有限数学示例

$3 x - 2 y + 4 z = 11$ , $2 x - 4 y + 5 z = 4$ , $x + 2 y - z = 10$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 3 & - 2 & 4 & 11 \\ 2 & - 4 & 5 & 4 \\ 1 & 2 & - 1 & 10 \end{array}]$

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 $\frac{1}{3}$ 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 $\frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{- 2}{3} & \frac{4}{3} & \frac{11}{3} \\ 2 & - 4 & 5 & 4 \\ 1 & 2 & - 1 & 10 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 2 & - 4 & 5 & 4 \\ 1 & 2 & - 1 & 10 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 2 - 2 \cdot 1 & - 4 - 2 (- \frac{2}{3}) & 5 - 2 (\frac{4}{3}) & 4 - 2 (\frac{11}{3}) \\ 1 & 2 & - 1 & 10 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 & - \frac{8}{3} & \frac{7}{3} & - \frac{10}{3} \\ 1 & 2 & - 1 & 10 \end{array}]$

解题步骤 2.3

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

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解题步骤 2.3.1

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

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 & - \frac{8}{3} & \frac{7}{3} & - \frac{10}{3} \\ 1 - 1 & 2 + \frac{2}{3} & - 1 - \frac{4}{3} & 10 - \frac{11}{3} \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 & - \frac{8}{3} & \frac{7}{3} & - \frac{10}{3} \\ 0 & \frac{8}{3} & - \frac{7}{3} & \frac{19}{3} \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $- \frac{3}{8}$ 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 $- \frac{3}{8}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ - \frac{3}{8} \cdot 0 & - \frac{3}{8} (- \frac{8}{3}) & - \frac{3}{8} \cdot \frac{7}{3} & - \frac{3}{8} (- \frac{10}{3}) \\ 0 & \frac{8}{3} & - \frac{7}{3} & \frac{19}{3} \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 & 1 & - \frac{7}{8} & \frac{5}{4} \\ 0 & \frac{8}{3} & - \frac{7}{3} & \frac{19}{3} \end{array}]$

解题步骤 2.5

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

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解题步骤 2.5.1

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

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 & 1 & - \frac{7}{8} & \frac{5}{4} \\ 0 - \frac{8}{3} \cdot 0 & \frac{8}{3} - \frac{8}{3} \cdot 1 & - \frac{7}{3} - \frac{8}{3} (- \frac{7}{8}) & \frac{19}{3} - \frac{8}{3} \cdot \frac{5}{4} \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 & 1 & - \frac{7}{8} & \frac{5}{4} \\ 0 & 0 & 0 & 3 \end{array}]$

解题步骤 2.6

Multiply each element of $R_{3}$ by $\frac{1}{3}$ to make the entry at $3, 4$ a $1$ .

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解题步骤 2.6.1

Multiply each element of $R_{3}$ by $\frac{1}{3}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 & 1 & - \frac{7}{8} & \frac{5}{4} \\ \frac{0}{3} & \frac{0}{3} & \frac{0}{3} & \frac{3}{3} \end{array}]$

解题步骤 2.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 & 1 & - \frac{7}{8} & \frac{5}{4} \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.7

Perform the row operation $R_{2} = R_{2} - \frac{5}{4} R_{3}$ to make the entry at $2, 4$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{2} = R_{2} - \frac{5}{4} R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 - \frac{5}{4} \cdot 0 & 1 - \frac{5}{4} \cdot 0 & - \frac{7}{8} - \frac{5}{4} \cdot 0 & \frac{5}{4} - \frac{5}{4} \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & \frac{11}{3} \\ 0 & 1 & - \frac{7}{8} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.8

Perform the row operation $R_{1} = R_{1} - \frac{11}{3} R_{3}$ to make the entry at $1, 4$ a $0$ .

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解题步骤 2.8.1

Perform the row operation $R_{1} = R_{1} - \frac{11}{3} R_{3}$ to make the entry at $1, 4$ a $0$ .

$[\begin{array}{c} 1 - \frac{11}{3} \cdot 0 & - \frac{2}{3} - \frac{11}{3} \cdot 0 & \frac{4}{3} - \frac{11}{3} \cdot 0 & \frac{11}{3} - \frac{11}{3} \cdot 1 \\ 0 & 1 & - \frac{7}{8} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{4}{3} & 0 \\ 0 & 1 & - \frac{7}{8} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.9

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

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解题步骤 2.9.1

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

$[\begin{array}{c} 1 + \frac{2}{3} \cdot 0 & - \frac{2}{3} + \frac{2}{3} \cdot 1 & \frac{4}{3} + \frac{2}{3} (- \frac{7}{8}) & 0 + \frac{2}{3} \cdot 0 \\ 0 & 1 & - \frac{7}{8} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 2.9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & 0 \\ 0 & 1 & - \frac{7}{8} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

解题步骤 3

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

$x + \frac{3}{4} z = 0$

$y - \frac{7}{8} z = 0$

$0 = 1$

解题步骤 4

The system is inconsistent so there is no solution.

$无解$

有限数学 示例

有限数学示例