有限数学示例

$- 21 x - 2 y + z = - 72$ , $12 x + y = 44$ , $- 24 x - 2 y + z = - 84$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} - 21 & - 2 & 1 & - 72 \\ 12 & 1 & 0 & 44 \\ - 24 & - 2 & 1 & - 84 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

Multiply each element of $R_{1}$ by $- \frac{1}{21}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{21}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{21} \cdot - 21 & - \frac{1}{21} \cdot - 2 & - \frac{1}{21} \cdot 1 & - \frac{1}{21} \cdot - 72 \\ 12 & 1 & 0 & 44 \\ - 24 & - 2 & 1 & - 84 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 12 & 1 & 0 & 44 \\ - 24 & - 2 & 1 & - 84 \end{array}]$

解题步骤 2.2

Perform the row operation $R_{2} = R_{2} - 12 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{2} = R_{2} - 12 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 12 - 12 \cdot 1 & 1 - 12 (\frac{2}{21}) & 0 - 12 (- \frac{1}{21}) & 44 - 12 (\frac{24}{7}) \\ - 24 & - 2 & 1 & - 84 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 0 & - \frac{1}{7} & \frac{4}{7} & \frac{20}{7} \\ - 24 & - 2 & 1 & - 84 \end{array}]$

解题步骤 2.3

Perform the row operation $R_{3} = R_{3} + 24 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 2.3.1

Perform the row operation $R_{3} = R_{3} + 24 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 0 & - \frac{1}{7} & \frac{4}{7} & \frac{20}{7} \\ - 24 + 24 \cdot 1 & - 2 + 24 (\frac{2}{21}) & 1 + 24 (- \frac{1}{21}) & - 84 + 24 (\frac{24}{7}) \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 0 & - \frac{1}{7} & \frac{4}{7} & \frac{20}{7} \\ 0 & \frac{2}{7} & - \frac{1}{7} & - \frac{12}{7} \end{array}]$

解题步骤 2.4

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

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解题步骤 2.4.1

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

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ - 7 \cdot 0 & - 7 (- \frac{1}{7}) & - 7 (\frac{4}{7}) & - 7 (\frac{20}{7}) \\ 0 & \frac{2}{7} & - \frac{1}{7} & - \frac{12}{7} \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 0 & 1 & - 4 & - 20 \\ 0 & \frac{2}{7} & - \frac{1}{7} & - \frac{12}{7} \end{array}]$

解题步骤 2.5

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

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解题步骤 2.5.1

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

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 0 & 1 & - 4 & - 20 \\ 0 - \frac{2}{7} \cdot 0 & \frac{2}{7} - \frac{2}{7} \cdot 1 & - \frac{1}{7} - \frac{2}{7} \cdot - 4 & - \frac{12}{7} - \frac{2}{7} \cdot - 20 \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 0 & 1 & - 4 & - 20 \\ 0 & 0 & 1 & 4 \end{array}]$

解题步骤 2.6

Perform the row operation $R_{2} = R_{2} + 4 R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.6.1

Perform the row operation $R_{2} = R_{2} + 4 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 0 + 4 \cdot 0 & 1 + 4 \cdot 0 & - 4 + 4 \cdot 1 & - 20 + 4 \cdot 4 \\ 0 & 0 & 1 & 4 \end{array}]$

解题步骤 2.6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{21} & - \frac{1}{21} & \frac{24}{7} \\ 0 & 1 & 0 & - 4 \\ 0 & 0 & 1 & 4 \end{array}]$

解题步骤 2.7

Perform the row operation $R_{1} = R_{1} + \frac{1}{21} R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{21} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{21} \cdot 0 & \frac{2}{21} + \frac{1}{21} \cdot 0 & - \frac{1}{21} + \frac{1}{21} \cdot 1 & \frac{24}{7} + \frac{1}{21} \cdot 4 \\ 0 & 1 & 0 & - 4 \\ 0 & 0 & 1 & 4 \end{array}]$

解题步骤 2.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{2}{21} & 0 & \frac{76}{21} \\ 0 & 1 & 0 & - 4 \\ 0 & 0 & 1 & 4 \end{array}]$

解题步骤 2.8

Perform the row operation $R_{1} = R_{1} - \frac{2}{21} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.8.1

Perform the row operation $R_{1} = R_{1} - \frac{2}{21} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{2}{21} \cdot 0 & \frac{2}{21} - \frac{2}{21} \cdot 1 & 0 - \frac{2}{21} \cdot 0 & \frac{76}{21} - \frac{2}{21} \cdot - 4 \\ 0 & 1 & 0 & - 4 \\ 0 & 0 & 1 & 4 \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x = 4$

$y = - 4$

$z = 4$

解题步骤 4

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

$(4, - 4, 4)$

有限数学 示例

有限数学示例