初级微积分示例

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

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

解题步骤 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

1

$1$ by its cofactor and add.

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解题步骤 1.1

Consider the corresponding sign chart.

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

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

解题步骤 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 1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

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

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

解题步骤 1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

解题步骤 1.5

The minor for

a_{21}

$a_{21}$ is the determinant with row

2

$2$ and column

1

$1$ deleted.

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

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

解题步骤 1.6

Multiply element

a_{21}

$a_{21}$ by its cofactor.

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

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

解题步骤 1.7

The minor for

a_{31}

$a_{31}$ is the determinant with row

3

$3$ and column

1

$1$ deleted.

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

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

解题步骤 1.8

Multiply element

a_{31}

$a_{31}$ by its cofactor.

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

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

解题步骤 1.9

The minor for

a_{41}

$a_{41}$ is the determinant with row

4

$4$ and column

1

$1$ deleted.

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

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

解题步骤 1.10

Multiply element

a_{41}

$a_{41}$ by its cofactor.

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

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

解题步骤 1.11

Add the terms together.

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

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

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

解题步骤 2

将

0

$0$ 乘以

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

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

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

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

解题步骤 3

计算

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

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

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解题步骤 3.1

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 3.1.1

Consider the corresponding sign chart.

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

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

解题步骤 3.1.2

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

-

$-$ position on the sign chart.

解题步骤 3.1.3

The minor for

a_{31}

$a_{31}$ is the determinant with row

3

$3$ and column

1

$1$ deleted.

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

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

解题步骤 3.1.4

Multiply element

a_{31}

$a_{31}$ by its cofactor.

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

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

解题步骤 3.1.5

The minor for

a_{32}

$a_{32}$ is the determinant with row

3

$3$ and column

2

$2$ deleted.

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

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

解题步骤 3.1.6

Multiply element

a_{32}

$a_{32}$ by its cofactor.

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

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

解题步骤 3.1.7

The minor for

a_{33}

$a_{33}$ is the determinant with row

3

$3$ and column

3

$3$ deleted.

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

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

解题步骤 3.1.8

Multiply element

a_{33}

$a_{33}$ by its cofactor.

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

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

解题步骤 3.1.9

Add the terms together.

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

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

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

解题步骤 3.2

将

0

$0$ 乘以

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

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

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

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

解题步骤 3.3

计算

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

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

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解题步骤 3.3.1

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。

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

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

解题步骤 3.3.2

化简行列式。

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解题步骤 3.3.2.1

化简每一项。

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解题步骤 3.3.2.1.1

将

3

$3$ 乘以

1

$1$ 。

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

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

解题步骤 3.3.2.1.2

将

- 2

$- 2$ 乘以

4

$4$ 。

1 (1 (3 - 8) - 1 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (1 (3 - 8) - 1 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 (1 (3 - 8) - 1 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

解题步骤 3.3.2.2

从

3

$3$ 中减去

8

$8$ 。

1 (1 \cdot - 5 - 1 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (1 \cdot - 5 - 1 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 (1 \cdot - 5 - 1 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

1 (1 \cdot - 5 - 1 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

解题步骤 3.4

计算

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

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

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解题步骤 3.4.1

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。

1 (1 \cdot - 5 - 1 (2 \cdot 3 - 1 \cdot 4) + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (1 \cdot - 5 - 1 (2 \cdot 3 - 1 \cdot 4) + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 3.4.2

化简行列式。

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解题步骤 3.4.2.1

化简每一项。

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解题步骤 3.4.2.1.1

将

2

$2$ 乘以

3

$3$ 。

1 (1 \cdot - 5 - 1 (6 - 1 \cdot 4) + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (1 \cdot - 5 - 1 (6 - 1 \cdot 4) + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 3.4.2.1.2

将

- 1

$- 1$ 乘以

4

$4$ 。

1 (1 \cdot - 5 - 1 (6 - 4) + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (1 \cdot - 5 - 1 (6 - 4) + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 (1 \cdot - 5 - 1 (6 - 4) + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (1 \cdot - 5 - 1 (6 - 4) + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 3.4.2.2

从

6

$6$ 中减去

4

$4$ 。

1 (1 \cdot - 5 - 1 \cdot 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (1 \cdot - 5 - 1 \cdot 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 (1 \cdot - 5 - 1 \cdot 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (1 \cdot - 5 - 1 \cdot 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 (1 \cdot - 5 - 1 \cdot 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (1 \cdot - 5 - 1 \cdot 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 3.5

化简行列式。

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解题步骤 3.5.1

化简每一项。

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解题步骤 3.5.1.1

将

- 5

$- 5$ 乘以

1

$1$ 。

1 (- 5 - 1 \cdot 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (- 5 - 1 \cdot 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 3.5.1.2

将

- 1

$- 1$ 乘以

2

$2$ 。

1 (- 5 - 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (- 5 - 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 (- 5 - 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (- 5 - 2 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 3.5.2

从

- 5

$- 5$ 中减去

2

$2$ 。

1 (- 7 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 (- 7 + 0) + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 3.5.3

将

- 7

$- 7$ 和

0

$0$ 相加。

1 \cdot - 7 + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 1 & 0 \end{array} | - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4

计算

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

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

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解题步骤 4.1

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 4.1.1

Consider the corresponding sign chart.

