有限数学示例

$3 x + 2 y + 4 z = 13$ , $6 x - 2 y - 5 z = - 24$ , $- 12 x - 2 y + 6 z = 16$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 3 & 2 & 4 & 13 \\ 6 & - 2 & - 5 & - 24 \\ - 12 & - 2 & 6 & 16 \end{array}]$

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

$[\begin{array}{c} \frac{3}{3} & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 6 & - 2 & - 5 & - 24 \\ - 12 & - 2 & 6 & 16 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 6 & - 2 & - 5 & - 24 \\ - 12 & - 2 & 6 & 16 \end{array}]$

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 6 - 6 \cdot 1 & - 2 - 6 (\frac{2}{3}) & - 5 - 6 (\frac{4}{3}) & - 24 - 6 (\frac{13}{3}) \\ - 12 & - 2 & 6 & 16 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 & - 6 & - 13 & - 50 \\ - 12 & - 2 & 6 & 16 \end{array}]$

解题步骤 2.3

Perform the row operation $R_{3} = R_{3} + 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 2.3.1

Perform the row operation $R_{3} = R_{3} + 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 & - 6 & - 13 & - 50 \\ - 12 + 12 \cdot 1 & - 2 + 12 (\frac{2}{3}) & 6 + 12 (\frac{4}{3}) & 16 + 12 (\frac{13}{3}) \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 & - 6 & - 13 & - 50 \\ 0 & 6 & 22 & 68 \end{array}]$

解题步骤 2.4

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

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot - 13 & - \frac{1}{6} \cdot - 50 \\ 0 & 6 & 22 & 68 \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 & 1 & \frac{13}{6} & \frac{25}{3} \\ 0 & 6 & 22 & 68 \end{array}]$

解题步骤 2.5

Perform the row operation $R_{3} = R_{3} - 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 2.5.1

Perform the row operation $R_{3} = R_{3} - 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 & 1 & \frac{13}{6} & \frac{25}{3} \\ 0 - 6 \cdot 0 & 6 - 6 \cdot 1 & 22 - 6 (\frac{13}{6}) & 68 - 6 (\frac{25}{3}) \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 & 1 & \frac{13}{6} & \frac{25}{3} \\ 0 & 0 & 9 & 18 \end{array}]$

解题步骤 2.6

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 2.6.1

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 & 1 & \frac{13}{6} & \frac{25}{3} \\ \frac{0}{9} & \frac{0}{9} & \frac{9}{9} & \frac{18}{9} \end{array}]$

解题步骤 2.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 & 1 & \frac{13}{6} & \frac{25}{3} \\ 0 & 0 & 1 & 2 \end{array}]$

解题步骤 2.7

Perform the row operation $R_{2} = R_{2} - \frac{13}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{2} = R_{2} - \frac{13}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 - \frac{13}{6} \cdot 0 & 1 - \frac{13}{6} \cdot 0 & \frac{13}{6} - \frac{13}{6} \cdot 1 & \frac{25}{3} - \frac{13}{6} \cdot 2 \\ 0 & 0 & 1 & 2 \end{array}]$

解题步骤 2.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{4}{3} & \frac{13}{3} \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 2 \end{array}]$

解题步骤 2.8

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

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解题步骤 2.8.1

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

$[\begin{array}{c} 1 - \frac{4}{3} \cdot 0 & \frac{2}{3} - \frac{4}{3} \cdot 0 & \frac{4}{3} - \frac{4}{3} \cdot 1 & \frac{13}{3} - \frac{4}{3} \cdot 2 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 2 \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

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

解题步骤 2.9

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

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解题步骤 2.9.1

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

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

解题步骤 2.9.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x = - 1$

$y = 4$

$z = 2$

解题步骤 4

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

$(- 1, 4, 2)$

有限数学 示例

有限数学示例