线性代数示例

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 0 & - 2 3 & - 12 & - 2 & - 6 - 2 & 10 & 2 & 5 - 1 & 6 & 1 & 3 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - 3 & 0 & - 2 \\ 3 & - 12 & - 2 & - 6 \\ - 2 & 10 & 2 & 5 \\ - 1 & 6 & 1 & 3 \end{array}]$

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} - 12 & - 2 & - 6 10 & 2 & 5 6 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} |$

解题步骤 1.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

1 ∣ ∣ ∣ ∣ \begin{matrix} - 12 & - 2 & - 6 10 & 2 & 5 6 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣

$1 | \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} |$

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} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} |$

解题步骤 1.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

3 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣

$3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} |$

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} 3 & - 12 & - 6 - 2 & 10 & 5 - 1 & 6 & 3 \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} 3 & - 12 & - 6 \\ - 2 & 10 & 5 \\ - 1 & 6 & 3 \end{array} |$

解题步骤 1.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

0 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 6 - 2 & 10 & 5 - 1 & 6 & 3 \end{matrix} ∣ ∣ ∣ ∣

$0 | \begin{array}{c} 3 & - 12 & - 6 \\ - 2 & 10 & 5 \\ - 1 & 6 & 3 \end{array} |$

解题步骤 1.1.9

The minor for

a_{14}

$a_{14}$ is the determinant with row

1

$1$ and column

4

$4$ deleted.

∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.1.10

Multiply element

a_{14}

$a_{14}$ by its cofactor.

2 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

$2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.1.11

Add the terms together.

1 ∣ ∣ ∣ ∣ \begin{matrix} - 12 & - 2 & - 6 10 & 2 & 5 6 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 3 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 0 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 6 - 2 & 10 & 5 - 1 & 6 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 2 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

$1 | \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} | + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 | \begin{array}{c} 3 & - 12 & - 6 \\ - 2 & 10 & 5 \\ - 1 & 6 & 3 \end{array} | + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

1 ∣ ∣ ∣ ∣ \begin{matrix} - 12 & - 2 & - 6 10 & 2 & 5 6 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 3 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 0 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 6 - 2 & 10 & 5 - 1 & 6 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 2 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

解题步骤 1.2

将

0

$0$ 乘以

∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 6 - 2 & 10 & 5 - 1 & 6 & 3 \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} 3 & - 12 & - 6 \\ - 2 & 10 & 5 \\ - 1 & 6 & 3 \end{array} |$ 。

1 ∣ ∣ ∣ ∣ \begin{matrix} - 12 & - 2 & - 6 10 & 2 & 5 6 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 3 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 0 + 2 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

$1 | \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} | + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3

计算

∣ ∣ ∣ ∣ \begin{matrix} - 12 & - 2 & - 6 10 & 2 & 5 6 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣

$| \begin{array}{c} - 12 & - 2 & - 6 \\ 10 & 2 & 5 \\ 6 & 1 & 3 \end{array} |$ 。

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解题步骤 1.3.1

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

1

$1$ by its cofactor and add.

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解题步骤 1.3.1.1

Consider the corresponding sign chart.

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

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

解题步骤 1.3.1.2

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

-

$-$ position on the sign chart.

解题步骤 1.3.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 2 & 5 1 & 3 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$

解题步骤 1.3.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

- 12 ∣ ∣ ∣ \begin{matrix} 2 & 5 1 & 3 \end{matrix} ∣ ∣ ∣

$- 12 | \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$

解题步骤 1.3.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 10 & 5 6 & 3 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} |$

解题步骤 1.3.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

2 ∣ ∣ ∣ \begin{matrix} 10 & 5 6 & 3 \end{matrix} ∣ ∣ ∣

$2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} |$

解题步骤 1.3.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

∣ ∣ ∣ \begin{matrix} 10 & 2 6 & 1 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$

解题步骤 1.3.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

- 6 ∣ ∣ ∣ \begin{matrix} 10 & 2 6 & 1 \end{matrix} ∣ ∣ ∣

$- 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$

解题步骤 1.3.1.9

Add the terms together.

