线性代数示例

$[\begin{array}{c} 2 & 6 & 7 & - 8 & - 9 & 6 & 5 & 0 & 1 & 11 \\ 1 & 5 & 7 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 8 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{6}{2} & \frac{7}{2} & \frac{- 8}{2} & \frac{- 9}{2} & \frac{6}{2} & \frac{5}{2} & \frac{0}{2} & \frac{1}{2} & \frac{11}{2} \\ 1 & 5 & 7 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 8 \end{array}]$

解题步骤 1.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 3 & \frac{7}{2} & - 4 & - \frac{9}{2} & 3 & \frac{5}{2} & 0 & \frac{1}{2} & \frac{11}{2} \\ 1 & 5 & 7 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 8 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 3 & \frac{7}{2} & - 4 & - \frac{9}{2} & 3 & \frac{5}{2} & 0 & \frac{1}{2} & \frac{11}{2} \\ 1 - 1 & 5 - 3 & 7 - \frac{7}{2} & 0 + 4 & 0 + \frac{9}{2} & 0 - 3 & 1 - \frac{5}{2} & - 2 - 0 & - 3 - \frac{1}{2} & - 8 - \frac{11}{2} \end{array}]$

解题步骤 1.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 3 & \frac{7}{2} & - 4 & - \frac{9}{2} & 3 & \frac{5}{2} & 0 & \frac{1}{2} & \frac{11}{2} \\ 0 & 2 & \frac{7}{2} & 4 & \frac{9}{2} & - 3 & - \frac{3}{2} & - 2 & - \frac{7}{2} & - \frac{27}{2} \end{array}]$

解题步骤 1.3

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

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解题步骤 1.3.1

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

$[\begin{array}{c} 1 & 3 & \frac{7}{2} & - 4 & - \frac{9}{2} & 3 & \frac{5}{2} & 0 & \frac{1}{2} & \frac{11}{2} \\ \frac{0}{2} & \frac{2}{2} & \frac{\frac{7}{2}}{2} & \frac{4}{2} & \frac{\frac{9}{2}}{2} & \frac{- 3}{2} & \frac{- \frac{3}{2}}{2} & \frac{- 2}{2} & \frac{- \frac{7}{2}}{2} & \frac{- \frac{27}{2}}{2} \end{array}]$

解题步骤 1.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 3 & \frac{7}{2} & - 4 & - \frac{9}{2} & 3 & \frac{5}{2} & 0 & \frac{1}{2} & \frac{11}{2} \\ 0 & 1 & \frac{7}{4} & 2 & \frac{9}{4} & - \frac{3}{2} & - \frac{3}{4} & - 1 & - \frac{7}{4} & - \frac{27}{4} \end{array}]$

解题步骤 1.4

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

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解题步骤 1.4.1

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

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & \frac{7}{2} - 3 (\frac{7}{4}) & - 4 - 3 \cdot 2 & - \frac{9}{2} - 3 (\frac{9}{4}) & 3 - 3 (- \frac{3}{2}) & \frac{5}{2} - 3 (- \frac{3}{4}) & 0 - 3 \cdot - 1 & \frac{1}{2} - 3 (- \frac{7}{4}) & \frac{11}{2} - 3 (- \frac{27}{4}) \\ 0 & 1 & \frac{7}{4} & 2 & \frac{9}{4} & - \frac{3}{2} & - \frac{3}{4} & - 1 & - \frac{7}{4} & - \frac{27}{4} \end{array}]$

解题步骤 1.4.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{7}{4} & - 10 & - \frac{45}{4} & \frac{15}{2} & \frac{19}{4} & 3 & \frac{23}{4} & \frac{103}{4} \\ 0 & 1 & \frac{7}{4} & 2 & \frac{9}{4} & - \frac{3}{2} & - \frac{3}{4} & - 1 & - \frac{7}{4} & - \frac{27}{4} \end{array}]$

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: $a_{11}$ and $a_{22}$

Pivot Columns: $1$ and $2$

线性代数 示例

线性代数示例