有限数学示例

$7 x - y - 6 z = 8$ , $2 x + y - z = 5$ , $5 x + y - 4 z = 9$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 7 & - 1 & - 6 & 8 \\ 2 & 1 & - 1 & 5 \\ 5 & 1 & - 4 & 9 \end{array}]$

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 $\frac{1}{7}$ 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 $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{7}{7} & \frac{- 1}{7} & \frac{- 6}{7} & \frac{8}{7} \\ 2 & 1 & - 1 & 5 \\ 5 & 1 & - 4 & 9 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 2 & 1 & - 1 & 5 \\ 5 & 1 & - 4 & 9 \end{array}]$

解题步骤 2.2

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 2.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 2 - 2 \cdot 1 & 1 - 2 (- \frac{1}{7}) & - 1 - 2 (- \frac{6}{7}) & 5 - 2 (\frac{8}{7}) \\ 5 & 1 & - 4 & 9 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 & \frac{9}{7} & \frac{5}{7} & \frac{19}{7} \\ 5 & 1 & - 4 & 9 \end{array}]$

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 $0$ .

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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 $0$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 & \frac{9}{7} & \frac{5}{7} & \frac{19}{7} \\ 5 - 5 \cdot 1 & 1 - 5 (- \frac{1}{7}) & - 4 - 5 (- \frac{6}{7}) & 9 - 5 (\frac{8}{7}) \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 & \frac{9}{7} & \frac{5}{7} & \frac{19}{7} \\ 0 & \frac{12}{7} & \frac{2}{7} & \frac{23}{7} \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $\frac{7}{9}$ 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 $\frac{7}{9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ \frac{7}{9} \cdot 0 & \frac{7}{9} \cdot \frac{9}{7} & \frac{7}{9} \cdot \frac{5}{7} & \frac{7}{9} \cdot \frac{19}{7} \\ 0 & \frac{12}{7} & \frac{2}{7} & \frac{23}{7} \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 & 1 & \frac{5}{9} & \frac{19}{9} \\ 0 & \frac{12}{7} & \frac{2}{7} & \frac{23}{7} \end{array}]$

解题步骤 2.5

Perform the row operation $R_{3} = R_{3} - \frac{12}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{12}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 & 1 & \frac{5}{9} & \frac{19}{9} \\ 0 - \frac{12}{7} \cdot 0 & \frac{12}{7} - \frac{12}{7} \cdot 1 & \frac{2}{7} - \frac{12}{7} \cdot \frac{5}{9} & \frac{23}{7} - \frac{12}{7} \cdot \frac{19}{9} \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 & 1 & \frac{5}{9} & \frac{19}{9} \\ 0 & 0 & - \frac{2}{3} & - \frac{1}{3} \end{array}]$

解题步骤 2.6

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

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解题步骤 2.6.1

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

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 & 1 & \frac{5}{9} & \frac{19}{9} \\ - \frac{3}{2} \cdot 0 & - \frac{3}{2} \cdot 0 & - \frac{3}{2} (- \frac{2}{3}) & - \frac{3}{2} (- \frac{1}{3}) \end{array}]$

解题步骤 2.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 & 1 & \frac{5}{9} & \frac{19}{9} \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

解题步骤 2.7

Perform the row operation $R_{2} = R_{2} - \frac{5}{9} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{2} = R_{2} - \frac{5}{9} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 - \frac{5}{9} \cdot 0 & 1 - \frac{5}{9} \cdot 0 & \frac{5}{9} - \frac{5}{9} \cdot 1 & \frac{19}{9} - \frac{5}{9} \cdot \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

解题步骤 2.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{6}{7} & \frac{8}{7} \\ 0 & 1 & 0 & \frac{11}{6} \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

解题步骤 2.8

Perform the row operation $R_{1} = R_{1} + \frac{6}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.8.1

Perform the row operation $R_{1} = R_{1} + \frac{6}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{6}{7} \cdot 0 & - \frac{1}{7} + \frac{6}{7} \cdot 0 & - \frac{6}{7} + \frac{6}{7} \cdot 1 & \frac{8}{7} + \frac{6}{7} \cdot \frac{1}{2} \\ 0 & 1 & 0 & \frac{11}{6} \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{7} & 0 & \frac{11}{7} \\ 0 & 1 & 0 & \frac{11}{6} \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

解题步骤 2.9

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

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解题步骤 2.9.1

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

$[\begin{array}{c} 1 + \frac{1}{7} \cdot 0 & - \frac{1}{7} + \frac{1}{7} \cdot 1 & 0 + \frac{1}{7} \cdot 0 & \frac{11}{7} + \frac{1}{7} \cdot \frac{11}{6} \\ 0 & 1 & 0 & \frac{11}{6} \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

解题步骤 2.9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & \frac{11}{6} \\ 0 & 1 & 0 & \frac{11}{6} \\ 0 & 0 & 1 & \frac{1}{2} \end{array}]$

解题步骤 3

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

$x = \frac{11}{6}$

$y = \frac{11}{6}$

$z = \frac{1}{2}$

解题步骤 4

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

$(\frac{11}{6}, \frac{11}{6}, \frac{1}{2})$

有限数学 示例

有限数学示例