有限数学示例

$x + 2 y = 3$ , $2 x + 6 y = 5$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 2 & 3 \\ 2 & 6 & 5 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

点击获取更多步骤...

解题步骤 2.1

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

点击获取更多步骤...

解题步骤 2.1.1

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

$[\begin{array}{c} 1 & 2 & 3 \\ 2 - 2 \cdot 1 & 6 - 2 \cdot 2 & 5 - 2 \cdot 3 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

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

解题步骤 2.2

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

点击获取更多步骤...

解题步骤 2.2.1

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

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

解题步骤 2.2.2

化简 $R_{2}$ 。

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

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 $0$ .

点击获取更多步骤...

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 $0$ .

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

解题步骤 2.3.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x = 4$

$y = - \frac{1}{2}$

解题步骤 4

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

$(4, - \frac{1}{2})$

有限数学 示例