有限数学示例

$x + y + z = 7$ , $x - y + 2 z = 7$ , $2 x + 3 z = 15$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 1 & 7 \\ 1 & - 1 & 2 & 7 \\ 2 & 0 & 3 & 15 \end{array}]$

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

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解题步骤 2.1.1

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

$[\begin{array}{c} 1 & 1 & 1 & 7 \\ 1 - 1 & - 1 - 1 & 2 - 1 & 7 - 7 \\ 2 & 0 & 3 & 15 \end{array}]$

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 1 & 1 & 7 \\ 0 & - 2 & 1 & 0 \\ 2 & 0 & 3 & 15 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 7 \\ 0 & - 2 & 1 & 0 \\ 2 - 2 \cdot 1 & 0 - 2 \cdot 1 & 3 - 2 \cdot 1 & 15 - 2 \cdot 7 \end{array}]$

解题步骤 2.2.2

化简 $R_{3}$ 。

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

解题步骤 2.3

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

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

解题步骤 2.3.2

化简 $R_{2}$ 。

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

解题步骤 2.4

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

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解题步骤 2.4.1

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

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

解题步骤 2.4.2

化简 $R_{3}$ 。

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

解题步骤 2.5

Perform the row operation $R_{1} = R_{1} - 7 R_{3}$ to make the entry at $1, 4$ a $0$ .

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解题步骤 2.5.1

Perform the row operation $R_{1} = R_{1} - 7 R_{3}$ to make the entry at $1, 4$ a $0$ .

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

解题步骤 2.5.2

化简 $R_{1}$ 。

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

解题步骤 2.6

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

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解题步骤 2.6.1

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

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

解题步骤 2.6.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x + \frac{3}{2} z = 0$

$y - \frac{1}{2} z = 0$

$0 = 1$

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