有限数学示例

$3 x - 6 y + 9 z = 12$ , $- x + 2 y - 3 z = - 4$ , $x + y + 2 z = 7$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 3 & - 6 & 9 & 12 \\ - 1 & 2 & - 3 & - 4 \\ 1 & 1 & 2 & 7 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{- 6}{3} & \frac{9}{3} & \frac{12}{3} \\ - 1 & 2 & - 3 & - 4 \\ 1 & 1 & 2 & 7 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

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

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & - 2 & 3 & 4 \\ - 1 + 1 \cdot 1 & 2 - 2 & - 3 + 1 \cdot 3 & - 4 + 1 \cdot 4 \\ 1 & 1 & 2 & 7 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

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

解题步骤 2.3

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

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解题步骤 2.3.1

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

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

解题步骤 2.3.2

化简 $R_{3}$ 。

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

解题步骤 2.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

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

解题步骤 2.5

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

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解题步骤 2.5.1

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

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

解题步骤 2.5.2

化简 $R_{2}$ 。

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

解题步骤 2.6

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 2.6.1

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

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

解题步骤 2.6.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x + \frac{7}{3} z = 6$

$y - \frac{1}{3} z = 1$

$0 = 0$

解题步骤 4

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

$(6 - \frac{7 z}{3}, 1 + \frac{z}{3}, z)$

有限数学 示例

有限数学示例