有限数学示例

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

解题步骤 1

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

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

解题步骤 1.2

化简 $R_{1}$ 。

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

解题步骤 2

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

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解题步骤 2.1

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

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & 0 \\ - \frac{1}{3} + \frac{1}{3} \cdot 1 & - \frac{2}{3} + \frac{1}{3} (- \frac{3}{2}) & 1 + \frac{1}{3} \cdot \frac{1}{2} & 0 + \frac{1}{3} \cdot 0 \\ 1 & - \frac{1}{3} & - \frac{2}{3} & 0 \end{array}]$

解题步骤 2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & 0 \\ 0 & - \frac{7}{6} & \frac{7}{6} & 0 \\ 1 & - \frac{1}{3} & - \frac{2}{3} & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & 0 \\ 0 & - \frac{7}{6} & \frac{7}{6} & 0 \\ 1 - 1 & - \frac{1}{3} + \frac{3}{2} & - \frac{2}{3} - \frac{1}{2} & 0 - 0 \end{array}]$

解题步骤 3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & 0 \\ 0 & - \frac{7}{6} & \frac{7}{6} & 0 \\ 0 & \frac{7}{6} & - \frac{7}{6} & 0 \end{array}]$

解题步骤 4

Multiply each element of $R_{2}$ by $- \frac{6}{7}$ 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 $- \frac{6}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & 0 \\ - \frac{6}{7} \cdot 0 & - \frac{6}{7} (- \frac{7}{6}) & - \frac{6}{7} \cdot \frac{7}{6} & - \frac{6}{7} \cdot 0 \\ 0 & \frac{7}{6} & - \frac{7}{6} & 0 \end{array}]$

解题步骤 4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & 0 \\ 0 & 1 & - 1 & 0 \\ 0 & \frac{7}{6} & - \frac{7}{6} & 0 \end{array}]$

解题步骤 5

Perform the row operation $R_{3} = R_{3} - \frac{7}{6} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 5.1

Perform the row operation $R_{3} = R_{3} - \frac{7}{6} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & 0 \\ 0 & 1 & - 1 & 0 \\ 0 - \frac{7}{6} \cdot 0 & \frac{7}{6} - \frac{7}{6} \cdot 1 & - \frac{7}{6} - \frac{7}{6} \cdot - 1 & 0 - \frac{7}{6} \cdot 0 \end{array}]$

解题步骤 5.2

化简 $R_{3}$ 。

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

解题步骤 6

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

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解题步骤 6.1

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

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

解题步骤 6.2

化简 $R_{1}$ 。

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

有限数学 示例

有限数学示例