有限数学示例

$6 x + 4 y = 8$ , $9 x + 6 y = 12$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 6 & 4 & 8 \\ 9 & 6 & 12 \end{array}]$

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 $\frac{1}{6}$ 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 $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{6}{6} & \frac{4}{6} & \frac{8}{6} \\ 9 & 6 & 12 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

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

解题步骤 2.2

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

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解题步骤 2.2.1

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

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

解题步骤 2.2.2

化简 $R_{2}$ 。

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

解题步骤 3

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

$x + \frac{2}{3} y = \frac{4}{3}$

$0 = 0$

解题步骤 4

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

$(\frac{4}{3} - \frac{2 y}{3}, y)$

有限数学 示例