有限数学示例

$120 x + 150 y + 180 x = 38700$ , $180 x + 80 y + 80 z = 42200$ , $5 x + 6 y + 4 z = 900$

解题步骤 1

将 $120 x$ 和 $180 x$ 相加。

$300 x + 150 y = 38700$

$180 x + 80 y + 80 z = 42200$

$5 x + 6 y + 4 z = 900$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 300 & 150 & 0 & 38700 \\ 180 & 80 & 80 & 42200 \\ 5 & 6 & 4 & 900 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

Multiply each element of $R_{1}$ by $\frac{1}{300}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{300}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{300}{300} & \frac{150}{300} & \frac{0}{300} & \frac{38700}{300} \\ 180 & 80 & 80 & 42200 \\ 5 & 6 & 4 & 900 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 180 & 80 & 80 & 42200 \\ 5 & 6 & 4 & 900 \end{array}]$

解题步骤 3.2

Perform the row operation $R_{2} = R_{2} - 180 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 3.2.1

Perform the row operation $R_{2} = R_{2} - 180 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 180 - 180 \cdot 1 & 80 - 180 (\frac{1}{2}) & 80 - 180 \cdot 0 & 42200 - 180 \cdot 129 \\ 5 & 6 & 4 & 900 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 & - 10 & 80 & 18980 \\ 5 & 6 & 4 & 900 \end{array}]$

解题步骤 3.3

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

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解题步骤 3.3.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 & - 10 & 80 & 18980 \\ 5 - 5 \cdot 1 & 6 - 5 (\frac{1}{2}) & 4 - 5 \cdot 0 & 900 - 5 \cdot 129 \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 & - 10 & 80 & 18980 \\ 0 & \frac{7}{2} & 4 & 255 \end{array}]$

解题步骤 3.4

Multiply each element of $R_{2}$ by $- \frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.4.1

Multiply each element of $R_{2}$ by $- \frac{1}{10}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ - \frac{1}{10} \cdot 0 & - \frac{1}{10} \cdot - 10 & - \frac{1}{10} \cdot 80 & - \frac{1}{10} \cdot 18980 \\ 0 & \frac{7}{2} & 4 & 255 \end{array}]$

解题步骤 3.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 & 1 & - 8 & - 1898 \\ 0 & \frac{7}{2} & 4 & 255 \end{array}]$

解题步骤 3.5

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

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解题步骤 3.5.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 & 1 & - 8 & - 1898 \\ 0 - \frac{7}{2} \cdot 0 & \frac{7}{2} - \frac{7}{2} \cdot 1 & 4 - \frac{7}{2} \cdot - 8 & 255 - \frac{7}{2} \cdot - 1898 \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 & 1 & - 8 & - 1898 \\ 0 & 0 & 32 & 6898 \end{array}]$

解题步骤 3.6

Multiply each element of $R_{3}$ by $\frac{1}{32}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 3.6.1

Multiply each element of $R_{3}$ by $\frac{1}{32}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 & 1 & - 8 & - 1898 \\ \frac{0}{32} & \frac{0}{32} & \frac{32}{32} & \frac{6898}{32} \end{array}]$

解题步骤 3.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 & 1 & - 8 & - 1898 \\ 0 & 0 & 1 & \frac{3449}{16} \end{array}]$

解题步骤 3.7

Perform the row operation $R_{2} = R_{2} + 8 R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 3.7.1

Perform the row operation $R_{2} = R_{2} + 8 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 + 8 \cdot 0 & 1 + 8 \cdot 0 & - 8 + 8 \cdot 1 & - 1898 + 8 (\frac{3449}{16}) \\ 0 & 0 & 1 & \frac{3449}{16} \end{array}]$

解题步骤 3.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{1}{2} & 0 & 129 \\ 0 & 1 & 0 & - \frac{347}{2} \\ 0 & 0 & 1 & \frac{3449}{16} \end{array}]$

解题步骤 3.8

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 3.8.1

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

$[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot 0 & 129 - \frac{1}{2} (- \frac{347}{2}) \\ 0 & 1 & 0 & - \frac{347}{2} \\ 0 & 0 & 1 & \frac{3449}{16} \end{array}]$

解题步骤 3.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & \frac{863}{4} \\ 0 & 1 & 0 & - \frac{347}{2} \\ 0 & 0 & 1 & \frac{3449}{16} \end{array}]$

解题步骤 4

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

$x = \frac{863}{4}$

$y = - \frac{347}{2}$

$z = \frac{3449}{16}$

解题步骤 5

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

$(\frac{863}{4}, - \frac{347}{2}, \frac{3449}{16})$

有限数学 示例

有限数学示例