有限数学示例

$x + 2 y - w = - 4$ , $3 x + 6 z + 9 w = 60$ , $x + 2 y - z = 4$ , $3 x - y + z + w = 8$

解题步骤 1

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

点击获取更多步骤...

解题步骤 1.1

移动 $2 y$ 。

$x - w + 2 y = - 4$

$3 x + 6 z + 9 w = 60$

$x + 2 y - z = 4$

$3 x - y + z + w = 8$

解题步骤 1.2

将 $x$ 和 $- w$ 重新排序。

$- w + x + 2 y = - 4$

$3 x + 6 z + 9 w = 60$

$x + 2 y - z = 4$

$3 x - y + z + w = 8$

解题步骤 1.3

移动 $6 z$ 。

$- w + x + 2 y = - 4$

$3 x + 9 w + 6 z = 60$

$x + 2 y - z = 4$

$3 x - y + z + w = 8$

解题步骤 1.4

将 $3 x$ 和 $9 w$ 重新排序。

$- w + x + 2 y = - 4$

$9 w + 3 x + 6 z = 60$

$x + 2 y - z = 4$

$3 x - y + z + w = 8$

解题步骤 1.5

移动 $z$ 。

$- w + x + 2 y = - 4$

$9 w + 3 x + 6 z = 60$

$x + 2 y - z = 4$

$3 x - y + w + z = 8$

解题步骤 1.6

移动 $- y$ 。

$- w + x + 2 y = - 4$

$9 w + 3 x + 6 z = 60$

$x + 2 y - z = 4$

$3 x + w - y + z = 8$

解题步骤 1.7

将 $3 x$ 和 $w$ 重新排序。

$- w + x + 2 y = - 4$

$9 w + 3 x + 6 z = 60$

$x + 2 y - z = 4$

$w + 3 x - y + z = 8$

$- w + x + 2 y = - 4$

$9 w + 3 x + 6 z = 60$

$x + 2 y - z = 4$

$w + 3 x - y + z = 8$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} - 1 & 1 & 2 & 0 & - 4 \\ 9 & 3 & 0 & 6 & 60 \\ 0 & 1 & 2 & - 1 & 4 \\ 1 & 3 & - 1 & 1 & 8 \end{array}]$

解题步骤 3

求行简化阶梯形矩阵。

点击获取更多步骤...

解题步骤 3.1

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

点击获取更多步骤...

解题步骤 3.1.1

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

$[\begin{array}{c} - - 1 & - 1 \cdot 1 & - 1 \cdot 2 & - 0 & - - 4 \\ 9 & 3 & 0 & 6 & 60 \\ 0 & 1 & 2 & - 1 & 4 \\ 1 & 3 & - 1 & 1 & 8 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & - 2 & 0 & 4 \\ 9 & 3 & 0 & 6 & 60 \\ 0 & 1 & 2 & - 1 & 4 \\ 1 & 3 & - 1 & 1 & 8 \end{array}]$

解题步骤 3.2

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

点击获取更多步骤...

解题步骤 3.2.1

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 0 & 4 \\ 9 - 9 \cdot 1 & 3 - 9 \cdot - 1 & 0 - 9 \cdot - 2 & 6 - 9 \cdot 0 & 60 - 9 \cdot 4 \\ 0 & 1 & 2 & - 1 & 4 \\ 1 & 3 & - 1 & 1 & 8 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & - 1 & - 2 & 0 & 4 \\ 0 & 12 & 18 & 6 & 24 \\ 0 & 1 & 2 & - 1 & 4 \\ 1 & 3 & - 1 & 1 & 8 \end{array}]$

解题步骤 3.3

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

点击获取更多步骤...

