线性代数示例

⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 1 1 & 4 & 2 3 & 3 & 4 \end{matrix} ⎤ ⎥ ⎦

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

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

1

$1$ 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_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 4 & 2 3 & 4 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 4 & 2 3 & 4 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 2 3 & 4 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

- 1 ∣ ∣ ∣ \begin{matrix} 1 & 2 3 & 4 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.1.9

Add the terms together.

0 ∣ ∣ ∣ \begin{matrix} 4 & 2 3 & 4 \end{matrix} ∣ ∣ ∣ - 1 ∣ ∣ ∣ \begin{matrix} 1 & 2 3 & 4 \end{matrix} ∣ ∣ ∣ + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

0 ∣ ∣ ∣ \begin{matrix} 4 & 2 3 & 4 \end{matrix} ∣ ∣ ∣ - 1 ∣ ∣ ∣ \begin{matrix} 1 & 2 3 & 4 \end{matrix} ∣ ∣ ∣ + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.2

将

0

$0$ 乘以

∣ ∣ ∣ \begin{matrix} 4 & 2 3 & 4 \end{matrix} ∣ ∣ ∣

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

0 - 1 ∣ ∣ ∣ \begin{matrix} 1 & 2 3 & 4 \end{matrix} ∣ ∣ ∣ + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.3

计算

∣ ∣ ∣ \begin{matrix} 1 & 2 3 & 4 \end{matrix} ∣ ∣ ∣

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

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解题步骤 1.3.1

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 - 1 (1 \cdot 4 - 3 \cdot 2) + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

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

4

$4$ 乘以

1

$1$ 。

0 - 1 (4 - 3 \cdot 2) + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.3.2.1.2

将

- 3

$- 3$ 乘以

2

$2$ 。

0 - 1 (4 - 6) + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

0 - 1 (4 - 6) + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.3.2.2

从

4

$4$ 中减去

6

$6$ 。

0 - 1 \cdot - 2 + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

0 - 1 \cdot - 2 + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

0 - 1 \cdot - 2 + 1 ∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

解题步骤 1.4

计算

∣ ∣ ∣ \begin{matrix} 1 & 4 3 & 3 \end{matrix} ∣ ∣ ∣

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

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

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

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

3

$3$ 乘以

1

$1$ 。

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

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

解题步骤 1.4.2.1.2

将

- 3

$- 3$ 乘以

4

$4$ 。

0 - 1 \cdot - 2 + 1 (3 - 12)

$0 - 1 \cdot - 2 + 1 (3 - 12)$

0 - 1 \cdot - 2 + 1 (3 - 12)

$0 - 1 \cdot - 2 + 1 (3 - 12)$

解题步骤 1.4.2.2

从

3

$3$ 中减去

12

$12$ 。

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

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

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

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

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

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

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

- 1

$- 1$ 乘以

- 2

$- 2$ 。

0 + 2 + 1 \cdot - 9

$0 + 2 + 1 \cdot - 9$

解题步骤 1.5.1.2

将

- 9

$- 9$ 乘以

1

$1$ 。

0 + 2 - 9

$0 + 2 - 9$

0 + 2 - 9

$0 + 2 - 9$

解题步骤 1.5.2

将

0

$0$ 和

2

$2$ 相加。

2 - 9

$2 - 9$

解题步骤 1.5.3

从

2

$2$ 中减去

9

$9$ 。

- 7

$- 7$

- 7

$- 7$

- 7

$- 7$

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} 0 & 1 & 1 & 1 & 0 & 0 1 & 4 & 2 & 0 & 1 & 0 3 & 3 & 4 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 4

求行简化阶梯形矩阵。

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解题步骤 4.1

Swap

R_{2}

$R_{2}$ with

R_{1}

$R_{1}$ to put a nonzero entry at

1, 1

$1, 1$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 3 & 3 & 4 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 4.2

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 4.2.1

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{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 3 - 3 \cdot 1 & 3 - 3 \cdot 4 & 4 - 3 \cdot 2 & 0 - 3 \cdot 0 & 0 - 3 \cdot 1 & 1 - 3 \cdot 0 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 4.2.2

化简

R_{3}

$R_{3}$ 。

⎡ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 0 & - 9 & - 2 & 0 & - 3 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 0 & - 9 & - 2 & 0 & - 3 & 1 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 4.3

Perform the row operation

R_{3} = R_{3} + 9 R_{2}

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

3, 2

$3, 2$ a

0

$0$ .

