有限数学示例

$[\begin{array}{c} 5 & 27 & 2 & 17 \\ 2 & 11 & 3 & - 5 \\ 1 & 5 & - 4 & 8 \end{array}]$

解题步骤 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

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

$[\begin{array}{c} \frac{5}{5} & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 2 & 11 & 3 & - 5 \\ 1 & 5 & - 4 & 8 \end{array}]$

解题步骤 1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 2 & 11 & 3 & - 5 \\ 1 & 5 & - 4 & 8 \end{array}]$

解题步骤 2

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

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解题步骤 2.1

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

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 2 - 2 \cdot 1 & 11 - 2 (\frac{27}{5}) & 3 - 2 (\frac{2}{5}) & - 5 - 2 (\frac{17}{5}) \\ 1 & 5 & - 4 & 8 \end{array}]$

解题步骤 2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 0 & \frac{1}{5} & \frac{11}{5} & - \frac{59}{5} \\ 1 & 5 & - 4 & 8 \end{array}]$

解题步骤 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 3.1

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

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 0 & \frac{1}{5} & \frac{11}{5} & - \frac{59}{5} \\ 1 - 1 & 5 - \frac{27}{5} & - 4 - \frac{2}{5} & 8 - \frac{17}{5} \end{array}]$

解题步骤 3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 0 & \frac{1}{5} & \frac{11}{5} & - \frac{59}{5} \\ 0 & - \frac{2}{5} & - \frac{22}{5} & \frac{23}{5} \end{array}]$

解题步骤 4

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

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 5 \cdot 0 & 5 (\frac{1}{5}) & 5 (\frac{11}{5}) & 5 (- \frac{59}{5}) \\ 0 & - \frac{2}{5} & - \frac{22}{5} & \frac{23}{5} \end{array}]$

解题步骤 4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 0 & 1 & 11 & - 59 \\ 0 & - \frac{2}{5} & - \frac{22}{5} & \frac{23}{5} \end{array}]$

解题步骤 5

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

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解题步骤 5.1

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

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 0 & 1 & 11 & - 59 \\ 0 + \frac{2}{5} \cdot 0 & - \frac{2}{5} + \frac{2}{5} \cdot 1 & - \frac{22}{5} + \frac{2}{5} \cdot 11 & \frac{23}{5} + \frac{2}{5} \cdot - 59 \end{array}]$

解题步骤 5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 0 & 1 & 11 & - 59 \\ 0 & 0 & 0 & - 19 \end{array}]$

解题步骤 6

Multiply each element of $R_{3}$ by $- \frac{1}{19}$ 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 $- \frac{1}{19}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 0 & 1 & 11 & - 59 \\ - \frac{1}{19} \cdot 0 & - \frac{1}{19} \cdot 0 & - \frac{1}{19} \cdot 0 & - \frac{1}{19} \cdot - 19 \end{array}]$

解题步骤 6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 0 & 1 & 11 & - 59 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 7

Perform the row operation $R_{2} = R_{2} + 59 R_{3}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 7.1

Perform the row operation $R_{2} = R_{2} + 59 R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & \frac{17}{5} \\ 0 + 59 \cdot 0 & 1 + 59 \cdot 0 & 11 + 59 \cdot 0 & - 59 + 59 \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 7.2

化简 $R_{2}$ 。

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

解题步骤 8

Perform the row operation $R_{1} = R_{1} - \frac{17}{5} R_{3}$ to make the entry at $1, 4$ a $0$ .

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解题步骤 8.1

Perform the row operation $R_{1} = R_{1} - \frac{17}{5} R_{3}$ to make the entry at $1, 4$ a $0$ .

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

解题步骤 8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{27}{5} & \frac{2}{5} & 0 \\ 0 & 1 & 11 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 9

Perform the row operation $R_{1} = R_{1} - \frac{27}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 9.1

Perform the row operation $R_{1} = R_{1} - \frac{27}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

解题步骤 9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & - 59 & 0 \\ 0 & 1 & 11 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

有限数学 示例

有限数学示例