有限数学示例

$[\begin{array}{c} 3 & 1 & 8 \\ 6 & 2 & 16 \\ - 9 & - 3 & - 24 \end{array}]$

解题步骤 1

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

$[\begin{array}{c} \frac{3}{3} & \frac{1}{3} & \frac{8}{3} \\ 6 & 2 & 16 \\ - 9 & - 3 & - 24 \end{array}]$

解题步骤 1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{8}{3} \\ 6 & 2 & 16 \\ - 9 & - 3 & - 24 \end{array}]$

解题步骤 2

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

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解题步骤 2.1

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

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{8}{3} \\ 6 - 6 \cdot 1 & 2 - 6 (\frac{1}{3}) & 16 - 6 (\frac{8}{3}) \\ - 9 & - 3 & - 24 \end{array}]$

解题步骤 2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{8}{3} \\ 0 & 0 & 0 \\ - 9 & - 3 & - 24 \end{array}]$

解题步骤 3

Perform the row operation $R_{3} = R_{3} + 9 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 3.1

Perform the row operation $R_{3} = R_{3} + 9 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{8}{3} \\ 0 & 0 & 0 \\ - 9 + 9 \cdot 1 & - 3 + 9 (\frac{1}{3}) & - 24 + 9 (\frac{8}{3}) \end{array}]$

解题步骤 3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{8}{3} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

有限数学 示例