有限数学示例

$x + y - 4 z = 5$ , $x - y - 3 z = 0$ , $\frac{2}{3} x - \frac{7}{3} z = 4$

解题步骤 1

Move variables to the left and constant terms to the right.

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解题步骤 1.1

化简每一项。

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解题步骤 1.1.1

组合 $\frac{2}{3}$ 和 $x$ 。

$x + y - 4 z = 5$

$x - y - 3 z = 0$

$\frac{2 x}{3} - \frac{7}{3} z = 4$

解题步骤 1.1.2

组合 $z$ 和 $\frac{7}{3}$ 。

$x + y - 4 z = 5$

$x - y - 3 z = 0$

$\frac{2 x}{3} - \frac{z \cdot 7}{3} = 4$

解题步骤 1.1.3

将 $7$ 移到 $z$ 的左侧。

$x + y - 4 z = 5$

$x - y - 3 z = 0$

$\frac{2 x}{3} - \frac{7 z}{3} = 4$

$x + y - 4 z = 5$

$x - y - 3 z = 0$

$\frac{2 x}{3} - \frac{7 z}{3} = 4$

解题步骤 1.2

重新排序项。

$x + y - 4 z = 5$

$x - y - 3 z = 0$

$\frac{2}{3} x - (\frac{7}{3} z) = 4$

解题步骤 1.3

去掉圆括号。

$x + y - 4 z = 5$

$x - y - 3 z = 0$

$\frac{2}{3} x - \frac{7}{3} z = 4$

$x + y - 4 z = 5$

$x - y - 3 z = 0$

$\frac{2}{3} x - \frac{7}{3} z = 4$

解题步骤 2

Write the system as a matrix.

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

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

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

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解题步骤 3.1.1

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

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

解题步骤 3.1.2

化简 $R_{2}$ 。

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

解题步骤 3.2

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

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解题步骤 3.2.1

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

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

解题步骤 3.2.2

化简 $R_{3}$ 。

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

解题步骤 3.3

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

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解题步骤 3.3.1

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

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

解题步骤 3.3.2

化简 $R_{2}$ 。

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

解题步骤 3.4

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

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解题步骤 3.4.1

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

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

解题步骤 3.4.2

化简 $R_{3}$ 。

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

解题步骤 3.5

Multiply each element of $R_{3}$ by $\frac{3}{7}$ to make the entry at $3, 4$ a $1$ .

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解题步骤 3.5.1

Multiply each element of $R_{3}$ by $\frac{3}{7}$ to make the entry at $3, 4$ a $1$ .

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

解题步骤 3.5.2

化简 $R_{3}$ 。

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

解题步骤 3.6

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

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解题步骤 3.6.1

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

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

解题步骤 3.6.2

化简 $R_{2}$ 。

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

解题步骤 3.7

Perform the row operation $R_{1} = R_{1} - 5 R_{3}$ to make the entry at $1, 4$ a $0$ .

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解题步骤 3.7.1

Perform the row operation $R_{1} = R_{1} - 5 R_{3}$ to make the entry at $1, 4$ a $0$ .

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

解题步骤 3.7.2

化简 $R_{1}$ 。

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

解题步骤 3.8

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

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解题步骤 3.8.1

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

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

解题步骤 3.8.2

化简 $R_{1}$ 。

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

解题步骤 4

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

$x - \frac{7}{2} z = 0$

$y - \frac{1}{2} z = 0$

$0 = 1$

解题步骤 5

The system is inconsistent so there is no solution.

$无解$

有限数学 示例

有限数学示例