线性代数示例

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

解题步骤 1

求行简化阶梯形矩阵。

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解题步骤 1.1

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

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解题步骤 1.1.1

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

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

解题步骤 1.1.2

化简 $R_{1}$ 。

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

解题步骤 1.2

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

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解题步骤 1.2.1

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

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

解题步骤 1.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 5 \\ 0 & \frac{7}{2} & \frac{5}{2} & 5 \\ - 4 & - 6 & 2 & - 2 \end{array}]$

解题步骤 1.3

Perform the row operation $R_{3} = R_{3} + 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 1.3.1

Perform the row operation $R_{3} = R_{3} + 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 5 \\ 0 & \frac{7}{2} & \frac{5}{2} & 5 \\ - 4 + 4 \cdot 1 & - 6 + 4 (\frac{3}{2}) & 2 + 4 (- \frac{1}{2}) & - 2 + 4 \cdot 5 \end{array}]$

解题步骤 1.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 5 \\ 0 & \frac{7}{2} & \frac{5}{2} & 5 \\ 0 & 0 & 0 & 18 \end{array}]$

解题步骤 1.4

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

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解题步骤 1.4.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 5 \\ \frac{2}{7} \cdot 0 & \frac{2}{7} \cdot \frac{7}{2} & \frac{2}{7} \cdot \frac{5}{2} & \frac{2}{7} \cdot 5 \\ 0 & 0 & 0 & 18 \end{array}]$

解题步骤 1.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 5 \\ 0 & 1 & \frac{5}{7} & \frac{10}{7} \\ 0 & 0 & 0 & 18 \end{array}]$

解题步骤 1.5

Multiply each element of $R_{3}$ by $\frac{1}{18}$ to make the entry at $3, 4$ a $1$ .

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解题步骤 1.5.1

Multiply each element of $R_{3}$ by $\frac{1}{18}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 5 \\ 0 & 1 & \frac{5}{7} & \frac{10}{7} \\ \frac{0}{18} & \frac{0}{18} & \frac{0}{18} & \frac{18}{18} \end{array}]$

解题步骤 1.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 5 \\ 0 & 1 & \frac{5}{7} & \frac{10}{7} \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 1.6

Perform the row operation $R_{2} = R_{2} - \frac{10}{7} R_{3}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 1.6.1

Perform the row operation $R_{2} = R_{2} - \frac{10}{7} R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 5 \\ 0 - \frac{10}{7} \cdot 0 & 1 - \frac{10}{7} \cdot 0 & \frac{5}{7} - \frac{10}{7} \cdot 0 & \frac{10}{7} - \frac{10}{7} \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 1.6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 5 \\ 0 & 1 & \frac{5}{7} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 1.7

Perform the row operation $R_{1} = R_{1} - 5 R_{3}$ to make the entry at $1, 4$ a $0$ .

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解题步骤 1.7.1

Perform the row operation $R_{1} = R_{1} - 5 R_{3}$ to make the entry at $1, 4$ a $0$ .

$[\begin{array}{c} 1 - 5 \cdot 0 & \frac{3}{2} - 5 \cdot 0 & - \frac{1}{2} - 5 \cdot 0 & 5 - 5 \cdot 1 \\ 0 & 1 & \frac{5}{7} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 1.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{3}{2} & - \frac{1}{2} & 0 \\ 0 & 1 & \frac{5}{7} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 1.8

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 1.8.1

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

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & - \frac{1}{2} - \frac{3}{2} \cdot \frac{5}{7} & 0 - \frac{3}{2} \cdot 0 \\ 0 & 1 & \frac{5}{7} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 1.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{11}{7} & 0 \\ 0 & 1 & \frac{5}{7} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2

The pivot positions are the locations with the leading $1$ in each row. The pivot columns are the columns that have a pivot position.

Pivot Positions: $a_{11}, a_{22},$ and $a_{34}$

Pivot Columns: $1, 2,$ and $4$

线性代数 示例

线性代数示例