}{3} \cdot \frac{1}{5} & 0 + \frac{1}{3} \cdot \frac{7}{25} & \frac{56}{3} + \frac{1}{3} \cdot \frac{11424}{5} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.12.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & \frac{1}{6} & 0 & 0 & 0 & \frac{392}{3} \\ 0 & 1 & \frac{4}{3} & 0 & \frac{16}{3} & \frac{1}{15} & \frac{7}{75} & \frac{11704}{15} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.13

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

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解题步骤 1.13.1

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

$[\begin{array}{c} 1 - \frac{1}{6} \cdot 0 & \frac{1}{2} - \frac{1}{6} \cdot 0 & \frac{4}{3} - \frac{1}{6} \cdot 0 & \frac{1}{6} - \frac{1}{6} \cdot 1 & 0 - \frac{1}{6} \cdot 10 & 0 - \frac{1}{6} \cdot \frac{1}{5} & 0 - \frac{1}{6} \cdot \frac{7}{25} & \frac{392}{3} - \frac{1}{6} \cdot \frac{11424}{5} \\ 0 & 1 & \frac{4}{3} & 0 & \frac{16}{3} & \frac{1}{15} & \frac{7}{75} & \frac{11704}{15} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.13.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & 0 & - \frac{5}{3} & - \frac{1}{30} & - \frac{7}{150} & - \frac{3752}{15} \\ 0 & 1 & \frac{4}{3} & 0 & \frac{16}{3} & \frac{1}{15} & \frac{7}{75} & \frac{11704}{15} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.14

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

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解题步骤 1.14.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & 0 & - \frac{5}{3} & - \frac{1}{30} & - \frac{7}{150} & - \frac{3752}{15} \\ 0 - \frac{4}{3} \cdot 0 & 1 - \frac{4}{3} \cdot 0 & \frac{4}{3} - \frac{4}{3} \cdot 1 & 0 - \frac{4}{3} \cdot 0 & \frac{16}{3} - \frac{4}{3} \cdot 1 & \frac{1}{15} - \frac{4}{3} (- \frac{1}{10}) & \frac{7}{75} - \frac{4}{3} \cdot \frac{1}{100} & \frac{11704}{15} - \frac{4}{3} \cdot \frac{448}{5} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.14.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{4}{3} & 0 & - \frac{5}{3} & - \frac{1}{30} & - \frac{7}{150} & - \frac{3752}{15} \\ 0 & 1 & 0 & 0 & 4 & \frac{1}{5} & \frac{2}{25} & \frac{3304}{5} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.15

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

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解题步骤 1.15.1

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

$[\begin{array}{c} 1 - \frac{4}{3} \cdot 0 & \frac{1}{2} - \frac{4}{3} \cdot 0 & \frac{4}{3} - \frac{4}{3} \cdot 1 & 0 - \frac{4}{3} \cdot 0 & - \frac{5}{3} - \frac{4}{3} \cdot 1 & - \frac{1}{30} - \frac{4}{3} (- \frac{1}{10}) & - \frac{7}{150} - \frac{4}{3} \cdot \frac{1}{100} & - \frac{3752}{15} - \frac{4}{3} \cdot \frac{448}{5} \\ 0 & 1 & 0 & 0 & 4 & \frac{1}{5} & \frac{2}{25} & \frac{3304}{5} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.15.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 0 & - 3 & \frac{1}{10} & - \frac{3}{50} & - \frac{1848}{5} \\ 0 & 1 & 0 & 0 & 4 & \frac{1}{5} & \frac{2}{25} & \frac{3304}{5} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.16

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

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解题步骤 1.16.1

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

$[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot 0 & 0 - \frac{1}{2} \cdot 0 & - 3 - \frac{1}{2} \cdot 4 & \frac{1}{10} - \frac{1}{2} \cdot \frac{1}{5} & - \frac{3}{50} - \frac{1}{2} \cdot \frac{2}{25} & - \frac{1848}{5} - \frac{1}{2} \cdot \frac{3304}{5} \\ 0 & 1 & 0 & 0 & 4 & \frac{1}{5} & \frac{2}{25} & \frac{3304}{5} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

解题步骤 1.16.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & 0 & - 5 & 0 & - \frac{1}{10} & - 700 \\ 0 & 1 & 0 & 0 & 4 & \frac{1}{5} & \frac{2}{25} & \frac{3304}{5} \\ 0 & 0 & 1 & 0 & 1 & - \frac{1}{10} & \frac{1}{100} & \frac{448}{5} \\ 0 & 0 & 0 & 1 & 10 & \frac{1}{5} & \frac{7}{25} & \frac{11424}{5} \end{array}]$

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