有限数学示例

$2 x + 7 y - 5 z = 0$ , $3 x - 4 y + 6 z = 1$ , $5 x + 3 y + z = 1$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 7 & - 5 & 0 \\ 3 & - 4 & 6 & 1 \\ 5 & 3 & 1 & 1 \end{array}]$

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

$[\begin{array}{c} \frac{2}{2} & \frac{7}{2} & \frac{- 5}{2} & \frac{0}{2} \\ 3 & - 4 & 6 & 1 \\ 5 & 3 & 1 & 1 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{7}{2} & - \frac{5}{2} & 0 \\ 3 & - 4 & 6 & 1 \\ 5 & 3 & 1 & 1 \end{array}]$

解题步骤 2.2

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 2.2.1

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

$[\begin{array}{c} 1 & \frac{7}{2} & - \frac{5}{2} & 0 \\ 3 - 3 \cdot 1 & - 4 - 3 (\frac{7}{2}) & 6 - 3 (- \frac{5}{2}) & 1 - 3 \cdot 0 \\ 5 & 3 & 1 & 1 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{7}{2} & - \frac{5}{2} & 0 \\ 0 & - \frac{29}{2} & \frac{27}{2} & 1 \\ 5 & 3 & 1 & 1 \end{array}]$

解题步骤 2.3

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 2.3.1

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

$[\begin{array}{c} 1 & \frac{7}{2} & - \frac{5}{2} & 0 \\ 0 & - \frac{29}{2} & \frac{27}{2} & 1 \\ 5 - 5 \cdot 1 & 3 - 5 (\frac{7}{2}) & 1 - 5 (- \frac{5}{2}) & 1 - 5 \cdot 0 \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{7}{2} & - \frac{5}{2} & 0 \\ 0 & - \frac{29}{2} & \frac{27}{2} & 1 \\ 0 & - \frac{29}{2} & \frac{27}{2} & 1 \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $- \frac{2}{29}$ 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 $- \frac{2}{29}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{7}{2} & - \frac{5}{2} & 0 \\ - \frac{2}{29} \cdot 0 & - \frac{2}{29} (- \frac{29}{2}) & - \frac{2}{29} \cdot \frac{27}{2} & - \frac{2}{29} \cdot 1 \\ 0 & - \frac{29}{2} & \frac{27}{2} & 1 \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{7}{2} & - \frac{5}{2} & 0 \\ 0 & 1 & - \frac{27}{29} & - \frac{2}{29} \\ 0 & - \frac{29}{2} & \frac{27}{2} & 1 \end{array}]$

解题步骤 2.5

Perform the row operation $R_{3} = R_{3} + \frac{29}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 2.5.1

Perform the row operation $R_{3} = R_{3} + \frac{29}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{2} & - \frac{5}{2} & 0 \\ 0 & 1 & - \frac{27}{29} & - \frac{2}{29} \\ 0 + \frac{29}{2} \cdot 0 & - \frac{29}{2} + \frac{29}{2} \cdot 1 & \frac{27}{2} + \frac{29}{2} (- \frac{27}{29}) & 1 + \frac{29}{2} (- \frac{2}{29}) \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{7}{2} & - \frac{5}{2} & 0 \\ 0 & 1 & - \frac{27}{29} & - \frac{2}{29} \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 2.6

Perform the row operation $R_{1} = R_{1} - \frac{7}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.6.1

Perform the row operation $R_{1} = R_{1} - \frac{7}{2} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{7}{2} \cdot 0 & \frac{7}{2} - \frac{7}{2} \cdot 1 & - \frac{5}{2} - \frac{7}{2} (- \frac{27}{29}) & 0 - \frac{7}{2} (- \frac{2}{29}) \\ 0 & 1 & - \frac{27}{29} & - \frac{2}{29} \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 2.6.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{22}{29} & \frac{7}{29} \\ 0 & 1 & - \frac{27}{29} & - \frac{2}{29} \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 3

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

$x + \frac{22}{29} z = \frac{7}{29}$

$y - \frac{27}{29} z = - \frac{2}{29}$

$0 = 0$

解题步骤 4

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

$(\frac{7}{29} - \frac{22 z}{29}, - \frac{2}{29} + \frac{27 z}{29}, z)$

有限数学 示例

有限数学示例