有限数学示例

$- 4 - 9 y - z = - 20$ , $- 8 x - 6 y - 2 z = 16$ , $6 x + 9 y + 2 = 22$

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

将所有不包含变量的项移到等式右边。

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解题步骤 1.1.1

在等式两边都加上 $4$ 。

$- 9 y - z = - 20 + 4$

$- 8 x - 6 y - 2 z = 16$

$6 x + 9 y + 2 = 22$

解题步骤 1.1.2

将 $- 20$ 和 $4$ 相加。

$- 9 y - z = - 16$

$- 8 x - 6 y - 2 z = 16$

$6 x + 9 y + 2 = 22$

$- 9 y - z = - 16$

$- 8 x - 6 y - 2 z = 16$

$6 x + 9 y + 2 = 22$

解题步骤 1.2

将所有不包含变量的项移到等式右边。

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解题步骤 1.2.1

从等式两边同时减去 $2$ 。

$- 9 y - z = - 16$

$- 8 x - 6 y - 2 z = 16$

$6 x + 9 y = 22 - 2$

解题步骤 1.2.2

从 $22$ 中减去 $2$ 。

$- 9 y - z = - 16$

$- 8 x - 6 y - 2 z = 16$

$6 x + 9 y = 20$

$- 9 y - z = - 16$

$- 8 x - 6 y - 2 z = 16$

$6 x + 9 y = 20$

$- 9 y - z = - 16$

$- 8 x - 6 y - 2 z = 16$

$6 x + 9 y = 20$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 0 & - 9 & - 1 & - 16 \\ - 8 & - 6 & - 2 & 16 \\ 6 & 9 & 0 & 20 \end{array}]$

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 $1, 1$ .

$[\begin{array}{c} - 8 & - 6 & - 2 & 16 \\ 0 & - 9 & - 1 & - 16 \\ 6 & 9 & 0 & 20 \end{array}]$

解题步骤 3.2

Multiply each element of $R_{1}$ by $- \frac{1}{8}$ 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 $- \frac{1}{8}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{8} \cdot - 8 & - \frac{1}{8} \cdot - 6 & - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 16 \\ 0 & - 9 & - 1 & - 16 \\ 6 & 9 & 0 & 20 \end{array}]$

解题步骤 3.2.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 & - 9 & - 1 & - 16 \\ 6 & 9 & 0 & 20 \end{array}]$

解题步骤 3.3

Perform the row operation $R_{3} = R_{3} - 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

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解题步骤 3.3.1

Perform the row operation $R_{3} = R_{3} - 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 & - 9 & - 1 & - 16 \\ 6 - 6 \cdot 1 & 9 - 6 (\frac{3}{4}) & 0 - 6 (\frac{1}{4}) & 20 - 6 \cdot - 2 \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 & - 9 & - 1 & - 16 \\ 0 & \frac{9}{2} & - \frac{3}{2} & 32 \end{array}]$

解题步骤 3.4

Multiply each element of $R_{2}$ by $- \frac{1}{9}$ 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 $- \frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ - \frac{1}{9} \cdot 0 & - \frac{1}{9} \cdot - 9 & - \frac{1}{9} \cdot - 1 & - \frac{1}{9} \cdot - 16 \\ 0 & \frac{9}{2} & - \frac{3}{2} & 32 \end{array}]$

解题步骤 3.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 & 1 & \frac{1}{9} & \frac{16}{9} \\ 0 & \frac{9}{2} & - \frac{3}{2} & 32 \end{array}]$

解题步骤 3.5

Perform the row operation $R_{3} = R_{3} - \frac{9}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{9}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 & 1 & \frac{1}{9} & \frac{16}{9} \\ 0 - \frac{9}{2} \cdot 0 & \frac{9}{2} - \frac{9}{2} \cdot 1 & - \frac{3}{2} - \frac{9}{2} \cdot \frac{1}{9} & 32 - \frac{9}{2} \cdot \frac{16}{9} \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 & 1 & \frac{1}{9} & \frac{16}{9} \\ 0 & 0 & - 2 & 24 \end{array}]$

解题步骤 3.6

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

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 & 1 & \frac{1}{9} & \frac{16}{9} \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 24 \end{array}]$

解题步骤 3.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 & 1 & \frac{1}{9} & \frac{16}{9} \\ 0 & 0 & 1 & - 12 \end{array}]$

解题步骤 3.7

Perform the row operation $R_{2} = R_{2} - \frac{1}{9} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 3.7.1

Perform the row operation $R_{2} = R_{2} - \frac{1}{9} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 - \frac{1}{9} \cdot 0 & 1 - \frac{1}{9} \cdot 0 & \frac{1}{9} - \frac{1}{9} \cdot 1 & \frac{16}{9} - \frac{1}{9} \cdot - 12 \\ 0 & 0 & 1 & - 12 \end{array}]$

解题步骤 3.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & - 2 \\ 0 & 1 & 0 & \frac{28}{9} \\ 0 & 0 & 1 & - 12 \end{array}]$

解题步骤 3.8

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

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解题步骤 3.8.1

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

$[\begin{array}{c} 1 - \frac{1}{4} \cdot 0 & \frac{3}{4} - \frac{1}{4} \cdot 0 & \frac{1}{4} - \frac{1}{4} \cdot 1 & - 2 - \frac{1}{4} \cdot - 12 \\ 0 & 1 & 0 & \frac{28}{9} \\ 0 & 0 & 1 & - 12 \end{array}]$

解题步骤 3.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{3}{4} & 0 & 1 \\ 0 & 1 & 0 & \frac{28}{9} \\ 0 & 0 & 1 & - 12 \end{array}]$

解题步骤 3.9

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.9.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & 0 - \frac{3}{4} \cdot 0 & 1 - \frac{3}{4} \cdot \frac{28}{9} \\ 0 & 1 & 0 & \frac{28}{9} \\ 0 & 0 & 1 & - 12 \end{array}]$

解题步骤 3.9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & - \frac{4}{3} \\ 0 & 1 & 0 & \frac{28}{9} \\ 0 & 0 & 1 & - 12 \end{array}]$

解题步骤 4

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

$x = - \frac{4}{3}$

$y = \frac{28}{9}$

$z = - 12$

解题步骤 5

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

$(- \frac{4}{3}, \frac{28}{9}, - 12)$

有限数学 示例

有限数学示例