有限数学示例

$2 x + 9 - z = 0$ , $3 x - y + 2 z = 1$ , $5 x + 8 y + z = 1$

解题步骤 1

从等式两边同时减去 $9$ 。

$2 x - z = - 9$

$3 x - y + 2 z = 1$

$5 x + 8 y + z = 1$

解题步骤 2

Write the system as a matrix.

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

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 $\frac{1}{2}$ 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 $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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

解题步骤 3.1.2

化简 $R_{1}$ 。

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

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 $0$ .

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

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 & - 1 & \frac{7}{2} & \frac{29}{2} \\ 5 & 8 & 1 & 1 \end{array}]$

解题步骤 3.3

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 3.3.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 & - 1 & \frac{7}{2} & \frac{29}{2} \\ 5 - 5 \cdot 1 & 8 - 5 \cdot 0 & 1 - 5 (- \frac{1}{2}) & 1 - 5 (- \frac{9}{2}) \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 & - 1 & \frac{7}{2} & \frac{29}{2} \\ 0 & 8 & \frac{7}{2} & \frac{47}{2} \end{array}]$

解题步骤 3.4

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 3.4.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ - 0 & - - 1 & - \frac{7}{2} & - \frac{29}{2} \\ 0 & 8 & \frac{7}{2} & \frac{47}{2} \end{array}]$

解题步骤 3.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{7}{2} & - \frac{29}{2} \\ 0 & 8 & \frac{7}{2} & \frac{47}{2} \end{array}]$

解题步骤 3.5

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

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解题步骤 3.5.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{7}{2} & - \frac{29}{2} \\ 0 - 8 \cdot 0 & 8 - 8 \cdot 1 & \frac{7}{2} - 8 (- \frac{7}{2}) & \frac{47}{2} - 8 (- \frac{29}{2}) \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{7}{2} & - \frac{29}{2} \\ 0 & 0 & \frac{63}{2} & \frac{279}{2} \end{array}]$

解题步骤 3.6

Multiply each element of $R_{3}$ by $\frac{2}{63}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 3.6.1

Multiply each element of $R_{3}$ by $\frac{2}{63}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{7}{2} & - \frac{29}{2} \\ \frac{2}{63} \cdot 0 & \frac{2}{63} \cdot 0 & \frac{2}{63} \cdot \frac{63}{2} & \frac{2}{63} \cdot \frac{279}{2} \end{array}]$

解题步骤 3.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 & 1 & - \frac{7}{2} & - \frac{29}{2} \\ 0 & 0 & 1 & \frac{31}{7} \end{array}]$

解题步骤 3.7

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

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解题步骤 3.7.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 + \frac{7}{2} \cdot 0 & 1 + \frac{7}{2} \cdot 0 & - \frac{7}{2} + \frac{7}{2} \cdot 1 & - \frac{29}{2} + \frac{7}{2} \cdot \frac{31}{7} \\ 0 & 0 & 1 & \frac{31}{7} \end{array}]$

解题步骤 3.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} & - \frac{9}{2} \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & \frac{31}{7} \end{array}]$

解题步骤 3.8

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

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解题步骤 3.8.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & 0 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & - \frac{9}{2} + \frac{1}{2} \cdot \frac{31}{7} \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & \frac{31}{7} \end{array}]$

解题步骤 3.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & - \frac{16}{7} \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & \frac{31}{7} \end{array}]$

解题步骤 4

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

$x = - \frac{16}{7}$

$y = 1$

$z = \frac{31}{7}$

解题步骤 5

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

$(- \frac{16}{7}, 1, \frac{31}{7})$

有限数学 示例

有限数学示例