有限数学示例

$[\begin{array}{c} 7 & 5 & - 1 & 0 \\ 9 & - 1 & 2 & 1 \\ 16 & 4 & 1 & 1 \end{array}]$

解题步骤 1

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

点击获取更多步骤...

解题步骤 1.1

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

$[\begin{array}{c} \frac{7}{7} & \frac{5}{7} & \frac{- 1}{7} & \frac{0}{7} \\ 9 & - 1 & 2 & 1 \\ 16 & 4 & 1 & 1 \end{array}]$

解题步骤 1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{5}{7} & - \frac{1}{7} & 0 \\ 9 & - 1 & 2 & 1 \\ 16 & 4 & 1 & 1 \end{array}]$

解题步骤 2

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

点击获取更多步骤...

解题步骤 2.1

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{7} & - \frac{1}{7} & 0 \\ 9 - 9 \cdot 1 & - 1 - 9 (\frac{5}{7}) & 2 - 9 (- \frac{1}{7}) & 1 - 9 \cdot 0 \\ 16 & 4 & 1 & 1 \end{array}]$

解题步骤 2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{7} & - \frac{1}{7} & 0 \\ 0 & - \frac{52}{7} & \frac{23}{7} & 1 \\ 16 & 4 & 1 & 1 \end{array}]$

解题步骤 3

Perform the row operation $R_{3} = R_{3} - 16 R_{1}$ to make the entry at $3, 1$ a $0$ .

点击获取更多步骤...

解题步骤 3.1

Perform the row operation $R_{3} = R_{3} - 16 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{7} & - \frac{1}{7} & 0 \\ 0 & - \frac{52}{7} & \frac{23}{7} & 1 \\ 16 - 16 \cdot 1 & 4 - 16 (\frac{5}{7}) & 1 - 16 (- \frac{1}{7}) & 1 - 16 \cdot 0 \end{array}]$

解题步骤 3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{5}{7} & - \frac{1}{7} & 0 \\ 0 & - \frac{52}{7} & \frac{23}{7} & 1 \\ 0 & - \frac{52}{7} & \frac{23}{7} & 1 \end{array}]$

解题步骤 4

Multiply each element of $R_{2}$ by $- \frac{7}{52}$ to make the entry at $2, 2$ a $1$ .

点击获取更多步骤...

解题步骤 4.1

Multiply each element of $R_{2}$ by $- \frac{7}{52}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{7} & - \frac{1}{7} & 0 \\ - \frac{7}{52} \cdot 0 & - \frac{7}{52} (- \frac{52}{7}) & - \frac{7}{52} \cdot \frac{23}{7} & - \frac{7}{52} \cdot 1 \\ 0 & - \frac{52}{7} & \frac{23}{7} & 1 \end{array}]$

解题步骤 4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{7} & - \frac{1}{7} & 0 \\ 0 & 1 & - \frac{23}{52} & - \frac{7}{52} \\ 0 & - \frac{52}{7} & \frac{23}{7} & 1 \end{array}]$

解题步骤 5

Perform the row operation $R_{3} = R_{3} + \frac{52}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

点击获取更多步骤...

解题步骤 5.1

Perform the row operation $R_{3} = R_{3} + \frac{52}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{7} & - \frac{1}{7} & 0 \\ 0 & 1 & - \frac{23}{52} & - \frac{7}{52} \\ 0 + \frac{52}{7} \cdot 0 & - \frac{52}{7} + \frac{52}{7} \cdot 1 & \frac{23}{7} + \frac{52}{7} (- \frac{23}{52}) & 1 + \frac{52}{7} (- \frac{7}{52}) \end{array}]$

解题步骤 5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{5}{7} & - \frac{1}{7} & 0 \\ 0 & 1 & - \frac{23}{52} & - \frac{7}{52} \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 6

Perform the row operation $R_{1} = R_{1} - \frac{5}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

点击获取更多步骤...

解题步骤 6.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{5}{7} \cdot 0 & \frac{5}{7} - \frac{5}{7} \cdot 1 & - \frac{1}{7} - \frac{5}{7} (- \frac{23}{52}) & 0 - \frac{5}{7} (- \frac{7}{52}) \\ 0 & 1 & - \frac{23}{52} & - \frac{7}{52} \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 6.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{9}{52} & \frac{5}{52} \\ 0 & 1 & - \frac{23}{52} & - \frac{7}{52} \\ 0 & 0 & 0 & 0 \end{array}]$

有限数学 示例

有限数学示例