有限数学示例

⎡ ⎢ ⎣ \begin{matrix} 3 & 3 & 0 1 & 0 & 3 0 & 2 & 0 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 1

Find the determinant.

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解题步骤 1.1

Choose the row or column with the most

0

$0$ elements. If there are no

0

$0$ elements choose any row or column. Multiply every element in row

3

$3$ by its cofactor and add.

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解题步骤 1.1.1

Consider the corresponding sign chart.

∣ ∣ ∣ ∣ \begin{matrix} + & - & + - & + & - + & - & + \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

解题步骤 1.1.2

The cofactor is the minor with the sign changed if the indices match a

-

$-$ position on the sign chart.

解题步骤 1.1.3

The minor for

a_{31}

$a_{31}$ is the determinant with row

3

$3$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 3 & 0 0 & 3 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 3 & 0 \\ 0 & 3 \end{array} |$

解题步骤 1.1.4

Multiply element

a_{31}

$a_{31}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 3 & 0 0 & 3 \end{matrix} ∣ ∣ ∣

$0 | \begin{array}{c} 3 & 0 \\ 0 & 3 \end{array} |$

解题步骤 1.1.5

The minor for

a_{32}

$a_{32}$ is the determinant with row

3

$3$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 3 & 0 1 & 3 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 3 & 0 \\ 1 & 3 \end{array} |$

解题步骤 1.1.6

Multiply element

a_{32}

$a_{32}$ by its cofactor.

- 2 ∣ ∣ ∣ \begin{matrix} 3 & 0 1 & 3 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.1.7

The minor for

a_{33}

$a_{33}$ is the determinant with row

3

$3$ and column

3

$3$ deleted.

∣ ∣ ∣ \begin{matrix} 3 & 3 1 & 0 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 3 & 3 \\ 1 & 0 \end{array} |$

解题步骤 1.1.8

Multiply element

a_{33}

$a_{33}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 3 & 3 1 & 0 \end{matrix} ∣ ∣ ∣

$0 | \begin{array}{c} 3 & 3 \\ 1 & 0 \end{array} |$

解题步骤 1.1.9

Add the terms together.

0 ∣ ∣ ∣ \begin{matrix} 3 & 0 0 & 3 \end{matrix} ∣ ∣ ∣ - 2 ∣ ∣ ∣ \begin{matrix} 3 & 0 1 & 3 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 3 & 3 1 & 0 \end{matrix} ∣ ∣ ∣

$0 | \begin{array}{c} 3 & 0 \\ 0 & 3 \end{array} | - 2 | \begin{array}{c} 3 & 0 \\ 1 & 3 \end{array} | + 0 | \begin{array}{c} 3 & 3 \\ 1 & 0 \end{array} |$

0 ∣ ∣ ∣ \begin{matrix} 3 & 0 0 & 3 \end{matrix} ∣ ∣ ∣ - 2 ∣ ∣ ∣ \begin{matrix} 3 & 0 1 & 3 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 3 & 3 1 & 0 \end{matrix} ∣ ∣ ∣

$0 | \begin{array}{c} 3 & 0 \\ 0 & 3 \end{array} | - 2 | \begin{array}{c} 3 & 0 \\ 1 & 3 \end{array} | + 0 | \begin{array}{c} 3 & 3 \\ 1 & 0 \end{array} |$

解题步骤 1.2

将

0

$0$ 乘以

∣ ∣ ∣ \begin{matrix} 3 & 0 0 & 3 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 3 & 0 \\ 0 & 3 \end{array} |$ 。

0 - 2 ∣ ∣ ∣ \begin{matrix} 3 & 0 1 & 3 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 3 & 3 1 & 0 \end{matrix} ∣ ∣ ∣

$0 - 2 | \begin{array}{c} 3 & 0 \\ 1 & 3 \end{array} | + 0 | \begin{array}{c} 3 & 3 \\ 1 & 0 \end{array} |$

解题步骤 1.3

将

0

$0$ 乘以

∣ ∣ ∣ \begin{matrix} 3 & 3 1 & 0 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 3 & 3 \\ 1 & 0 \end{array} |$ 。

0 - 2 ∣ ∣ ∣ \begin{matrix} 3 & 0 1 & 3 \end{matrix} ∣ ∣ ∣ + 0

$0 - 2 | \begin{array}{c} 3 & 0 \\ 1 & 3 \end{array} | + 0$

解题步骤 1.4

计算

∣ ∣ ∣ \begin{matrix} 3 & 0 1 & 3 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 3 & 0 \\ 1 & 3 \end{array} |$ 。

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解题步骤 1.4.1

可以使用公式

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 求

2 \times 2

$2 \times 2$ 矩阵的行列式。

0 - 2 (3 \cdot 3 - 1 \cdot 0) + 0

$0 - 2 (3 \cdot 3 - 1 \cdot 0) + 0$

解题步骤 1.4.2

化简行列式。

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解题步骤 1.4.2.1

将

3

$3$ 乘以

3

$3$ 。

0 - 2 (9 - 1 \cdot 0) + 0

$0 - 2 (9 - 1 \cdot 0) + 0$

解题步骤 1.4.2.2

从

9

$9$ 中减去

0

$0$ 。

0 - 2 \cdot 9 + 0

$0 - 2 \cdot 9 + 0$

0 - 2 \cdot 9 + 0

$0 - 2 \cdot 9 + 0$

0 - 2 \cdot 9 + 0

$0 - 2 \cdot 9 + 0$

解题步骤 1.5

化简行列式。

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解题步骤 1.5.1

将

- 2

$- 2$ 乘以

9

$9$ 。

0 - 18 + 0

$0 - 18 + 0$

解题步骤 1.5.2

从

0

$0$ 中减去

18

$18$ 。

- 18 + 0

$- 18 + 0$

解题步骤 1.5.3

将

- 18

$- 18$ 和

0

$0$ 相加。

- 18

$- 18$

- 18

$- 18$

- 18

$- 18$

解题步骤 2

Since the determinant is non-zero, the inverse exists.

