有限数学示例

$x + 2 y = 6$ , $5 x + 10 y = 30$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 2 & 6 \\ 5 & 10 & 30 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 2.1.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 2 & 6 \\ 5 - 5 \cdot 1 & 10 - 5 \cdot 2 & 30 - 5 \cdot 6 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

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

解题步骤 3

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

$x + 2 y = 6$

$0 = 0$

解题步骤 4

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

$(6 - 2 y, y)$

有限数学 示例