有限数学示例

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

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

将所有包含变量的项移到等式左边。

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解题步骤 1.1.1

从等式两边同时减去 $3 x$ 。

$y - 3 x = z - 2$

$z = 3 x + 4$

$y = 5 z$

解题步骤 1.1.2

从等式两边同时减去 $z$ 。

$y - 3 x - z = - 2$

$z = 3 x + 4$

$y = 5 z$

$y - 3 x - z = - 2$

$z = 3 x + 4$

$y = 5 z$

解题步骤 1.2

将 $y$ 和 $- 3 x$ 重新排序。

$- 3 x + y - z = - 2$

$z = 3 x + 4$

$y = 5 z$

解题步骤 1.3

从等式两边同时减去 $3 x$ 。

$- 3 x + y - z = - 2$

$z - 3 x = 4$

$y = 5 z$

解题步骤 1.4

将 $z$ 和 $- 3 x$ 重新排序。

$- 3 x + y - z = - 2$

$- 3 x + z = 4$

$y = 5 z$

解题步骤 1.5

从等式两边同时减去 $5 z$ 。

$- 3 x + y - z = - 2$

$- 3 x + z = 4$

$y - 5 z = 0$

$- 3 x + y - z = - 2$

$- 3 x + z = 4$

$y - 5 z = 0$

解题步骤 2

Write the system as a matrix.

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

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

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

解题步骤 3.1.2

化简 $R_{1}$ 。

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

解题步骤 3.2

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

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解题步骤 3.2.1

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

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

解题步骤 3.2.2

化简 $R_{2}$ 。

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

解题步骤 3.3

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

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ - 0 & - - 1 & - 1 \cdot 2 & - 1 \cdot 6 \\ 0 & 1 & - 5 & 0 \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

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

解题步骤 3.4

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

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解题步骤 3.4.1

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

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - 2 & - 6 \\ 0 - 0 & 1 - 1 & - 5 + 2 & 0 + 6 \end{array}]$

解题步骤 3.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - 2 & - 6 \\ 0 & 0 & - 3 & 6 \end{array}]$

解题步骤 3.5

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

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解题步骤 3.5.1

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

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - 2 & - 6 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 6 \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - 2 & - 6 \\ 0 & 0 & 1 & - 2 \end{array}]$

解题步骤 3.6

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

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解题步骤 3.6.1

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

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ 0 + 2 \cdot 0 & 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 6 + 2 \cdot - 2 \\ 0 & 0 & 1 & - 2 \end{array}]$

解题步骤 3.6.2

化简 $R_{2}$ 。

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

解题步骤 3.7

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

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解题步骤 3.7.1

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

$[\begin{array}{c} 1 - \frac{1}{3} \cdot 0 & - \frac{1}{3} - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & \frac{2}{3} - \frac{1}{3} \cdot - 2 \\ 0 & 1 & 0 & - 10 \\ 0 & 0 & 1 & - 2 \end{array}]$

解题步骤 3.7.2

化简 $R_{1}$ 。

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

解题步骤 3.8

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

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解题步骤 3.8.1

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

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

解题步骤 3.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & - 2 \\ 0 & 1 & 0 & - 10 \\ 0 & 0 & 1 & - 2 \end{array}]$

解题步骤 4

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

$x = - 2$

$y = - 10$

$z = - 2$

解题步骤 5

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

$(- 2, - 10, - 2)$

有限数学 示例

有限数学示例