初级微积分示例

$- x - 5 y - 5 z = 2$ , $4 x - 5 y + 4 z = 19$ , $x + 5 y - z = - 20$

解题步骤 1

Write the system as a matrix.

$[\begin{array}{c} - 1 & - 5 & - 5 & 2 \\ 4 & - 5 & 4 & 19 \\ 1 & 5 & - 1 & - 20 \end{array}]$

解题步骤 2

求行简化阶梯形矩阵。

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解题步骤 2.1

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

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解题步骤 2.1.1

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

$[\begin{array}{c} - - 1 & - - 5 & - - 5 & - 1 \cdot 2 \\ 4 & - 5 & 4 & 19 \\ 1 & 5 & - 1 & - 20 \end{array}]$

解题步骤 2.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ 4 & - 5 & 4 & 19 \\ 1 & 5 & - 1 & - 20 \end{array}]$

解题步骤 2.2

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 2.2.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ 4 - 4 \cdot 1 & - 5 - 4 \cdot 5 & 4 - 4 \cdot 5 & 19 - 4 \cdot - 2 \\ 1 & 5 & - 1 & - 20 \end{array}]$

解题步骤 2.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ 0 & - 25 & - 16 & 27 \\ 1 & 5 & - 1 & - 20 \end{array}]$

解题步骤 2.3

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

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解题步骤 2.3.1

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

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ 0 & - 25 & - 16 & 27 \\ 1 - 1 & 5 - 5 & - 1 - 5 & - 20 + 2 \end{array}]$

解题步骤 2.3.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ 0 & - 25 & - 16 & 27 \\ 0 & 0 & - 6 & - 18 \end{array}]$

解题步骤 2.4

Multiply each element of $R_{2}$ by $- \frac{1}{25}$ to make the entry at $2, 2$ a $1$ .

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解题步骤 2.4.1

Multiply each element of $R_{2}$ by $- \frac{1}{25}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ - \frac{1}{25} \cdot 0 & - \frac{1}{25} \cdot - 25 & - \frac{1}{25} \cdot - 16 & - \frac{1}{25} \cdot 27 \\ 0 & 0 & - 6 & - 18 \end{array}]$

解题步骤 2.4.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ 0 & 1 & \frac{16}{25} & - \frac{27}{25} \\ 0 & 0 & - 6 & - 18 \end{array}]$

解题步骤 2.5

Multiply each element of $R_{3}$ by $- \frac{1}{6}$ 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 $- \frac{1}{6}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ 0 & 1 & \frac{16}{25} & - \frac{27}{25} \\ - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot - 18 \end{array}]$

解题步骤 2.5.2

化简 $R_{3}$ 。

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ 0 & 1 & \frac{16}{25} & - \frac{27}{25} \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 2.6

Perform the row operation $R_{2} = R_{2} - \frac{16}{25} R_{3}$ to make the entry at $2, 3$ a $0$ .

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解题步骤 2.6.1

Perform the row operation $R_{2} = R_{2} - \frac{16}{25} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 5 & 5 & - 2 \\ 0 - \frac{16}{25} \cdot 0 & 1 - \frac{16}{25} \cdot 0 & \frac{16}{25} - \frac{16}{25} \cdot 1 & - \frac{27}{25} - \frac{16}{25} \cdot 3 \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 2.6.2

化简 $R_{2}$ 。

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

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 $0$ .

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

解题步骤 2.7.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & 5 & 0 & - 17 \\ 0 & 1 & 0 & - 3 \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 2.8

Perform the row operation $R_{1} = R_{1} - 5 R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 2.8.1

Perform the row operation $R_{1} = R_{1} - 5 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 5 \cdot 0 & 5 - 5 \cdot 1 & 0 - 5 \cdot 0 & - 17 - 5 \cdot - 3 \\ 0 & 1 & 0 & - 3 \\ 0 & 0 & 1 & 3 \end{array}]$

解题步骤 2.8.2

化简 $R_{1}$ 。

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

解题步骤 3

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

$x = - 2$

$y = - 3$

$z = 3$

解题步骤 4

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

$(- 2, - 3, 3)$

初级微积分 示例

初级微积分示例