示例

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

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 $\frac{1}{4}$ 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 $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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

解题步骤 1.1.2

化简 $R_{1}$ 。

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

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 $0$ .

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

解题步骤 1.2.2

化简 $R_{2}$ 。

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

解题步骤 1.3

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 1.3.1

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

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

解题步骤 1.3.2

化简 $R_{2}$ 。

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

解题步骤 1.4

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

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解题步骤 1.4.1

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

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

解题步骤 1.4.2

化简 $R_{1}$ 。

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

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 $2$

解题步骤 3

The rank is the number of pivot columns.

$2$

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