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

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

解题步骤 4.1.2

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

-

$-$ position on the sign chart.

解题步骤 4.1.3

The minor for

a_{31}

$a_{31}$ is the determinant with row

3

$3$ and column

1

$1$ deleted.

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

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

解题步骤 4.1.4

Multiply element

a_{31}

$a_{31}$ by its cofactor.

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

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

解题步骤 4.1.5

The minor for

a_{32}

$a_{32}$ is the determinant with row

3

$3$ and column

2

$2$ deleted.

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

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

解题步骤 4.1.6

Multiply element

a_{32}

$a_{32}$ by its cofactor.

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

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

解题步骤 4.1.7

The minor for

a_{33}

$a_{33}$ is the determinant with row

3

$3$ and column

3

$3$ deleted.

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

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

解题步骤 4.1.8

Multiply element

a_{33}

$a_{33}$ by its cofactor.

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

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

解题步骤 4.1.9

Add the terms together.

1 \cdot - 7 + 0 + 2 (1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} | - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 2 & 1 \end{array} |) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} | - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 2 & 1 \end{array} |) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 (1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} | - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 2 & 1 \end{array} |) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

解题步骤 4.2

将

0

$0$ 乘以

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

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

1 \cdot - 7 + 0 + 2 (1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} | - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 | \begin{array}{c} 2 & 3 \\ 1 & 4 \end{array} | - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.3

计算

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

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

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解题步骤 4.3.1

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。

1 \cdot - 7 + 0 + 2 (1 (2 \cdot 4 - 1 \cdot 3) - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 (2 \cdot 4 - 1 \cdot 3) - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.3.2

化简行列式。

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解题步骤 4.3.2.1

化简每一项。

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解题步骤 4.3.2.1.1

将

2

$2$ 乘以

4

$4$ 。

1 \cdot - 7 + 0 + 2 (1 (8 - 1 \cdot 3) - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 (8 - 1 \cdot 3) - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.3.2.1.2

将

- 1

$- 1$ 乘以

3

$3$ 。

1 \cdot - 7 + 0 + 2 (1 (8 - 3) - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 (8 - 3) - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 (1 (8 - 3) - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 (8 - 3) - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.3.2.2

从

8

$8$ 中减去

3

$3$ 。

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 | \begin{array}{c} 1 & 3 \\ 2 & 4 \end{array} | + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.4

计算

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

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

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解题步骤 4.4.1

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。

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 (1 \cdot 4 - 2 \cdot 3) + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 (1 \cdot 4 - 2 \cdot 3) + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.4.2

化简行列式。

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解题步骤 4.4.2.1

化简每一项。

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解题步骤 4.4.2.1.1

将

4

$4$ 乘以

1

$1$ 。

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 (4 - 2 \cdot 3) + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 (4 - 2 \cdot 3) + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.4.2.1.2

将

- 2

$- 2$ 乘以

3

$3$ 。

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 (4 - 6) + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 (4 - 6) + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 (4 - 6) + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 (4 - 6) + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.4.2.2

从

4

$4$ 中减去

6

$6$ 。

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 \cdot - 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 \cdot - 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 \cdot - 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 \cdot - 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 \cdot - 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (1 \cdot 5 - 1 \cdot - 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.5

化简行列式。

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解题步骤 4.5.1

化简每一项。

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解题步骤 4.5.1.1

将

5

$5$ 乘以

1

$1$ 。

1 \cdot - 7 + 0 + 2 (5 - 1 \cdot - 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (5 - 1 \cdot - 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.5.1.2

将

- 1

$- 1$ 乘以

- 2

$- 2$ 。

1 \cdot - 7 + 0 + 2 (5 + 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (5 + 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 (5 + 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (5 + 2 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.5.2

将

5

$5$ 和

2

$2$ 相加。

1 \cdot - 7 + 0 + 2 (7 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 (7 + 0) - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 4.5.3

将

7

$7$ 和

0

$0$ 相加。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 | \begin{array}{c} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 2 & 3 \end{array} |$

解题步骤 5

计算

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

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

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解题步骤 5.1

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 5.1.1

Consider the corresponding sign chart.

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

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

解题步骤 5.1.2

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

-

$-$ position on the sign chart.

解题步骤 5.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

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

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

解题步骤 5.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

解题步骤 5.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

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

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

解题步骤 5.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

解题步骤 5.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

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

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

解题步骤 5.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

解题步骤 5.1.9

Add the terms together.

