线性代数示例

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

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 $1$ .

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

解题步骤 1.1.2

化简 $R_{1}$ 。

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

解题步骤 1.2

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

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解题步骤 1.2.1

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

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

解题步骤 1.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{4} \\ 0 & \frac{29}{4} \end{array}]$

解题步骤 1.3

Multiply each element of $R_{2}$ by $\frac{4}{29}$ 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 $\frac{4}{29}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{4} \\ \frac{4}{29} \cdot 0 & \frac{4}{29} \cdot \frac{29}{4} \end{array}]$

解题步骤 1.3.2

化简 $R_{2}$ 。

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

解题步骤 1.4

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

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解题步骤 1.4.1

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

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

解题步骤 1.4.2

化简 $R_{1}$ 。

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

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

Pivot Columns: $1$ and $2$

解题步骤 3

The rank is the number of pivot columns.

$2$

线性代数 示例

线性代数示例