有限数学示例

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

解题步骤 1

Multiply each element of $R_{1}$ by $\frac{6}{5}$ 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 $\frac{6}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{6}{5} \cdot \frac{5}{6} & \frac{6}{5} \cdot \frac{1}{5} \\ \frac{1}{6} & \frac{4}{5} \end{array}]$

解题步骤 1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{6}{25} \\ \frac{1}{6} & \frac{4}{5} \end{array}]$

解题步骤 2

Perform the row operation $R_{2} = R_{2} - \frac{1}{6} R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 2.1

Perform the row operation $R_{2} = R_{2} - \frac{1}{6} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{25} \\ \frac{1}{6} - \frac{1}{6} \cdot 1 & \frac{4}{5} - \frac{1}{6} \cdot \frac{6}{25} \end{array}]$

解题步骤 2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{6}{25} \\ 0 & \frac{19}{25} \end{array}]$

解题步骤 3

Multiply each element of $R_{2}$ by $\frac{25}{19}$ 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 $\frac{25}{19}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{6}{25} \\ \frac{25}{19} \cdot 0 & \frac{25}{19} \cdot \frac{19}{25} \end{array}]$

解题步骤 3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{6}{25} \\ 0 & 1 \end{array}]$

解题步骤 4

Perform the row operation $R_{1} = R_{1} - \frac{6}{25} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 4.1

Perform the row operation $R_{1} = R_{1} - \frac{6}{25} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{6}{25} \cdot 0 & \frac{6}{25} - \frac{6}{25} \cdot 1 \\ 0 & 1 \end{array}]$

解题步骤 4.2

化简 $R_{1}$ 。

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

有限数学 示例