有限数学示例

$x \cdot 1 + 2 x \cdot 2 + 8 x \cdot 3 = 2$ , $x \cdot 1 + x \cdot 2 + 4 x \cdot 3 = 1$

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

化简每一项。

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解题步骤 1.1.1

将 $x$ 乘以 $1$ 。

$x + 2 x \cdot 2 + 8 x \cdot 3 = 2$

$x \cdot 1 + x \cdot 2 + 4 x \cdot 3 = 1$

解题步骤 1.1.2

将 $2$ 乘以 $2$ 。

$x + 4 x + 8 x \cdot 3 = 2$

$x \cdot 1 + x \cdot 2 + 4 x \cdot 3 = 1$

解题步骤 1.1.3

将 $3$ 乘以 $8$ 。

$x + 4 x + 24 x = 2$

$x \cdot 1 + x \cdot 2 + 4 x \cdot 3 = 1$

$x + 4 x + 24 x = 2$

$x \cdot 1 + x \cdot 2 + 4 x \cdot 3 = 1$

解题步骤 1.2

将 $x$ 和 $4 x$ 相加。

$5 x + 24 x = 2$

$x \cdot 1 + x \cdot 2 + 4 x \cdot 3 = 1$

解题步骤 1.3

将 $5 x$ 和 $24 x$ 相加。

$29 x = 2$

$x \cdot 1 + x \cdot 2 + 4 x \cdot 3 = 1$

解题步骤 1.4

化简每一项。

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解题步骤 1.4.1

将 $x$ 乘以 $1$ 。

$29 x = 2$

$x + x \cdot 2 + 4 x \cdot 3 = 1$

解题步骤 1.4.2

将 $2$ 移到 $x$ 的左侧。

$29 x = 2$

$x + 2 \cdot x + 4 x \cdot 3 = 1$

解题步骤 1.4.3

将 $3$ 乘以 $4$ 。

$29 x = 2$

$x + 2 x + 12 x = 1$

$29 x = 2$

$x + 2 x + 12 x = 1$

解题步骤 1.5

将 $x$ 和 $2 x$ 相加。

$29 x = 2$

$3 x + 12 x = 1$

解题步骤 1.6

将 $3 x$ 和 $12 x$ 相加。

$29 x = 2$

$15 x = 1$

$29 x = 2$

$15 x = 1$

解题步骤 2

Write the system as a matrix.

$[\begin{array}{c} 29 & 2 \\ 15 & 1 \end{array}]$

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 $\frac{1}{29}$ 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 $\frac{1}{29}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{29}{29} & \frac{2}{29} \\ 15 & 1 \end{array}]$

解题步骤 3.1.2

化简 $R_{1}$ 。

$[\begin{array}{c} 1 & \frac{2}{29} \\ 15 & 1 \end{array}]$

$[\begin{array}{c} 1 & \frac{2}{29} \\ 15 & 1 \end{array}]$

解题步骤 3.2

Perform the row operation $R_{2} = R_{2} - 15 R_{1}$ to make the entry at $2, 1$ a $0$ .

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解题步骤 3.2.1

Perform the row operation $R_{2} = R_{2} - 15 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{29} \\ 15 - 15 \cdot 1 & 1 - 15 (\frac{2}{29}) \end{array}]$

解题步骤 3.2.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{29} \\ 0 & - \frac{1}{29} \end{array}]$

$[\begin{array}{c} 1 & \frac{2}{29} \\ 0 & - \frac{1}{29} \end{array}]$

解题步骤 3.3

Multiply each element of $R_{2}$ by $- 29$ to make the entry at $2, 2$ a $1$ .

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解题步骤 3.3.1

Multiply each element of $R_{2}$ by $- 29$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{2}{29} \\ - 29 \cdot 0 & - 29 (- \frac{1}{29}) \end{array}]$

解题步骤 3.3.2

化简 $R_{2}$ 。

$[\begin{array}{c} 1 & \frac{2}{29} \\ 0 & 1 \end{array}]$

$[\begin{array}{c} 1 & \frac{2}{29} \\ 0 & 1 \end{array}]$

解题步骤 3.4

Perform the row operation $R_{1} = R_{1} - \frac{2}{29} R_{2}$ to make the entry at $1, 2$ a $0$ .

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解题步骤 3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{2}{29} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{2}{29} \cdot 0 & \frac{2}{29} - \frac{2}{29} \cdot 1 \\ 0 & 1 \end{array}]$

解题步骤 3.4.2

化简 $R_{1}$ 。

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

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

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

解题步骤 4

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

$x = 0$

$0 = 1$

解题步骤 5

The system is inconsistent so there is no solution.

$无解$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$