1 (- 12 ∣ ∣ ∣ \begin{matrix} 2 & 5 1 & 3 \end{matrix} ∣ ∣ ∣ + 2 ∣ ∣ ∣ \begin{matrix} 10 & 5 6 & 3 \end{matrix} ∣ ∣ ∣ - 6 ∣ ∣ ∣ \begin{matrix} 10 & 2 6 & 1 \end{matrix} ∣ ∣ ∣) + 3 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 0 + 2 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

$1 (- 12 | \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} | + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

1 (- 12 ∣ ∣ ∣ \begin{matrix} 2 & 5 1 & 3 \end{matrix} ∣ ∣ ∣ + 2 ∣ ∣ ∣ \begin{matrix} 10 & 5 6 & 3 \end{matrix} ∣ ∣ ∣ - 6 ∣ ∣ ∣ \begin{matrix} 10 & 2 6 & 1 \end{matrix} ∣ ∣ ∣) + 3 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 0 + 2 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

解题步骤 1.3.2

计算

∣ ∣ ∣ \begin{matrix} 2 & 5 1 & 3 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$ 。

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解题步骤 1.3.2.1

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。

1 (- 12 (2 \cdot 3 - 1 \cdot 5) + 2 ∣ ∣ ∣ \begin{matrix} 10 & 5 6 & 3 \end{matrix} ∣ ∣ ∣ - 6 ∣ ∣ ∣ \begin{matrix} 10 & 2 6 & 1 \end{matrix} ∣ ∣ ∣) + 3 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 0 + 2 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

$1 (- 12 (2 \cdot 3 - 1 \cdot 5) + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.2.2

化简行列式。

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解题步骤 1.3.2.2.1

化简每一项。

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解题步骤 1.3.2.2.1.1

将

2

$2$ 乘以

3

$3$ 。

1 (- 12 (6 - 1 \cdot 5) + 2 ∣ ∣ ∣ \begin{matrix} 10 & 5 6 & 3 \end{matrix} ∣ ∣ ∣ - 6 ∣ ∣ ∣ \begin{matrix} 10 & 2 6 & 1 \end{matrix} ∣ ∣ ∣) + 3 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 0 + 2 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

$1 (- 12 (6 - 1 \cdot 5) + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.2.2.1.2

将

- 1

$- 1$ 乘以

5

$5$ 。

1 (- 12 (6 - 5) + 2 ∣ ∣ ∣ \begin{matrix} 10 & 5 6 & 3 \end{matrix} ∣ ∣ ∣ - 6 ∣ ∣ ∣ \begin{matrix} 10 & 2 6 & 1 \end{matrix} ∣ ∣ ∣) + 3 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & - 6 - 2 & 2 & 5 - 1 & 1 & 3 \end{matrix} ∣ ∣ ∣ ∣ + 0 + 2 ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 12 & - 2 - 2 & 10 & 2 - 1 & 6 & 1 \end{matrix} ∣ ∣ ∣ ∣

$1 (- 12 (6 - 5) + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.2.2.2

从

$6$ 中减去

$5$ 。

$1 (- 12 \cdot 1 + 2 | \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} | - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.3

计算

$| \begin{array}{c} 10 & 5 \\ 6 & 3 \end{array} |$ 。

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解题步骤 1.3.3.1

可以使用公式

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

$2 \times 2$ 矩阵的行列式。

$1 (- 12 \cdot 1 + 2 (10 \cdot 3 - 6 \cdot 5) - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.3.2

化简行列式。

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解题步骤 1.3.3.2.1

化简每一项。

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解题步骤 1.3.3.2.1.1

将

$10$ 乘以

$3$ 。

$1 (- 12 \cdot 1 + 2 (30 - 6 \cdot 5) - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.3.2.1.2

将

$- 6$ 乘以

$5$ 。

$1 (- 12 \cdot 1 + 2 (30 - 30) - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.3.2.2

从

$30$ 中减去

$30$ 。

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.4

计算

$| \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$ 。

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解题步骤 1.3.4.1

可以使用公式

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

$2 \times 2$ 矩阵的行列式。

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 (10 \cdot 1 - 6 \cdot 2)) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.4.2

化简行列式。

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解题步骤 1.3.4.2.1

化简每一项。

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解题步骤 1.3.4.2.1.1

将

$10$ 乘以

$1$ 。

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 (10 - 6 \cdot 2)) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.4.2.1.2

将

$- 6$ 乘以

$2$ 。

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 (10 - 12)) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.4.2.2

从

$10$ 中减去

$12$ 。

$1 (- 12 \cdot 1 + 2 \cdot 0 - 6 \cdot - 2) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.5

