有限数学示例

$x + 7 y = 2$ , $5 x + 2 y = 15$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 7 & 2 \\ 5 & 2 & 15 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 7 & 2 \\ 5 - 5 \cdot 1 & 2 - 5 \cdot 7 & 15 - 5 \cdot 2 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 7 & 2 \\ 0 & - 33 & 5 \end{array}]$

解题步骤 2.2

Multiply each element of $R_{2}$ by $- \frac{1}{33}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{33}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 7 & 2 \\ - \frac{1}{33} \cdot 0 & - \frac{1}{33} \cdot - 33 & - \frac{1}{33} \cdot 5 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 7 & 2 \\ 0 & 1 & - \frac{5}{33} \end{array}]$

解题步骤 2.3

Perform the row operation $R_{1} = R_{1} - 7 R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.3.1

Perform the row operation $R_{1} = R_{1} - 7 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 7 \cdot 0 & 7 - 7 \cdot 1 & 2 - 7 (- \frac{5}{33}) \\ 0 & 1 & - \frac{5}{33} \end{array}]$

解题步骤 2.3.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & \frac{101}{33} \\ 0 & 1 & - \frac{5}{33} \end{array}]$

解题步骤 3

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

$x = \frac{101}{33}$

$y = - \frac{5}{33}$

解题步骤 4

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

$(\frac{101}{33}, - \frac{5}{33})$

有限数学 示例