有限数学示例

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

解题步骤 1

Write the system as a matrix.

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

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 $- \frac{1}{2}$ 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 $- \frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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

解题步骤 2.1.2

化简 $R_{1}$ 。

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

解题步骤 2.2

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

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解题步骤 2.2.1

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

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

解题步骤 2.2.2

化简 $R_{2}$ 。

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

解题步骤 3

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

$x - 3 y = - 2$

$0 = 0$

解题步骤 4

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

$(- 2 + 3 y, y)$

有限数学 示例