有限数学示例

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

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

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 column

2

$2$ by its cofactor and add.

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解题步骤 1.1.1

Consider the corresponding sign chart.

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

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

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_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

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

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

解题步骤 1.1.4

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

解题步骤 1.1.5

The minor for

a_{22}

$a_{22}$ is the determinant with row

2

$2$ and column

2

$2$ deleted.

| \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} |

$| \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} |$

解题步骤 1.1.6

Multiply element

a_{22}

$a_{22}$ by its cofactor.

1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} |

$1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} |$

解题步骤 1.1.7

The minor for

a_{32}

$a_{32}$ is the determinant with row

3

$3$ and column

2

$2$ deleted.

| \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |

$| \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |$

解题步骤 1.1.8

Multiply element

a_{32}

$a_{32}$ by its cofactor.

0 | \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |

$0 | \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |$

解题步骤 1.1.9

Add the terms together.

- 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} | + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0 | \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |

$- 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} | + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0 | \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |$

- 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} | + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0 | \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |

$- 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} | + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0 | \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |$

解题步骤 1.2

将

0

$0$ 乘以

| \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |

$| \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |$ 。

- 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} | + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0

$- 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} | + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0$

解题步骤 1.3

计算

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

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

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解题步骤 1.3.1

可以使用公式

| \begin{array}{c} a & b \\ c & d \end{array} | = 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。

- 3 (1 \cdot 2 - 3 \cdot 2) + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0

$- 3 (1 \cdot 2 - 3 \cdot 2) + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0$

解题步骤 1.3.2

化简行列式。

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解题步骤 1.3.2.1

化简每一项。

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解题步骤 1.3.2.1.1

将

2

$2$ 乘以

1

$1$ 。

- 3 (2 - 3 \cdot 2) + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0

$- 3 (2 - 3 \cdot 2) + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0$

解题步骤 1.3.2.1.2

将

- 3

$- 3$ 乘以

2

$2$ 。

- 3 (2 - 6) + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0

$- 3 (2 - 6) + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0$

- 3 (2 - 6) + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0

$- 3 (2 - 6) + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0$

解题步骤 1.3.2.2

从

2

$2$ 中减去

6

$6$ 。

- 3 \cdot - 4 + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0

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

- 3 \cdot - 4 + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0

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

- 3 \cdot - 4 + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0

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

解题步骤 1.4

计算

| \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} |

$| \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} |$ 。

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解题步骤 1.4.1

可以使用公式

| \begin{array}{c} a & b \\ c & d \end{array} | = 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。

- 3 \cdot - 4 + 1 (4 \cdot 2 - 3 \cdot 4) + 0

$- 3 \cdot - 4 + 1 (4 \cdot 2 - 3 \cdot 4) + 0$

解题步骤 1.4.2

化简行列式。

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解题步骤 1.4.2.1

化简每一项。

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解题步骤 1.4.2.1.1

将

4

$4$ 乘以

2

$2$ 。

- 3 \cdot - 4 + 1 (8 - 3 \cdot 4) + 0

$- 3 \cdot - 4 + 1 (8 - 3 \cdot 4) + 0$

解题步骤 1.4.2.1.2

将

- 3

$- 3$ 乘以

4

$4$ 。

- 3 \cdot - 4 + 1 (8 - 12) + 0

$- 3 \cdot - 4 + 1 (8 - 12) + 0$

- 3 \cdot - 4 + 1 (8 - 12) + 0

$- 3 \cdot - 4 + 1 (8 - 12) + 0$

解题步骤 1.4.2.2

从

8

$8$ 中减去

12

$12$ 。

- 3 \cdot - 4 + 1 \cdot - 4 + 0

$- 3 \cdot - 4 + 1 \cdot - 4 + 0$

- 3 \cdot - 4 + 1 \cdot - 4 + 0

$- 3 \cdot - 4 + 1 \cdot - 4 + 0$

- 3 \cdot - 4 + 1 \cdot - 4 + 0

$- 3 \cdot - 4 + 1 \cdot - 4 + 0$

解题步骤 1.5

化简行列式。

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解题步骤 1.5.1

化简每一项。

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解题步骤 1.5.1.1

将

- 3

$- 3$ 乘以

- 4

$- 4$ 。

12 + 1 \cdot - 4 + 0

$12 + 1 \cdot - 4 + 0$

解题步骤 1.5.1.2

将

- 4

$- 4$ 乘以

1

$1$ 。

12 - 4 + 0

$12 - 4 + 0$

12 - 4 + 0

$12 - 4 + 0$

解题步骤 1.5.2

从

12

$12$ 中减去

4

$4$ 。

8 + 0

$8 + 0$

解题步骤 1.5.3

将

8

$8$ 和

0

$0$ 相加。

8

$8$

8

$8$

8

$8$

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{array}{c} 4 & 3 & 4 & 1 & 0 & 0 \\ 1 & 1 & 2 & 0 & 1 & 0 \\ 3 & 0 & 2 & 0 & 0 & 1 \end{array}]

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

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}{4}

$\frac{1}{4}$ 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}{4}

$\frac{1}{4}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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

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

解题步骤 4.1.2

化简

R_{1}

$R_{1}$ 。

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

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

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

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

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

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

解题步骤 4.2.2

化简

R_{2}

$R_{2}$ 。

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

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

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

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

解题步骤 4.3

Perform the row operation

R_{3} = R_{3} - 3 R_{1}

$R_{3} = R_{3} - 3 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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解题步骤 4.3.1

