有限数学示例

$5 x + 2 y + 2 z = 11$ , $5 x + 3 y + 2 z = 12$ , $5 x + 3 y + 3 z = 14$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 5 & 2 & 2 & 11 \\ 5 & 3 & 2 & 12 \\ 5 & 3 & 3 & 14 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

点击获取更多步骤...

解题步骤 2.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

点击获取更多步骤...

解题步骤 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{2}{5} & \frac{2}{5} & \frac{11}{5} \\ 5 & 3 & 2 & 12 \\ 5 & 3 & 3 & 14 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{2}{5} & \frac{11}{5} \\ 5 & 3 & 2 & 12 \\ 5 & 3 & 3 & 14 \end{array}]$

解题步骤 2.2

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

点击获取更多步骤...

解题步骤 2.2.1

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

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{2}{5} & \frac{11}{5} \\ 5 - 5 \cdot 1 & 3 - 5 (\frac{2}{5}) & 2 - 5 (\frac{2}{5}) & 12 - 5 (\frac{11}{5}) \\ 5 & 3 & 3 & 14 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{2}{5} & \frac{11}{5} \\ 0 & 1 & 0 & 1 \\ 5 & 3 & 3 & 14 \end{array}]$

解题步骤 2.3

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

点击获取更多步骤...

解题步骤 2.3.1

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{2}{5} & \frac{11}{5} \\ 0 & 1 & 0 & 1 \\ 5 - 5 \cdot 1 & 3 - 5 (\frac{2}{5}) & 3 - 5 (\frac{2}{5}) & 14 - 5 (\frac{11}{5}) \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{2}{5} & \frac{11}{5} \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 3 \end{array}]$

解题步骤 2.4

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

点击获取更多步骤...

解题步骤 2.4.1

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

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

解题步骤 2.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{2}{5} & \frac{11}{5} \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}]$

解题步骤 2.5

Perform the row operation $R_{1} = R_{1} - \frac{2}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

点击获取更多步骤...

解题步骤 2.5.1

Perform the row operation $R_{1} = R_{1} - \frac{2}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{2}{5} \cdot 0 & \frac{2}{5} - \frac{2}{5} \cdot 0 & \frac{2}{5} - \frac{2}{5} \cdot 1 & \frac{11}{5} - \frac{2}{5} \cdot 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}]$

解题步骤 2.5.2

化简 $R_{1}$ 。

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

解题步骤 2.6

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

点击获取更多步骤...

解题步骤 2.6.1

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

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

解题步骤 2.6.2

化简 $R_{1}$ 。

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

解题步骤 3

Use the result matrix to declare the final solution to the system of equations.

$x = 1$

$y = 1$

$z = 2$

解题步骤 4

The solution is the set of ordered pairs that make the system true.

$(1, 1, 2)$

有限数学 示例

有限数学示例