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 | \begin{array}{c} 1 & 4 \\ 2 & 3 \end{array} | - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 | \begin{array}{c} 1 & 4 \\ 2 & 3 \end{array} | - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 | \begin{array}{c} 1 & 4 \\ 2 & 3 \end{array} | - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 | \begin{array}{c} 1 & 4 \\ 2 & 3 \end{array} | - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

解题步骤 5.2

计算

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

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

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解题步骤 5.2.1

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。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 (1 \cdot 3 - 2 \cdot 4) - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 (1 \cdot 3 - 2 \cdot 4) - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

解题步骤 5.2.2

化简行列式。

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解题步骤 5.2.2.1

化简每一项。

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解题步骤 5.2.2.1.1

将

3

$3$ 乘以

1

$1$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 (3 - 2 \cdot 4) - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 (3 - 2 \cdot 4) - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

解题步骤 5.2.2.1.2

将

- 2

$- 2$ 乘以

4

$4$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 (3 - 8) - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 (3 - 8) - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 (3 - 8) - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 (3 - 8) - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

解题步骤 5.2.2.2

从

3

$3$ 中减去

8

$8$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 | \begin{array}{c} 2 & 4 \\ 1 & 3 \end{array} | + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

解题步骤 5.3

计算

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

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

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解题步骤 5.3.1

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。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 (2 \cdot 3 - 1 \cdot 4) + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 (2 \cdot 3 - 1 \cdot 4) + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

解题步骤 5.3.2

化简行列式。

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解题步骤 5.3.2.1

化简每一项。

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解题步骤 5.3.2.1.1

将

2

$2$ 乘以

3

$3$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 (6 - 1 \cdot 4) + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 (6 - 1 \cdot 4) + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

解题步骤 5.3.2.1.2

将

- 1

$- 1$ 乘以

4

$4$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 (6 - 4) + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 (6 - 4) + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 (6 - 4) + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 (6 - 4) + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

解题步骤 5.3.2.2

从

6

$6$ 中减去

4

$4$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} |)$

解题步骤 5.4

计算

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

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

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解题步骤 5.4.1

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。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 (2 \cdot 2 - 1 \cdot 1))

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 (2 \cdot 2 - 1 \cdot 1))$

解题步骤 5.4.2

化简行列式。

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解题步骤 5.4.2.1

化简每一项。

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解题步骤 5.4.2.1.1

将

2

$2$ 乘以

2

$2$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 (4 - 1 \cdot 1))

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 (4 - 1 \cdot 1))$

解题步骤 5.4.2.1.2

将

- 1

$- 1$ 乘以

1

$1$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 (4 - 1))

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 (4 - 1))$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 (4 - 1))

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 (4 - 1))$

解题步骤 5.4.2.2

从

4

$4$ 中减去

1

$1$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 \cdot 3)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 \cdot 3)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 \cdot 3)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 \cdot 3)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 \cdot 3)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (1 \cdot - 5 - 2 \cdot 2 + 3 \cdot 3)$

解题步骤 5.5

化简行列式。

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解题步骤 5.5.1

化简每一项。

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解题步骤 5.5.1.1

将

- 5

$- 5$ 乘以

1

$1$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 5 - 2 \cdot 2 + 3 \cdot 3)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 5 - 2 \cdot 2 + 3 \cdot 3)$

解题步骤 5.5.1.2

将

- 2

$- 2$ 乘以

2

$2$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 5 - 4 + 3 \cdot 3)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 5 - 4 + 3 \cdot 3)$

解题步骤 5.5.1.3

将

3

$3$ 乘以

3

$3$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 5 - 4 + 9)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 5 - 4 + 9)$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 5 - 4 + 9)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 5 - 4 + 9)$

解题步骤 5.5.2

从

- 5

$- 5$ 中减去

4

$4$ 。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 9 + 9)

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 (- 9 + 9)$

解题步骤 5.5.3

将

- 9

$- 9$ 和

9

$9$ 相加。

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 \cdot 0

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 \cdot 0$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 \cdot 0

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 \cdot 0$

1 \cdot - 7 + 0 + 2 \cdot 7 - 2 \cdot 0

$1 \cdot - 7 + 0 + 2 \cdot 7 - 2 \cdot 0$

解题步骤 6

化简行列式。

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解题步骤 6.1

化简每一项。

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解题步骤 6.1.1

将

- 7

$- 7$ 乘以

1

$1$ 。

- 7 + 0 + 2 \cdot 7 - 2 \cdot 0

$- 7 + 0 + 2 \cdot 7 - 2 \cdot 0$

解题步骤 6.1.2

将

2

$2$ 乘以

7

$7$ 。

- 7 + 0 + 14 - 2 \cdot 0

$- 7 + 0 + 14 - 2 \cdot 0$

解题步骤 6.1.3

将

- 2

$- 2$ 乘以

0

$0$ 。

- 7 + 0 + 14 + 0

$- 7 + 0 + 14 + 0$

- 7 + 0 + 14 + 0

$- 7 + 0 + 14 + 0$

解题步骤 6.2

将

- 7

$- 7$ 和

0

$0$ 相加。

- 7 + 14 + 0

$- 7 + 14 + 0$

解题步骤 6.3

将

- 7

$- 7$ 和

14

$14$ 相加。

7 + 0

$7 + 0$

解题步骤 6.4

将

7

$7$ 和

0

$0$ 相加。

7

$7$

7

$7$

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