线性代数示例

$[\begin{array}{c} 12 & - 15 & 9 \\ - 2 & 3 & 4 \\ 8 & - 10 & 6 \end{array}]$

解题步骤 1

Multiply each element of $R_{1}$ by $\frac{1}{12}$ 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 $\frac{1}{12}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{12}{12} & \frac{- 15}{12} & \frac{9}{12} \\ - 2 & 3 & 4 \\ 8 & - 10 & 6 \end{array}]$

解题步骤 1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{5}{4} & \frac{3}{4} \\ - 2 & 3 & 4 \\ 8 & - 10 & 6 \end{array}]$

解题步骤 2

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

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解题步骤 2.1

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

$[\begin{array}{c} 1 & - \frac{5}{4} & \frac{3}{4} \\ - 2 + 2 \cdot 1 & 3 + 2 (- \frac{5}{4}) & 4 + 2 (\frac{3}{4}) \\ 8 & - 10 & 6 \end{array}]$

解题步骤 2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{5}{4} & \frac{3}{4} \\ 0 & \frac{1}{2} & \frac{11}{2} \\ 8 & - 10 & 6 \end{array}]$

解题步骤 3

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 3.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{4} & \frac{3}{4} \\ 0 & \frac{1}{2} & \frac{11}{2} \\ 8 - 8 \cdot 1 & - 10 - 8 (- \frac{5}{4}) & 6 - 8 (\frac{3}{4}) \end{array}]$

解题步骤 3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{5}{4} & \frac{3}{4} \\ 0 & \frac{1}{2} & \frac{11}{2} \\ 0 & 0 & 0 \end{array}]$

解题步骤 4

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

$[\begin{array}{c} 1 & - \frac{5}{4} & \frac{3}{4} \\ 2 \cdot 0 & 2 (\frac{1}{2}) & 2 (\frac{11}{2}) \\ 0 & 0 & 0 \end{array}]$

解题步骤 4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{5}{4} & \frac{3}{4} \\ 0 & 1 & 11 \\ 0 & 0 & 0 \end{array}]$

解题步骤 5

Perform the row operation $R_{1} = R_{1} + \frac{5}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 5.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{5}{4} \cdot 0 & - \frac{5}{4} + \frac{5}{4} \cdot 1 & \frac{3}{4} + \frac{5}{4} \cdot 11 \\ 0 & 1 & 11 \\ 0 & 0 & 0 \end{array}]$

解题步骤 5.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{29}{2} \\ 0 & 1 & 11 \\ 0 & 0 & 0 \end{array}]$

线性代数 示例

线性代数示例