有限数学示例

$z = 3$ , $y - 6 x = 8$ , $x = 3 y - 5 z$

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 $y$ 和 $- 6 x$ 重新排序。

$z = 3$

$- 6 x + y = 8$

$x = 3 y - 5 z$

解题步骤 1.2

将所有包含变量的项移到等式左边。

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解题步骤 1.2.1

从等式两边同时减去 $3 y$ 。

$z = 3$

$- 6 x + y = 8$

$x - 3 y = - 5 z$

解题步骤 1.2.2

在等式两边都加上 $5 z$ 。

$z = 3$

$- 6 x + y = 8$

$x - 3 y + 5 z = 0$

$z = 3$

$- 6 x + y = 8$

$x - 3 y + 5 z = 0$

$z = 3$

$- 6 x + y = 8$

$x - 3 y + 5 z = 0$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 0 & 0 & 1 & 3 \\ - 6 & 1 & 0 & 8 \\ 1 & - 3 & 5 & 0 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} - 6 & 1 & 0 & 8 \\ 0 & 0 & 1 & 3 \\ 1 & - 3 & 5 & 0 \end{array}]$

解题步骤 3.2

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

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解题步骤 3.2.1

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot 1 & - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot 8 \\ 0 & 0 & 1 & 3 \\ 1 & - 3 & 5 & 0 \end{array}]$

解题步骤 3.2.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{6} & 0 & - \frac{4}{3} \\ 0 & 0 & 1 & 3 \\ 1 & - 3 & 5 & 0 \end{array}]$

解题步骤 3.3

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 3.3.1

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

$[\begin{array}{c} 1 & - \frac{1}{6} & 0 & - \frac{4}{3} \\ 0 & 0 & 1 & 3 \\ 1 - 1 & - 3 + \frac{1}{6} & 5 - 0 & 0 + \frac{4}{3} \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{6} & 0 & - \frac{4}{3} \\ 0 & 0 & 1 & 3 \\ 0 & - \frac{17}{6} & 5 & \frac{4}{3} \end{array}]$

解题步骤 3.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & 0 & - \frac{4}{3} \\ 0 & - \frac{17}{6} & 5 & \frac{4}{3} \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 3.5

Multiply each element of $R_{2}$ by $- \frac{6}{17}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.5.1

Multiply each element of $R_{2}$ by $- \frac{6}{17}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & 0 & - \frac{4}{3} \\ - \frac{6}{17} \cdot 0 & - \frac{6}{17} (- \frac{17}{6}) & - \frac{6}{17} \cdot 5 & - \frac{6}{17} \cdot \frac{4}{3} \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 3.5.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{6} & 0 & - \frac{4}{3} \\ 0 & 1 & - \frac{30}{17} & - \frac{8}{17} \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 3.6

Perform the row operation $R_{2} = R_{2} + \frac{30}{17} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 3.6.1

Perform the row operation $R_{2} = R_{2} + \frac{30}{17} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{6} & 0 & - \frac{4}{3} \\ 0 + \frac{30}{17} \cdot 0 & 1 + \frac{30}{17} \cdot 0 & - \frac{30}{17} + \frac{30}{17} \cdot 1 & - \frac{8}{17} + \frac{30}{17} \cdot 3 \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 3.6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{6} & 0 & - \frac{4}{3} \\ 0 & 1 & 0 & \frac{82}{17} \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 3.7

Perform the row operation $R_{1} = R_{1} + \frac{1}{6} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.7.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{6} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{6} \cdot 0 & - \frac{1}{6} + \frac{1}{6} \cdot 1 & 0 + \frac{1}{6} \cdot 0 & - \frac{4}{3} + \frac{1}{6} \cdot \frac{82}{17} \\ 0 & 1 & 0 & \frac{82}{17} \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 3.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & - \frac{9}{17} \\ 0 & 1 & 0 & \frac{82}{17} \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 4

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

$x = - \frac{9}{17}$

$y = \frac{82}{17}$

$z = 3$

解题步骤 5

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

$(- \frac{9}{17}, \frac{82}{17}, 3)$

有限数学 示例

有限数学示例