线性代数示例

[\begin{array}{c} 7 & 5 & 0 \\ 4 & 5 & 8 \\ 0 & - 1 & 5 \end{array}]

$[\begin{array}{c} 7 & 5 & 0 \\ 4 & 5 & 8 \\ 0 & - 1 & 5 \end{array}]$

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 row

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} 5 & 8 \\ - 1 & 5 \end{array} |

$| \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} |$

解题步骤 1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

7 | \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} |

$7 | \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} |$

解题步骤 1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

| \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} |

$| \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} |$

解题步骤 1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

- 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} |

$- 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} |$

解题步骤 1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

| \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |

$| \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |$

解题步骤 1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

0 | \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |

$0 | \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |$

解题步骤 1.9

Add the terms together.

7 | \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} | - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0 | \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |

$7 | \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} | - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0 | \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |$

7 | \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} | - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0 | \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |

$7 | \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} | - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0 | \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |$

解题步骤 2

将

0

$0$ 乘以

| \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |

$| \begin{array}{c} 4 & 5 \\ 0 & - 1 \end{array} |$ 。

7 | \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} | - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 | \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} | - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

解题步骤 3

计算

| \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} |

$| \begin{array}{c} 5 & 8 \\ - 1 & 5 \end{array} |$ 。

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解题步骤 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。

7 (5 \cdot 5 - (- 1 \cdot 8)) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 (5 \cdot 5 - (- 1 \cdot 8)) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

解题步骤 3.2

化简行列式。

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解题步骤 3.2.1

化简每一项。

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解题步骤 3.2.1.1

将

5

$5$ 乘以

5

$5$ 。

7 (25 - (- 1 \cdot 8)) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 (25 - (- 1 \cdot 8)) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

解题步骤 3.2.1.2

乘以

- (- 1 \cdot 8)

$- (- 1 \cdot 8)$ 。

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解题步骤 3.2.1.2.1

将

- 1

$- 1$ 乘以

8

$8$ 。

7 (25 - - 8) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 (25 - - 8) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

解题步骤 3.2.1.2.2

将

- 1

$- 1$ 乘以

- 8

$- 8$ 。

7 (25 + 8) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 (25 + 8) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

7 (25 + 8) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 (25 + 8) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

7 (25 + 8) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 (25 + 8) - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

解题步骤 3.2.2

将

25

$25$ 和

8

$8$ 相加。

7 \cdot 33 - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 \cdot 33 - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

7 \cdot 33 - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 \cdot 33 - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

7 \cdot 33 - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0

$7 \cdot 33 - 5 | \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} | + 0$

解题步骤 4

计算

| \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} |

$| \begin{array}{c} 4 & 8 \\ 0 & 5 \end{array} |$ 。

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解题步骤 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。

7 \cdot 33 - 5 (4 \cdot 5 + 0 \cdot 8) + 0

$7 \cdot 33 - 5 (4 \cdot 5 + 0 \cdot 8) + 0$

解题步骤 4.2

化简行列式。

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解题步骤 4.2.1

化简每一项。

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解题步骤 4.2.1.1

将

4

$4$ 乘以

5

$5$ 。

7 \cdot 33 - 5 (20 + 0 \cdot 8) + 0

$7 \cdot 33 - 5 (20 + 0 \cdot 8) + 0$

解题步骤 4.2.1.2

将

0

$0$ 乘以

8

$8$ 。

7 \cdot 33 - 5 (20 + 0) + 0

$7 \cdot 33 - 5 (20 + 0) + 0$

7 \cdot 33 - 5 (20 + 0) + 0

$7 \cdot 33 - 5 (20 + 0) + 0$

解题步骤 4.2.2

将

20

$20$ 和

0

$0$ 相加。

7 \cdot 33 - 5 \cdot 20 + 0

$7 \cdot 33 - 5 \cdot 20 + 0$

7 \cdot 33 - 5 \cdot 20 + 0

$7 \cdot 33 - 5 \cdot 20 + 0$

7 \cdot 33 - 5 \cdot 20 + 0

$7 \cdot 33 - 5 \cdot 20 + 0$

解题步骤 5

化简行列式。

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解题步骤 5.1

化简每一项。

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解题步骤 5.1.1

将

7

$7$ 乘以

33

$33$ 。

231 - 5 \cdot 20 + 0

$231 - 5 \cdot 20 + 0$

解题步骤 5.1.2

将

- 5

$- 5$ 乘以

20

$20$ 。

231 - 100 + 0

$231 - 100 + 0$

231 - 100 + 0

$231 - 100 + 0$

解题步骤 5.2

从

231

$231$ 中减去

100

$100$ 。

131 + 0

$131 + 0$

解题步骤 5.3

将

131

$131$ 和

0

$0$ 相加。

131

$131$

131

$131$

线性代数 示例

线性代数示例