Perform the row operation

R_{3} = R_{3} - 3 R_{1}

$R_{3} = R_{3} - 3 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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

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

解题步骤 4.3.2

化简

R_{3}

$R_{3}$ 。

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

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

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

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

解题步骤 4.4

Multiply each element of

R_{2}

$R_{2}$ by

4

$4$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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解题步骤 4.4.1

Multiply each element of

R_{2}

$R_{2}$ by

4

$4$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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

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

解题步骤 4.4.2

化简

R_{2}

$R_{2}$ 。

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

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

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

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

解题步骤 4.5

Perform the row operation

R_{3} = R_{3} + \frac{9}{4} R_{2}

$R_{3} = R_{3} + \frac{9}{4} R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

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解题步骤 4.5.1

Perform the row operation

R_{3} = R_{3} + \frac{9}{4} R_{2}

$R_{3} = R_{3} + \frac{9}{4} R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 + \frac{9}{4} \cdot 0 & - \frac{9}{4} + \frac{9}{4} \cdot 1 & - 1 + \frac{9}{4} \cdot 4 & - \frac{3}{4} + \frac{9}{4} \cdot - 1 & 0 + \frac{9}{4} \cdot 4 & 1 + \frac{9}{4} \cdot 0 \end{array}]

$[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 + \frac{9}{4} \cdot 0 & - \frac{9}{4} + \frac{9}{4} \cdot 1 & - 1 + \frac{9}{4} \cdot 4 & - \frac{3}{4} + \frac{9}{4} \cdot - 1 & 0 + \frac{9}{4} \cdot 4 & 1 + \frac{9}{4} \cdot 0 \end{array}]$

解题步骤 4.5.2

化简

R_{3}

$R_{3}$ 。

[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 & 0 & 8 & - 3 & 9 & 1 \end{array}]

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

[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 & 0 & 8 & - 3 & 9 & 1 \end{array}]

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

解题步骤 4.6

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{8}

$\frac{1}{8}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

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解题步骤 4.6.1

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{8}

$\frac{1}{8}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ \frac{0}{8} & \frac{0}{8} & \frac{8}{8} & \frac{- 3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]

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

解题步骤 4.6.2

化简

R_{3}

$R_{3}$ 。

[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]

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

[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]

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

解题步骤 4.7

Perform the row operation

R_{2} = R_{2} - 4 R_{3}

$R_{2} = R_{2} - 4 R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

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解题步骤 4.7.1

Perform the row operation

R_{2} = R_{2} - 4 R_{3}

$R_{2} = R_{2} - 4 R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 - 4 \cdot 0 & 1 - 4 \cdot 0 & 4 - 4 \cdot 1 & - 1 - 4 (- \frac{3}{8}) & 4 - 4 (\frac{9}{8}) & 0 - 4 (\frac{1}{8}) \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]

$[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 - 4 \cdot 0 & 1 - 4 \cdot 0 & 4 - 4 \cdot 1 & - 1 - 4 (- \frac{3}{8}) & 4 - 4 (\frac{9}{8}) & 0 - 4 (\frac{1}{8}) \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]$

解题步骤 4.7.2

化简

R_{2}

$R_{2}$ 。

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

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

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

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

解题步骤 4.8

Perform the row operation

R_{1} = R_{1} - R_{3}

$R_{1} = R_{1} - R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

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解题步骤 4.8.1

Perform the row operation

R_{1} = R_{1} - R_{3}

$R_{1} = R_{1} - R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

[\begin{array}{c} 1 - 0 & \frac{3}{4} - 0 & 1 - 1 & \frac{1}{4} + \frac{3}{8} & 0 - \frac{9}{8} & 0 - \frac{1}{8} \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]

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

解题步骤 4.8.2

化简

R_{1}

$R_{1}$ 。

[\begin{array}{c} 1 & \frac{3}{4} & 0 & \frac{5}{8} & - \frac{9}{8} & - \frac{1}{8} \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]

$[\begin{array}{c} 1 & \frac{3}{4} & 0 & \frac{5}{8} & - \frac{9}{8} & - \frac{1}{8} \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]$

[\begin{array}{c} 1 & \frac{3}{4} & 0 & \frac{5}{8} & - \frac{9}{8} & - \frac{1}{8} \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]

解题步骤 4.9

Perform the row operation

R_{1} = R_{1} - \frac{3}{4} R_{2}

$R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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解题步骤 4.9.1

Perform the row operation

R_{1} = R_{1} - \frac{3}{4} R_{2}

$R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & 0 - \frac{3}{4} \cdot 0 & \frac{5}{8} - \frac{3}{4} \cdot \frac{1}{2} & - \frac{9}{8} - \frac{3}{4} (- \frac{1}{2}) & - \frac{1}{8} - \frac{3}{4} (- \frac{1}{2}) \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]

$[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & 0 - \frac{3}{4} \cdot 0 & \frac{5}{8} - \frac{3}{4} \cdot \frac{1}{2} & - \frac{9}{8} - \frac{3}{4} (- \frac{1}{2}) & - \frac{1}{8} - \frac{3}{4} (- \frac{1}{2}) \\ 0 & 1 & 0 & \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ 0 & 0 & 1 & - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]$

解题步骤 4.9.2

化简

R_{1}

$R_{1}$ 。

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

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

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

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

解题步骤 5

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

[\begin{array}{c} \frac{1}{4} & - \frac{3}{4} & \frac{1}{4} \\ \frac{1}{2} & - \frac{1}{2} & - \frac{1}{2} \\ - \frac{3}{8} & \frac{9}{8} & \frac{1}{8} \end{array}]

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

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