有限数学示例

$2 x - 3 y + z - 7 = 0$ , $x + y - z = 1$ , $2 x + 2 y - 3 z + 4 = 0$

解题步骤 1

Move variables to the left and constant terms to the right.

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解题步骤 1.1

在等式两边都加上 $7$ 。

$2 x - 3 y + z = 7$

$x + y - z = 1$

$2 x + 2 y - 3 z + 4 = 0$

解题步骤 1.2

从等式两边同时减去 $4$ 。

$2 x - 3 y + z = 7$

$x + y - z = 1$

$2 x + 2 y - 3 z = - 4$

$2 x - 3 y + z = 7$

$x + y - z = 1$

$2 x + 2 y - 3 z = - 4$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 2 & - 3 & 1 & 7 \\ 1 & 1 & - 1 & 1 \\ 2 & 2 & - 3 & - 4 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

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解题步骤 3.1

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 3.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{- 3}{2} & \frac{1}{2} & \frac{7}{2} \\ 1 & 1 & - 1 & 1 \\ 2 & 2 & - 3 & - 4 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 1 & 1 & - 1 & 1 \\ 2 & 2 & - 3 & - 4 \end{array}]$

解题步骤 3.2

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 3.2.1

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

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 1 - 1 & 1 + \frac{3}{2} & - 1 - \frac{1}{2} & 1 - \frac{7}{2} \\ 2 & 2 & - 3 & - 4 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 & \frac{5}{2} & - \frac{3}{2} & - \frac{5}{2} \\ 2 & 2 & - 3 & - 4 \end{array}]$

解题步骤 3.3

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 3.3.1

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

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 & \frac{5}{2} & - \frac{3}{2} & - \frac{5}{2} \\ 2 - 2 \cdot 1 & 2 - 2 (- \frac{3}{2}) & - 3 - 2 (\frac{1}{2}) & - 4 - 2 (\frac{7}{2}) \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 & \frac{5}{2} & - \frac{3}{2} & - \frac{5}{2} \\ 0 & 5 & - 4 & - 11 \end{array}]$

解题步骤 3.4

Multiply each element of $R_{2}$ by $\frac{2}{5}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.4.1

Multiply each element of $R_{2}$ by $\frac{2}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ \frac{2}{5} \cdot 0 & \frac{2}{5} \cdot \frac{5}{2} & \frac{2}{5} (- \frac{3}{2}) & \frac{2}{5} (- \frac{5}{2}) \\ 0 & 5 & - 4 & - 11 \end{array}]$

解题步骤 3.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 & 1 & - \frac{3}{5} & - 1 \\ 0 & 5 & - 4 & - 11 \end{array}]$

解题步骤 3.5

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 3.5.1

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

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 & 1 & - \frac{3}{5} & - 1 \\ 0 - 5 \cdot 0 & 5 - 5 \cdot 1 & - 4 - 5 (- \frac{3}{5}) & - 11 - 5 \cdot - 1 \end{array}]$

解题步骤 3.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 & 1 & - \frac{3}{5} & - 1 \\ 0 & 0 & - 1 & - 6 \end{array}]$

解题步骤 3.6

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

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解题步骤 3.6.1

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

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 & 1 & - \frac{3}{5} & - 1 \\ - 0 & - 0 & - - 1 & - - 6 \end{array}]$

解题步骤 3.6.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 & 1 & - \frac{3}{5} & - 1 \\ 0 & 0 & 1 & 6 \end{array}]$

解题步骤 3.7

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

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解题步骤 3.7.1

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

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 + \frac{3}{5} \cdot 0 & 1 + \frac{3}{5} \cdot 0 & - \frac{3}{5} + \frac{3}{5} \cdot 1 & - 1 + \frac{3}{5} \cdot 6 \\ 0 & 0 & 1 & 6 \end{array}]$

解题步骤 3.7.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & \frac{1}{2} & \frac{7}{2} \\ 0 & 1 & 0 & \frac{13}{5} \\ 0 & 0 & 1 & 6 \end{array}]$

解题步骤 3.8

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

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解题步骤 3.8.1

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

$[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & - \frac{3}{2} - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & \frac{7}{2} - \frac{1}{2} \cdot 6 \\ 0 & 1 & 0 & \frac{13}{5} \\ 0 & 0 & 1 & 6 \end{array}]$

解题步骤 3.8.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - \frac{3}{2} & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{13}{5} \\ 0 & 0 & 1 & 6 \end{array}]$

解题步骤 3.9

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

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解题步骤 3.9.1

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

$[\begin{array}{c} 1 + \frac{3}{2} \cdot 0 & - \frac{3}{2} + \frac{3}{2} \cdot 1 & 0 + \frac{3}{2} \cdot 0 & \frac{1}{2} + \frac{3}{2} \cdot \frac{13}{5} \\ 0 & 1 & 0 & \frac{13}{5} \\ 0 & 0 & 1 & 6 \end{array}]$

解题步骤 3.9.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 0 & 0 & \frac{22}{5} \\ 0 & 1 & 0 & \frac{13}{5} \\ 0 & 0 & 1 & 6 \end{array}]$

解题步骤 4

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

$x = \frac{22}{5}$

$y = \frac{13}{5}$

$z = 6$

解题步骤 5

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

$(\frac{22}{5}, \frac{13}{5}, 6)$

有限数学 示例

有限数学示例