有限数学示例

$3 x - 6 y + 6 z = - 3$ , $- 6 + 12 y - 12 z = 6$ , $6 x - 12 y + 12 z = - 6$

解题步骤 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

在等式两边都加上 $6$ 。

$3 x - 6 y + 6 z = - 3$

$12 y - 12 z = 6 + 6$

$6 x - 12 y + 12 z = - 6$

解题步骤 1.2

将 $6$ 和 $6$ 相加。

$3 x - 6 y + 6 z = - 3$

$12 y - 12 z = 12$

$6 x - 12 y + 12 z = - 6$

$3 x - 6 y + 6 z = - 3$

$12 y - 12 z = 12$

$6 x - 12 y + 12 z = - 6$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 3 & - 6 & 6 & - 3 \\ 0 & 12 & - 12 & 12 \\ 6 & - 12 & 12 & - 6 \end{array}]$

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

$[\begin{array}{c} \frac{3}{3} & \frac{- 6}{3} & \frac{6}{3} & \frac{- 3}{3} \\ 0 & 12 & - 12 & 12 \\ 6 & - 12 & 12 & - 6 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 2 & 2 & - 1 \\ 0 & 12 & - 12 & 12 \\ 6 & - 12 & 12 & - 6 \end{array}]$

解题步骤 3.2

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 3.2.1

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

$[\begin{array}{c} 1 & - 2 & 2 & - 1 \\ 0 & 12 & - 12 & 12 \\ 6 - 6 \cdot 1 & - 12 - 6 \cdot - 2 & 12 - 6 \cdot 2 & - 6 - 6 \cdot - 1 \end{array}]$

解题步骤 3.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 2 & 2 & - 1 \\ 0 & 12 & - 12 & 12 \\ 0 & 0 & 0 & 0 \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $\frac{1}{12}$ 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 $\frac{1}{12}$ to make the entry at $2, 2$ a $1$ .

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

解题步骤 3.3.2

化简 $R_{2}$ 。

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

解题步骤 3.4

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

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解题步骤 3.4.1

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

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

解题步骤 3.4.2

化简 $R_{1}$ 。

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

解题步骤 4

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

$x = 1$

$y - z = 1$

$0 = 0$

解题步骤 5

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

$(1, 1 + z, z)$

有限数学 示例

有限数学示例