有限数学示例

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

解题步骤 1

Write the system as a matrix.

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

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

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解题步骤 2.1.1

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

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

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 2 & - 5 \\ 0 & 4 & 12 \\ 2 & 1 & 3 \end{array}]$

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & - 2 & - 5 \\ 0 & 4 & 12 \\ 2 - 2 \cdot 1 & 1 - 2 \cdot - 2 & 3 - 2 \cdot - 5 \end{array}]$

解题步骤 2.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 2 & - 5 \\ 0 & 4 & 12 \\ 0 & 5 & 13 \end{array}]$

解题步骤 2.3

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 2 & - 5 \\ \frac{0}{4} & \frac{4}{4} & \frac{12}{4} \\ 0 & 5 & 13 \end{array}]$

解题步骤 2.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 2 & - 5 \\ 0 & 1 & 3 \\ 0 & 5 & 13 \end{array}]$

解题步骤 2.4

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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解题步骤 2.4.1

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 2 & - 5 \\ 0 & 1 & 3 \\ 0 - 5 \cdot 0 & 5 - 5 \cdot 1 & 13 - 5 \cdot 3 \end{array}]$

解题步骤 2.4.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 2 & - 5 \\ 0 & 1 & 3 \\ 0 & 0 & - 2 \end{array}]$

解题步骤 2.5

Multiply each element of $R_{3}$ by $- \frac{1}{2}$ to make the entry at $3, 3$ a $1$ .

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解题步骤 2.5.1

Multiply each element of $R_{3}$ by $- \frac{1}{2}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 2 & - 5 \\ 0 & 1 & 3 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - 2 & - 5 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array}]$

解题步骤 2.6

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

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解题步骤 2.6.1

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

$[\begin{array}{c} 1 & - 2 & - 5 \\ 0 - 3 \cdot 0 & 1 - 3 \cdot 0 & 3 - 3 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

解题步骤 2.6.2

化简 $R_{2}$ 。

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

解题步骤 2.7

Perform the row operation $R_{1} = R_{1} + 5 R_{3}$ to make the entry at $1, 3$ a $0$ .

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解题步骤 2.7.1

Perform the row operation $R_{1} = R_{1} + 5 R_{3}$ to make the entry at $1, 3$ a $0$ .

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

解题步骤 2.7.2

化简 $R_{1}$ 。

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

解题步骤 2.8

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 2.8.1

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

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

解题步骤 2.8.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x = 0$

$y = 0$

$0 = 1$

解题步骤 4

The system is inconsistent so there is no solution.

$无解$

有限数学 示例

有限数学示例