化简行列式。

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解题步骤 1.3.5.1

化简每一项。

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解题步骤 1.3.5.1.1

将

$- 12$ 乘以

$1$ 。

$1 (- 12 + 2 \cdot 0 - 6 \cdot - 2) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.5.1.2

将

$2$ 乘以

$0$ 。

$1 (- 12 + 0 - 6 \cdot - 2) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.5.1.3

将

$- 6$ 乘以

$- 2$ 。

$1 (- 12 + 0 + 12) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.5.2

将

$- 12$ 和

$0$ 相加。

$1 (- 12 + 12) + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.3.5.3

将

$- 12$ 和

$12$ 相加。

$1 \cdot 0 + 3 | \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} | + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4

计算

$| \begin{array}{c} 3 & - 2 & - 6 \\ - 2 & 2 & 5 \\ - 1 & 1 & 3 \end{array} |$ 。

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解题步骤 1.4.1

Choose the row or column with the most

$0$ elements. If there are no

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

$1$ by its cofactor and add.

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解题步骤 1.4.1.1

Consider the corresponding sign chart.

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

解题步骤 1.4.1.2

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

$-$ position on the sign chart.

解题步骤 1.4.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$

解题步骤 1.4.1.4

Multiply element

$a_{11}$ by its cofactor.

$3 | \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$

解题步骤 1.4.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} |$

解题步骤 1.4.1.6

Multiply element

$a_{12}$ by its cofactor.

$2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} |$

解题步骤 1.4.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

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

解题步骤 1.4.1.8

Multiply element

$a_{13}$ by its cofactor.

$- 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |$

解题步骤 1.4.1.9

Add the terms together.

$1 \cdot 0 + 3 (3 | \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} | + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.2

计算

$| \begin{array}{c} 2 & 5 \\ 1 & 3 \end{array} |$ 。

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解题步骤 1.4.2.1

可以使用公式

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

$2 \times 2$ 矩阵的行列式。

$1 \cdot 0 + 3 (3 (2 \cdot 3 - 1 \cdot 5) + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.2.2

化简行列式。

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解题步骤 1.4.2.2.1

化简每一项。

点击获取更多步骤...

解题步骤 1.4.2.2.1.1

将

$2$ 乘以

$3$ 。

$1 \cdot 0 + 3 (3 (6 - 1 \cdot 5) + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.2.2.1.2

将

$- 1$ 乘以

$5$ 。

$1 \cdot 0 + 3 (3 (6 - 5) + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.2.2.2

从

$6$ 中减去

$5$ 。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 | \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} | - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.3

计算

$| \begin{array}{c} - 2 & 5 \\ - 1 & 3 \end{array} |$ 。

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解题步骤 1.4.3.1

可以使用公式

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

$2 \times 2$ 矩阵的行列式。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 (- 2 \cdot 3 - (- 1 \cdot 5)) - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.3.2

化简行列式。

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解题步骤 1.4.3.2.1

化简每一项。

点击获取更多步骤...

解题步骤 1.4.3.2.1.1

将

$- 2$ 乘以

$3$ 。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 (- 6 - (- 1 \cdot 5)) - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.3.2.1.2

乘以

$- (- 1 \cdot 5)$ 。

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解题步骤 1.4.3.2.1.2.1

将

$- 1$ 乘以

$5$ 。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 (- 6 - - 5) - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.3.2.1.2.2

将

$- 1$ 乘以

$- 5$ 。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 (- 6 + 5) - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.3.2.2

将

$- 6$ 和

$5$ 相加。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.4

计算

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

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解题步骤 1.4.4.1

可以使用公式

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

$2 \times 2$ 矩阵的行列式。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 (- 2 \cdot 1 - (- 1 \cdot 2))) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.4.2

化简行列式。

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解题步骤 1.4.4.2.1

化简每一项。

点击获取更多步骤...

解题步骤 1.4.4.2.1.1

将

$- 2$ 乘以

$1$ 。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 (- 2 - (- 1 \cdot 2))) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.4.2.1.2

乘以

$- (- 1 \cdot 2)$ 。

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解题步骤 1.4.4.2.1.2.1

将

$- 1$ 乘以

$2$ 。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 (- 2 - - 2)) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.4.2.1.2.2

将

$- 1$ 乘以

$- 2$ 。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 (- 2 + 2)) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.4.2.2

将

$- 2$ 和

$2$ 相加。

$1 \cdot 0 + 3 (3 \cdot 1 + 2 \cdot - 1 - 6 \cdot 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.5

化简行列式。

点击获取更多步骤...

