有限数学示例

$6 x - y + 7 x = 8$ , $- 7 y + 7 z = 14$ , $7 x + 7 y + 5 z = 8$

解题步骤 1

将 $6 x$ 和 $7 x$ 相加。

$13 x - y = 8$

$- 7 y + 7 z = 14$

$7 x + 7 y + 5 z = 8$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 13 & - 1 & 0 & 8 \\ 0 & - 7 & 7 & 14 \\ 7 & 7 & 5 & 8 \end{array}]$

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 $\frac{1}{13}$ 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 $\frac{1}{13}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{13}{13} & \frac{- 1}{13} & \frac{0}{13} & \frac{8}{13} \\ 0 & - 7 & 7 & 14 \\ 7 & 7 & 5 & 8 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 & - 7 & 7 & 14 \\ 7 & 7 & 5 & 8 \end{array}]$

解题步骤 3.2

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 3.2.1

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 & - 7 & 7 & 14 \\ 7 - 7 \cdot 1 & 7 - 7 (- \frac{1}{13}) & 5 - 7 \cdot 0 & 8 - 7 (\frac{8}{13}) \end{array}]$

解题步骤 3.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 & - 7 & 7 & 14 \\ 0 & \frac{98}{13} & 5 & \frac{48}{13} \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $- \frac{1}{7}$ 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 $- \frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ - \frac{1}{7} \cdot 0 & - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot 7 & - \frac{1}{7} \cdot 14 \\ 0 & \frac{98}{13} & 5 & \frac{48}{13} \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 & 1 & - 1 & - 2 \\ 0 & \frac{98}{13} & 5 & \frac{48}{13} \end{array}]$

解题步骤 3.4

Perform the row operation $R_{3} = R_{3} - \frac{98}{13} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 3.4.1

Perform the row operation $R_{3} = R_{3} - \frac{98}{13} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 & 1 & - 1 & - 2 \\ 0 - \frac{98}{13} \cdot 0 & \frac{98}{13} - \frac{98}{13} \cdot 1 & 5 - \frac{98}{13} \cdot - 1 & \frac{48}{13} - \frac{98}{13} \cdot - 2 \end{array}]$

解题步骤 3.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & \frac{163}{13} & \frac{244}{13} \end{array}]$

解题步骤 3.5

Multiply each element of $R_{3}$ by $\frac{13}{163}$ 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 $\frac{13}{163}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 & 1 & - 1 & - 2 \\ \frac{13}{163} \cdot 0 & \frac{13}{163} \cdot 0 & \frac{13}{163} \cdot \frac{163}{13} & \frac{13}{163} \cdot \frac{244}{13} \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 & 1 & - 1 & - 2 \\ 0 & 0 & 1 & \frac{244}{163} \end{array}]$

解题步骤 3.6

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

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解题步骤 3.6.1

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

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & - 2 + \frac{244}{163} \\ 0 & 0 & 1 & \frac{244}{163} \end{array}]$

解题步骤 3.6.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{1}{13} & 0 & \frac{8}{13} \\ 0 & 1 & 0 & - \frac{82}{163} \\ 0 & 0 & 1 & \frac{244}{163} \end{array}]$

解题步骤 3.7

Perform the row operation $R_{1} = R_{1} + \frac{1}{13} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.7.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{13} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{13} \cdot 0 & - \frac{1}{13} + \frac{1}{13} \cdot 1 & 0 + \frac{1}{13} \cdot 0 & \frac{8}{13} + \frac{1}{13} (- \frac{82}{163}) \\ 0 & 1 & 0 & - \frac{82}{163} \\ 0 & 0 & 1 & \frac{244}{163} \end{array}]$

解题步骤 3.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & \frac{94}{163} \\ 0 & 1 & 0 & - \frac{82}{163} \\ 0 & 0 & 1 & \frac{244}{163} \end{array}]$

解题步骤 4

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

$x = \frac{94}{163}$

$y = - \frac{82}{163}$

$z = \frac{244}{163}$

解题步骤 5

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

$(\frac{94}{163}, - \frac{82}{163}, \frac{244}{163})$

有限数学 示例

有限数学示例