有限数学示例

$[\begin{array}{c} 12 & 3 & 1 & 0 & 0 & 15 \\ 1 & 4 & 0 & 1 & 0 & 12 \\ - 6 & - 24 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1

Multiply each element of $R_{1}$ by $\frac{1}{12}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 1.1

Multiply each element of $R_{1}$ by $\frac{1}{12}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{12}{12} & \frac{3}{12} & \frac{1}{12} & \frac{0}{12} & \frac{0}{12} & \frac{15}{12} \\ 1 & 4 & 0 & 1 & 0 & 12 \\ - 6 & - 24 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 1 & 4 & 0 & 1 & 0 & 12 \\ - 6 & - 24 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 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 2.1

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

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 1 - 1 & 4 - \frac{1}{4} & 0 - \frac{1}{12} & 1 - 0 & 0 - 0 & 12 - \frac{5}{4} \\ - 6 & - 24 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 & \frac{15}{4} & - \frac{1}{12} & 1 & 0 & \frac{43}{4} \\ - 6 & - 24 & 0 & 0 & 1 & 0 \end{array}]$

解题步骤 3

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

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解题步骤 3.1

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

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 & \frac{15}{4} & - \frac{1}{12} & 1 & 0 & \frac{43}{4} \\ - 6 + 6 \cdot 1 & - 24 + 6 (\frac{1}{4}) & 0 + 6 (\frac{1}{12}) & 0 + 6 \cdot 0 & 1 + 6 \cdot 0 & 0 + 6 (\frac{5}{4}) \end{array}]$

解题步骤 3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 & \frac{15}{4} & - \frac{1}{12} & 1 & 0 & \frac{43}{4} \\ 0 & - \frac{45}{2} & \frac{1}{2} & 0 & 1 & \frac{15}{2} \end{array}]$

解题步骤 4

Multiply each element of $R_{2}$ by $\frac{4}{15}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 4.1

Multiply each element of $R_{2}$ by $\frac{4}{15}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ \frac{4}{15} \cdot 0 & \frac{4}{15} \cdot \frac{15}{4} & \frac{4}{15} (- \frac{1}{12}) & \frac{4}{15} \cdot 1 & \frac{4}{15} \cdot 0 & \frac{4}{15} \cdot \frac{43}{4} \\ 0 & - \frac{45}{2} & \frac{1}{2} & 0 & 1 & \frac{15}{2} \end{array}]$

解题步骤 4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 & 1 & - \frac{1}{45} & \frac{4}{15} & 0 & \frac{43}{15} \\ 0 & - \frac{45}{2} & \frac{1}{2} & 0 & 1 & \frac{15}{2} \end{array}]$

解题步骤 5

Perform the row operation $R_{3} = R_{3} + \frac{45}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 5.1

Perform the row operation $R_{3} = R_{3} + \frac{45}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 & 1 & - \frac{1}{45} & \frac{4}{15} & 0 & \frac{43}{15} \\ 0 + \frac{45}{2} \cdot 0 & - \frac{45}{2} + \frac{45}{2} \cdot 1 & \frac{1}{2} + \frac{45}{2} (- \frac{1}{45}) & 0 + \frac{45}{2} \cdot \frac{4}{15} & 1 + \frac{45}{2} \cdot 0 & \frac{15}{2} + \frac{45}{2} \cdot \frac{43}{15} \end{array}]$

解题步骤 5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 & 1 & - \frac{1}{45} & \frac{4}{15} & 0 & \frac{43}{15} \\ 0 & 0 & 0 & 6 & 1 & 72 \end{array}]$

解题步骤 6

Multiply each element of $R_{3}$ by $\frac{1}{6}$ to make the entry at $3, 4$ a $1$ .

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解题步骤 6.1

Multiply each element of $R_{3}$ by $\frac{1}{6}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 & 1 & - \frac{1}{45} & \frac{4}{15} & 0 & \frac{43}{15} \\ \frac{0}{6} & \frac{0}{6} & \frac{0}{6} & \frac{6}{6} & \frac{1}{6} & \frac{72}{6} \end{array}]$

解题步骤 6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 & 1 & - \frac{1}{45} & \frac{4}{15} & 0 & \frac{43}{15} \\ 0 & 0 & 0 & 1 & \frac{1}{6} & 12 \end{array}]$

解题步骤 7

Perform the row operation $R_{2} = R_{2} - \frac{4}{15} R_{3}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 7.1

Perform the row operation $R_{2} = R_{2} - \frac{4}{15} R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 - \frac{4}{15} \cdot 0 & 1 - \frac{4}{15} \cdot 0 & - \frac{1}{45} - \frac{4}{15} \cdot 0 & \frac{4}{15} - \frac{4}{15} \cdot 1 & 0 - \frac{4}{15} \cdot \frac{1}{6} & \frac{43}{15} - \frac{4}{15} \cdot 12 \\ 0 & 0 & 0 & 1 & \frac{1}{6} & 12 \end{array}]$

解题步骤 7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{1}{12} & 0 & 0 & \frac{5}{4} \\ 0 & 1 & - \frac{1}{45} & 0 & - \frac{2}{45} & - \frac{1}{3} \\ 0 & 0 & 0 & 1 & \frac{1}{6} & 12 \end{array}]$

解题步骤 8

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

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解题步骤 8.1

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

$[\begin{array}{c} 1 - \frac{1}{4} \cdot 0 & \frac{1}{4} - \frac{1}{4} \cdot 1 & \frac{1}{12} - \frac{1}{4} (- \frac{1}{45}) & 0 - \frac{1}{4} \cdot 0 & 0 - \frac{1}{4} (- \frac{2}{45}) & \frac{5}{4} - \frac{1}{4} (- \frac{1}{3}) \\ 0 & 1 & - \frac{1}{45} & 0 & - \frac{2}{45} & - \frac{1}{3} \\ 0 & 0 & 0 & 1 & \frac{1}{6} & 12 \end{array}]$

解题步骤 8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{4}{45} & 0 & \frac{1}{90} & \frac{4}{3} \\ 0 & 1 & - \frac{1}{45} & 0 & - \frac{2}{45} & - \frac{1}{3} \\ 0 & 0 & 0 & 1 & \frac{1}{6} & 12 \end{array}]$

有限数学 示例

有限数学示例