有限数学示例

$[\begin{array}{c} 1 & 2 \\ 2 & 1 \end{array}]$

解题步骤 1

求行简化阶梯形矩阵。

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解题步骤 1.1

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 1.1.1

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

$[\begin{array}{c} 1 & 2 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot 2 \end{array}]$

解题步骤 1.1.2

化简 $R_{2}$ 。

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

解题步骤 1.2

Multiply each element of $R_{2}$ by $- \frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 1.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

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

解题步骤 1.2.2

化简 $R_{2}$ 。

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

解题步骤 1.3

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

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解题步骤 1.3.1

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

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

解题步骤 1.3.2

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 $2$

有限数学 示例