解题步骤 3

Set up a

3 \times 6

$3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

⎡ ⎢ ⎣ \begin{matrix} 3 & 3 & 0 & 1 & 0 & 0 1 & 0 & 3 & 0 & 1 & 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 4

求行简化阶梯形矩阵。

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解题步骤 4.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{3}

$\frac{1}{3}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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解题步骤 4.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{3}

$\frac{1}{3}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} \frac{3}{3} & \frac{3}{3} & \frac{0}{3} & \frac{1}{3} & \frac{0}{3} & \frac{0}{3} 1 & 0 & 3 & 0 & 1 & 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.1.2

化简

R_{1}

$R_{1}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 1 & 0 & 3 & 0 & 1 & 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 1 & 0 & 3 & 0 & 1 & 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.2

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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解题步骤 4.2.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 1 - 1 & 0 - 1 & 3 - 0 & 0 - \frac{1}{3} & 1 - 0 & 0 - 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.2.2

化简

R_{2}

$R_{2}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & - 1 & 3 & - \frac{1}{3} & 1 & 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & - 1 & 3 & - \frac{1}{3} & 1 & 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.3

Multiply each element of

R_{2}

$R_{2}$ by

- 1

$- 1$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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解题步骤 4.3.1

Multiply each element of

R_{2}

$R_{2}$ by

- 1

$- 1$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 - 0 & - - 1 & - 1 \cdot 3 & - - \frac{1}{3} & - 1 \cdot 1 & - 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.3.2

化简

R_{2}

$R_{2}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 0 & 2 & 0 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.4

Perform the row operation

R_{3} = R_{3} - 2 R_{2}

$R_{3} = R_{3} - 2 R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

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解题步骤 4.4.1

Perform the row operation

R_{3} = R_{3} - 2 R_{2}

$R_{3} = R_{3} - 2 R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 0 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot - 3 & 0 - 2 (\frac{1}{3}) & 0 - 2 \cdot - 1 & 1 - 2 \cdot 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

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

解题步骤 4.4.2

化简

R_{3}

$R_{3}$ 。

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 0 & 0 & 6 & - \frac{2}{3} & 2 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 \\ 0 & 0 & 6 & - \frac{2}{3} & 2 & 1 \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 0 & 0 & 6 & - \frac{2}{3} & 2 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 \\ 0 & 0 & 6 & - \frac{2}{3} & 2 & 1 \end{array}]$

解题步骤 4.5

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{6}

$\frac{1}{6}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

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解题步骤 4.5.1

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{6}

$\frac{1}{6}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 \frac{0}{6} & \frac{0}{6} & \frac{6}{6} & \frac{- \frac{2}{3}}{6} & \frac{2}{6} & \frac{1}{6} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 \\ \frac{0}{6} & \frac{0}{6} & \frac{6}{6} & \frac{- \frac{2}{3}}{6} & \frac{2}{6} & \frac{1}{6} \end{array}]$

解题步骤 4.5.2

化简

R_{3}

$R_{3}$ 。

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 0 & 0 & 1 & - \frac{1}{9} & \frac{1}{3} & \frac{1}{6} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 \\ 0 & 0 & 1 & - \frac{1}{9} & \frac{1}{3} & \frac{1}{6} \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 0 & 0 & 1 & - \frac{1}{9} & \frac{1}{3} & \frac{1}{6} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 \\ 0 & 1 & - 3 & \frac{1}{3} & - 1 & 0 \\ 0 & 0 & 1 & - \frac{1}{9} & \frac{1}{3} & \frac{1}{6} \end{array}]$

解题步骤 4.6

Perform the row operation

R_{2} = R_{2} + 3 R_{3}

$R_{2} = R_{2} + 3 R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

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解题步骤 4.6.1

Perform the row operation

R_{2} = R_{2} + 3 R_{3}

$R_{2} = R_{2} + 3 R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 + 3 \cdot 0 & 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & \frac{1}{3} + 3 (- \frac{1}{9}) & - 1 + 3 (\frac{1}{3}) & 0 + 3 (\frac{1}{6}) 0 & 0 & 1 & - \frac{1}{9} & \frac{1}{3} & \frac{1}{6} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 \\ 0 + 3 \cdot 0 & 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & \frac{1}{3} + 3 (- \frac{1}{9}) & - 1 + 3 (\frac{1}{3}) & 0 + 3 (\frac{1}{6}) \\ 0 & 0 & 1 & - \frac{1}{9} & \frac{1}{3} & \frac{1}{6} \end{array}]$

解题步骤 4.6.2

化简

R_{2}

$R_{2}$ 。

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 1 & 0 & \frac{1}{3} & 0 & 0 0 & 1 & 0 & 0 & 0 & \frac{1}{2} 0 & 0 & 1 & - \frac{1}{9} & \frac{1}{3} & \frac{1}{6} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

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

解题步骤 4.7

Perform the row operation

$R_{1} = R_{1} - R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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解题步骤 4.7.1

Perform the row operation

$R_{1} = R_{1} - R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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

解题步骤 4.7.2

化简

$R_{1}$ 。

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

解题步骤 5

The right half of the reduced row echelon form is the inverse.

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

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有限数学 示例

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