有限数学示例

$[\begin{array}{c} 0 & 1 & \frac{5}{7} & - \frac{1}{7} & 0 & \frac{16}{7} \\ 1 & 0 & - \frac{3}{7} & \frac{2}{7} & 0 & \frac{30}{7} \\ 0 & 0 & \frac{13}{7} & \frac{3}{7} & 1 & \frac{220}{7} \end{array}]$

解题步骤 1

求行简化阶梯形矩阵。

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解题步骤 1.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 1 & 0 & - \frac{3}{7} & \frac{2}{7} & 0 & \frac{30}{7} \\ 0 & 1 & \frac{5}{7} & - \frac{1}{7} & 0 & \frac{16}{7} \\ 0 & 0 & \frac{13}{7} & \frac{3}{7} & 1 & \frac{220}{7} \end{array}]$

解题步骤 1.2

Multiply each element of $R_{3}$ by $\frac{7}{13}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 1.2.1

Multiply each element of $R_{3}$ by $\frac{7}{13}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & - \frac{3}{7} & \frac{2}{7} & 0 & \frac{30}{7} \\ 0 & 1 & \frac{5}{7} & - \frac{1}{7} & 0 & \frac{16}{7} \\ \frac{7}{13} \cdot 0 & \frac{7}{13} \cdot 0 & \frac{7}{13} \cdot \frac{13}{7} & \frac{7}{13} \cdot \frac{3}{7} & \frac{7}{13} \cdot 1 & \frac{7}{13} \cdot \frac{220}{7} \end{array}]$

解题步骤 1.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{3}{7} & \frac{2}{7} & 0 & \frac{30}{7} \\ 0 & 1 & \frac{5}{7} & - \frac{1}{7} & 0 & \frac{16}{7} \\ 0 & 0 & 1 & \frac{3}{13} & \frac{7}{13} & \frac{220}{13} \end{array}]$

解题步骤 1.3

Perform the row operation $R_{2} = R_{2} - \frac{5}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 1.3.1

Perform the row operation $R_{2} = R_{2} - \frac{5}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 0 & - \frac{3}{7} & \frac{2}{7} & 0 & \frac{30}{7} \\ 0 - \frac{5}{7} \cdot 0 & 1 - \frac{5}{7} \cdot 0 & \frac{5}{7} - \frac{5}{7} \cdot 1 & - \frac{1}{7} - \frac{5}{7} \cdot \frac{3}{13} & 0 - \frac{5}{7} \cdot \frac{7}{13} & \frac{16}{7} - \frac{5}{7} \cdot \frac{220}{13} \\ 0 & 0 & 1 & \frac{3}{13} & \frac{7}{13} & \frac{220}{13} \end{array}]$

解题步骤 1.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{3}{7} & \frac{2}{7} & 0 & \frac{30}{7} \\ 0 & 1 & 0 & - \frac{4}{13} & - \frac{5}{13} & - \frac{892}{91} \\ 0 & 0 & 1 & \frac{3}{13} & \frac{7}{13} & \frac{220}{13} \end{array}]$

解题步骤 1.4

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

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解题步骤 1.4.1

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

$[\begin{array}{c} 1 + \frac{3}{7} \cdot 0 & 0 + \frac{3}{7} \cdot 0 & - \frac{3}{7} + \frac{3}{7} \cdot 1 & \frac{2}{7} + \frac{3}{7} \cdot \frac{3}{13} & 0 + \frac{3}{7} \cdot \frac{7}{13} & \frac{30}{7} + \frac{3}{7} \cdot \frac{220}{13} \\ 0 & 1 & 0 & - \frac{4}{13} & - \frac{5}{13} & - \frac{892}{91} \\ 0 & 0 & 1 & \frac{3}{13} & \frac{7}{13} & \frac{220}{13} \end{array}]$

解题步骤 1.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & \frac{5}{13} & \frac{3}{13} & \frac{150}{13} \\ 0 & 1 & 0 & - \frac{4}{13} & - \frac{5}{13} & - \frac{892}{91} \\ 0 & 0 & 1 & \frac{3}{13} & \frac{7}{13} & \frac{220}{13} \end{array}]$

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: $a_{11}, a_{22},$ and $a_{33}$

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

有限数学 示例

有限数学示例