有限数学示例

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

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 $0$ .

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

解题步骤 1.2

化简 $R_{2}$ 。

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

解题步骤 2

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 2.1

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

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

解题步骤 2.2

化简 $R_{3}$ 。

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

解题步骤 3

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

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

解题步骤 3.2

化简 $R_{2}$ 。

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

解题步骤 4

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

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解题步骤 4.1

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

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

解题步骤 4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 1 & 0 \\ 0 & 1 & - \frac{5}{4} \\ 0 & 0 & - \frac{7}{2} \end{array}]$

解题步骤 5

Multiply each element of $R_{3}$ by $- \frac{2}{7}$ 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 $- \frac{2}{7}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 0 \\ 0 & 1 & - \frac{5}{4} \\ - \frac{2}{7} \cdot 0 & - \frac{2}{7} \cdot 0 & - \frac{2}{7} (- \frac{7}{2}) \end{array}]$

解题步骤 5.2

化简 $R_{3}$ 。

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

解题步骤 6

Perform the row operation $R_{2} = R_{2} + \frac{5}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 6.1

Perform the row operation $R_{2} = R_{2} + \frac{5}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 1 & 0 \\ 0 + \frac{5}{4} \cdot 0 & 1 + \frac{5}{4} \cdot 0 & - \frac{5}{4} + \frac{5}{4} \cdot 1 \\ 0 & 0 & 1 \end{array}]$

解题步骤 6.2

化简 $R_{2}$ 。

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

解题步骤 7

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

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解题步骤 7.1

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

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

解题步骤 7.2

化简 $R_{1}$ 。

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

有限数学 示例

有限数学示例