有限数学示例

$(2 x + 5 y + 3 z) = 1$ , $(x + 2 y + 2 z) = 0$ , $(4 x + 9 y + 7 z) = 1$

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

去掉圆括号。

$2 x + 5 y + 3 z = 1$

$(x + 2 y + 2 z) = 0$

$(4 x + 9 y + 7 z) = 1$

解题步骤 1.2

去掉圆括号。

$2 x + 5 y + 3 z = 1$

$x + 2 y + 2 z = 0$

$(4 x + 9 y + 7 z) = 1$

解题步骤 1.3

去掉圆括号。

$2 x + 5 y + 3 z = 1$

$x + 2 y + 2 z = 0$

$4 x + 9 y + 7 z = 1$

$2 x + 5 y + 3 z = 1$

$x + 2 y + 2 z = 0$

$4 x + 9 y + 7 z = 1$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 2 & 5 & 3 & 1 \\ 1 & 2 & 2 & 0 \\ 4 & 9 & 7 & 1 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{5}{2} & \frac{3}{2} & \frac{1}{2} \\ 1 & 2 & 2 & 0 \\ 4 & 9 & 7 & 1 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & \frac{3}{2} & \frac{1}{2} \\ 1 & 2 & 2 & 0 \\ 4 & 9 & 7 & 1 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{5}{2} & \frac{3}{2} & \frac{1}{2} \\ 1 - 1 & 2 - \frac{5}{2} & 2 - \frac{3}{2} & 0 - \frac{1}{2} \\ 4 & 9 & 7 & 1 \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{5}{2} & \frac{3}{2} & \frac{1}{2} \\ 0 & - \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} \\ 4 & 9 & 7 & 1 \end{array}]$

解题步骤 3.3

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

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解题步骤 3.3.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & \frac{3}{2} & \frac{1}{2} \\ 0 & - \frac{1}{2} & \frac{1}{2} & - \frac{1}{2} \\ 4 - 4 \cdot 1 & 9 - 4 (\frac{5}{2}) & 7 - 4 (\frac{3}{2}) & 1 - 4 (\frac{1}{2}) \end{array}]$

解题步骤 3.3.2

化简 $R_{3}$ 。

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

解题步骤 3.4

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

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

解题步骤 3.4.2

化简 $R_{2}$ 。

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

解题步骤 3.5

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

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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 $0$ .

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

解题步骤 3.5.2

化简 $R_{3}$ 。

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

解题步骤 3.6

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

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解题步骤 3.6.1

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

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

解题步骤 3.6.2

化简 $R_{1}$ 。

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

解题步骤 4

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

$x + 4 z = - 2$

$y - z = 1$

$0 = 0$

解题步骤 5

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

$(- 2 - 4 z, 1 + z, z)$

有限数学 示例

有限数学示例