有限数学示例

$x - 3 z = - 2$ , $2 x + 2 y + z = 4$ , $3 x + y - 2 z = 5$

解题步骤 1

Write the system as a matrix.

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

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

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解题步骤 2.1.1

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

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

解题步骤 2.1.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 0 & - 3 & - 2 \\ 0 & 2 & 7 & 8 \\ 3 & 1 & - 2 & 5 \end{array}]$

解题步骤 2.2

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

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解题步骤 2.2.1

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

$[\begin{array}{c} 1 & 0 & - 3 & - 2 \\ 0 & 2 & 7 & 8 \\ 3 - 3 \cdot 1 & 1 - 3 \cdot 0 & - 2 - 3 \cdot - 3 & 5 - 3 \cdot - 2 \end{array}]$

解题步骤 2.2.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 0 & - 3 & - 2 \\ 0 & 2 & 7 & 8 \\ 0 & 1 & 7 & 11 \end{array}]$

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 $1$ .

$[\begin{array}{c} 1 & 0 & - 3 & - 2 \\ \frac{0}{2} & \frac{2}{2} & \frac{7}{2} & \frac{8}{2} \\ 0 & 1 & 7 & 11 \end{array}]$

解题步骤 2.3.2

化简 $R_{2}$ 。

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

解题步骤 2.4

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

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解题步骤 2.4.1

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

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

解题步骤 2.4.2

化简 $R_{3}$ 。

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

解题步骤 2.5

Multiply each element of $R_{3}$ by $\frac{2}{7}$ 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 $\frac{2}{7}$ to make the entry at $3, 3$ a $1$ .

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

解题步骤 2.5.2

化简 $R_{3}$ 。

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

解题步骤 2.6

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.6.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{3}$ to make the entry at $2, 3$ a $0$ .

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

解题步骤 2.6.2

化简 $R_{2}$ 。

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

解题步骤 2.7

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

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解题步骤 2.7.1

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

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

解题步骤 2.7.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x = 4$

$y = - 3$

$z = 2$

解题步骤 4

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

$(4, - 3, 2)$

有限数学 示例

有限数学示例