有限数学示例

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ 3 & 9 & - 21 & 0 \\ 2 & 4 & - 10 & - 2 \end{array}]$

解题步骤 1

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 1.1

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

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ 3 - 3 \cdot 1 & 9 - 3 \cdot 5 & - 21 - 3 \cdot - 12 & 0 - 3 \cdot 1 \\ 2 & 4 & - 10 & - 2 \end{array}]$

解题步骤 1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ 0 & - 6 & 15 & - 3 \\ 2 & 4 & - 10 & - 2 \end{array}]$

解题步骤 2

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

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解题步骤 2.1

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

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ 0 & - 6 & 15 & - 3 \\ 2 - 2 \cdot 1 & 4 - 2 \cdot 5 & - 10 - 2 \cdot - 12 & - 2 - 2 \cdot 1 \end{array}]$

解题步骤 2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ 0 & - 6 & 15 & - 3 \\ 0 & - 6 & 14 & - 4 \end{array}]$

解题步骤 3

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.1

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot 15 & - \frac{1}{6} \cdot - 3 \\ 0 & - 6 & 14 & - 4 \end{array}]$

解题步骤 3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ 0 & 1 & - \frac{5}{2} & \frac{1}{2} \\ 0 & - 6 & 14 & - 4 \end{array}]$

解题步骤 4

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

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解题步骤 4.1

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

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ 0 & 1 & - \frac{5}{2} & \frac{1}{2} \\ 0 + 6 \cdot 0 & - 6 + 6 \cdot 1 & 14 + 6 (- \frac{5}{2}) & - 4 + 6 (\frac{1}{2}) \end{array}]$

解题步骤 4.2

化简 $R_{3}$ 。

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

解题步骤 5

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

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解题步骤 5.1

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

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

解题步骤 5.2

化简 $R_{3}$ 。

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

解题步骤 6

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

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解题步骤 6.1

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

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ 0 + \frac{5}{2} \cdot 0 & 1 + \frac{5}{2} \cdot 0 & - \frac{5}{2} + \frac{5}{2} \cdot 1 & \frac{1}{2} + \frac{5}{2} \cdot 1 \\ 0 & 0 & 1 & 1 \end{array}]$

解题步骤 6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 5 & - 12 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1 \end{array}]$

解题步骤 7

Perform the row operation $R_{1} = R_{1} + 12 R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 7.1

Perform the row operation $R_{1} = R_{1} + 12 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + 12 \cdot 0 & 5 + 12 \cdot 0 & - 12 + 12 \cdot 1 & 1 + 12 \cdot 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1 \end{array}]$

解题步骤 7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 5 & 0 & 13 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1 \end{array}]$

解题步骤 8

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

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解题步骤 8.1

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

$[\begin{array}{c} 1 - 5 \cdot 0 & 5 - 5 \cdot 1 & 0 - 5 \cdot 0 & 13 - 5 \cdot 3 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1 \end{array}]$

解题步骤 8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & - 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1 \end{array}]$

有限数学 示例

有限数学示例