解题步骤 3.3.1

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

$[\begin{array}{c} 1 & - 1 & - 2 & 0 & 4 \\ 0 & 12 & 18 & 6 & 24 \\ 0 & 1 & 2 & - 1 & 4 \\ 1 - 1 & 3 + 1 & - 1 + 2 & 1 - 0 & 8 - 4 \end{array}]$

解题步骤 3.3.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & - 1 & - 2 & 0 & 4 \\ 0 & 12 & 18 & 6 & 24 \\ 0 & 1 & 2 & - 1 & 4 \\ 0 & 4 & 1 & 1 & 4 \end{array}]$

解题步骤 3.4

Multiply each element of $R_{2}$ by $\frac{1}{12}$ to make the entry at $2, 2$ a $1$ .

点击获取更多步骤...

解题步骤 3.4.1

Multiply each element of $R_{2}$ by $\frac{1}{12}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 0 & 4 \\ \frac{0}{12} & \frac{12}{12} & \frac{18}{12} & \frac{6}{12} & \frac{24}{12} \\ 0 & 1 & 2 & - 1 & 4 \\ 0 & 4 & 1 & 1 & 4 \end{array}]$

解题步骤 3.4.2

化简 $R_{2}$ 。

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

解题步骤 3.5

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

点击获取更多步骤...

解题步骤 3.5.1

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

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

解题步骤 3.5.2

化简 $R_{3}$ 。

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

解题步骤 3.6

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

点击获取更多步骤...

解题步骤 3.6.1

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

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

解题步骤 3.6.2

化简 $R_{4}$ 。

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

解题步骤 3.7

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

点击获取更多步骤...

解题步骤 3.7.1

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

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

解题步骤 3.7.2

化简 $R_{3}$ 。

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

解题步骤 3.8

Perform the row operation $R_{4} = R_{4} + 5 R_{3}$ to make the entry at $4, 3$ a $0$ .

点击获取更多步骤...

解题步骤 3.8.1

Perform the row operation $R_{4} = R_{4} + 5 R_{3}$ to make the entry at $4, 3$ a $0$ .

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

解题步骤 3.8.2

化简 $R_{4}$ 。

$[\begin{array}{c} 1 & - 1 & - 2 & 0 & 4 \\ 0 & 1 & \frac{3}{2} & \frac{1}{2} & 2 \\ 0 & 0 & 1 & - 3 & 4 \\ 0 & 0 & 0 & - 16 & 16 \end{array}]$

解题步骤 3.9

Multiply each element of $R_{4}$ by $- \frac{1}{16}$ to make the entry at $4, 4$ a $1$ .

点击获取更多步骤...

解题步骤 3.9.1

Multiply each element of $R_{4}$ by $- \frac{1}{16}$ to make the entry at $4, 4$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 2 & 0 & 4 \\ 0 & 1 & \frac{3}{2} & \frac{1}{2} & 2 \\ 0 & 0 & 1 & - 3 & 4 \\ - \frac{1}{16} \cdot 0 & - \frac{1}{16} \cdot 0 & - \frac{1}{16} \cdot 0 & - \frac{1}{16} \cdot - 16 & - \frac{1}{16} \cdot 16 \end{array}]$

解题步骤 3.9.2

化简 $R_{4}$ 。

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

解题步骤 3.10

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

点击获取更多步骤...

解题步骤 3.10.1

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

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

解题步骤 3.10.2

化简 $R_{3}$ 。

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

解题步骤 3.11

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

点击获取更多步骤...

解题步骤 3.11.1

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

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

解题步骤 3.11.2

化简 $R_{2}$ 。

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

解题步骤 3.12

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

点击获取更多步骤...

解题步骤 3.12.1

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

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

解题步骤 3.12.2

化简 $R_{2}$ 。

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

解题步骤 3.13

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

点击获取更多步骤...

解题步骤 3.13.1

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

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

解题步骤 3.13.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & - 1 & 0 & 0 & 6 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & - 1 \end{array}]$

解题步骤 3.14

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

点击获取更多步骤...

解题步骤 3.14.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 0 + 0 & 6 + 1 \cdot 1 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & - 1 \end{array}]$

解题步骤 3.14.2

化简 $R_{1}$ 。

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

解题步骤 4

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

$w = 7$

$x = 1$

$y = 1$

$z = - 1$

解题步骤 5

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

$(7, 1, 1, - 1)$

有限数学 示例

有限数学示例