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解题步骤 4.3.1

Perform the row operation

R_{3} = R_{3} + 9 R_{2}

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

3, 2

$3, 2$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 0 + 9 \cdot 0 & - 9 + 9 \cdot 1 & - 2 + 9 \cdot 1 & 0 + 9 \cdot 1 & - 3 + 9 \cdot 0 & 1 + 9 \cdot 0 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 4.3.2

化简

R_{3}

$R_{3}$ 。

⎡ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 0 & 0 & 7 & 9 & - 3 & 1 \end{matrix} ⎤ ⎥ ⎦

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

⎡ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 0 & 0 & 7 & 9 & - 3 & 1 \end{matrix} ⎤ ⎥ ⎦

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

解题步骤 4.4

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{7}

$\frac{1}{7}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

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解题步骤 4.4.1

Multiply each element of

R_{3}

$R_{3}$ by

\frac{1}{7}

$\frac{1}{7}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 \frac{0}{7} & \frac{0}{7} & \frac{7}{7} & \frac{9}{7} & \frac{- 3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.4.2

化简

R_{3}

$R_{3}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 1 & 1 & 0 & 0 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.5

Perform the row operation

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

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

2, 3

$2, 3$ a

0

$0$ .

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解题步骤 4.5.1

Perform the row operation

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

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

2, 3

$2, 3$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 - 0 & 1 - 0 & 1 - 1 & 1 - \frac{9}{7} & 0 + \frac{3}{7} & 0 - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.5.2

化简

R_{2}

$R_{2}$ 。

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 2 & 0 & 1 & 0 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎦

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

解题步骤 4.6

Perform the row operation

R_{1} = R_{1} - 2 R_{3}

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

1, 3

$1, 3$ a

0

$0$ .

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解题步骤 4.6.1

Perform the row operation

R_{1} = R_{1} - 2 R_{3}

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

1, 3

$1, 3$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 - 2 \cdot 0 & 4 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 (\frac{9}{7}) & 1 - 2 (- \frac{3}{7}) & 0 - 2 (\frac{1}{7}) 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

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

解题步骤 4.6.2

化简

R_{1}

$R_{1}$ 。

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 & - \frac{18}{7} & \frac{13}{7} & - \frac{2}{7} 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 4 & 0 & - \frac{18}{7} & \frac{13}{7} & - \frac{2}{7} \\ 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} \\ 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 4 & 0 & - \frac{18}{7} & \frac{13}{7} & - \frac{2}{7} 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

解题步骤 4.7

Perform the row operation

R_{1} = R_{1} - 4 R_{2}

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

1, 2

$1, 2$ a

0

$0$ .

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解题步骤 4.7.1

Perform the row operation

R_{1} = R_{1} - 4 R_{2}

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

1, 2

$1, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 - 4 \cdot 0 & 4 - 4 \cdot 1 & 0 - 4 \cdot 0 & - \frac{18}{7} - 4 (- \frac{2}{7}) & \frac{13}{7} - 4 (\frac{3}{7}) & - \frac{2}{7} - 4 (- \frac{1}{7}) 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 - 4 \cdot 0 & 4 - 4 \cdot 1 & 0 - 4 \cdot 0 & - \frac{18}{7} - 4 (- \frac{2}{7}) & \frac{13}{7} - 4 (\frac{3}{7}) & - \frac{2}{7} - 4 (- \frac{1}{7}) \\ 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} \\ 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{array}]$

解题步骤 4.7.2

化简

R_{1}

$R_{1}$ 。

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & - \frac{10}{7} & \frac{1}{7} & \frac{2}{7} 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & 0 & 0 & - \frac{10}{7} & \frac{1}{7} & \frac{2}{7} \\ 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} \\ 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{array}]$

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & - \frac{10}{7} & \frac{1}{7} & \frac{2}{7} 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & - \frac{10}{7} & \frac{1}{7} & \frac{2}{7} 0 & 1 & 0 & - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} 0 & 0 & 1 & \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

解题步骤 5

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

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - \frac{10}{7} & \frac{1}{7} & \frac{2}{7} - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} - \frac{10}{7} & \frac{1}{7} & \frac{2}{7} \\ - \frac{2}{7} & \frac{3}{7} & - \frac{1}{7} \\ \frac{9}{7} & - \frac{3}{7} & \frac{1}{7} \end{array}]$

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