有限数学示例

$2 x + 4 y = - 6$ , $- 3 x - 5 y = 12$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 4 & - 6 \\ - 3 & - 5 & 12 \end{array}]$

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 $\frac{1}{2}$ 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 $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{4}{2} & \frac{- 6}{2} \\ - 3 & - 5 & 12 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 2 & - 3 \\ - 3 & - 5 & 12 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 2 & - 3 \\ - 3 + 3 \cdot 1 & - 5 + 3 \cdot 2 & 12 + 3 \cdot - 3 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

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

解题步骤 2.3

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 2.3.1

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

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

解题步骤 2.3.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & - 9 \\ 0 & 1 & 3 \end{array}]$

解题步骤 3

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

$x = - 9$

$y = 3$

解题步骤 4

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

$(- 9, 3)$

有限数学 示例