解题步骤 1.4.5.1

化简每一项。

点击获取更多步骤...

解题步骤 1.4.5.1.1

将

$3$ 乘以

$1$ 。

$1 \cdot 0 + 3 (3 + 2 \cdot - 1 - 6 \cdot 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.5.1.2

将

$2$ 乘以

$- 1$ 。

$1 \cdot 0 + 3 (3 - 2 - 6 \cdot 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.5.1.3

将

$- 6$ 乘以

$0$ 。

$1 \cdot 0 + 3 (3 - 2 + 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.5.2

从

$3$ 中减去

$2$ 。

$1 \cdot 0 + 3 (1 + 0) + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.4.5.3

将

$1$ 和

$0$ 相加。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 | \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$

解题步骤 1.5

计算

$| \begin{array}{c} 3 & - 12 & - 2 \\ - 2 & 10 & 2 \\ - 1 & 6 & 1 \end{array} |$ 。

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解题步骤 1.5.1

Choose the row or column with the most

$0$ elements. If there are no

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

$1$ by its cofactor and add.

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解题步骤 1.5.1.1

Consider the corresponding sign chart.

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

解题步骤 1.5.1.2

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

$-$ position on the sign chart.

解题步骤 1.5.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$

解题步骤 1.5.1.4

Multiply element

$a_{11}$ by its cofactor.

$3 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$

解题步骤 1.5.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

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

解题步骤 1.5.1.6

Multiply element

$a_{12}$ by its cofactor.

$12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} |$

解题步骤 1.5.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |$

解题步骤 1.5.1.8

Multiply element

$a_{13}$ by its cofactor.

$- 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |$

解题步骤 1.5.1.9

Add the terms together.

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 | \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} | + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.2

计算

$| \begin{array}{c} 10 & 2 \\ 6 & 1 \end{array} |$ 。

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解题步骤 1.5.2.1

可以使用公式

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

$2 \times 2$ 矩阵的行列式。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 (10 \cdot 1 - 6 \cdot 2) + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.2.2

化简行列式。

点击获取更多步骤...

解题步骤 1.5.2.2.1

化简每一项。

点击获取更多步骤...

解题步骤 1.5.2.2.1.1

将

$10$ 乘以

$1$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 (10 - 6 \cdot 2) + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.2.2.1.2

将

$- 6$ 乘以

$2$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 (10 - 12) + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.2.2.2

从

$10$ 中减去

$12$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 | \begin{array}{c} - 2 & 2 \\ - 1 & 1 \end{array} | - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.3

计算

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

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解题步骤 1.5.3.1

可以使用公式

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

$2 \times 2$ 矩阵的行列式。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 (- 2 \cdot 1 - (- 1 \cdot 2)) - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.3.2

化简行列式。

点击获取更多步骤...

解题步骤 1.5.3.2.1

化简每一项。

点击获取更多步骤...

解题步骤 1.5.3.2.1.1

将

$- 2$ 乘以

$1$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 (- 2 - (- 1 \cdot 2)) - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.3.2.1.2

乘以

$- (- 1 \cdot 2)$ 。

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解题步骤 1.5.3.2.1.2.1

将

$- 1$ 乘以

$2$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 (- 2 - - 2) - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.3.2.1.2.2

将

$- 1$ 乘以

$- 2$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 (- 2 + 2) - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.3.2.2

将

$- 2$ 和

$2$ 相加。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 | \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |)$

解题步骤 1.5.4

计算

$| \begin{array}{c} - 2 & 10 \\ - 1 & 6 \end{array} |$ 。

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解题步骤 1.5.4.1

可以使用公式

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

$2 \times 2$ 矩阵的行列式。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 (- 2 \cdot 6 - (- 1 \cdot 10)))$

解题步骤 1.5.4.2

化简行列式。

点击获取更多步骤...

解题步骤 1.5.4.2.1

化简每一项。

点击获取更多步骤...

解题步骤 1.5.4.2.1.1

将

$- 2$ 乘以

$6$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 (- 12 - (- 1 \cdot 10)))$

解题步骤 1.5.4.2.1.2

乘以

$- (- 1 \cdot 10)$ 。

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解题步骤 1.5.4.2.1.2.1

将

$- 1$ 乘以

$10$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 (- 12 - - 10))$

解题步骤 1.5.4.2.1.2.2

将

$- 1$ 乘以

$- 10$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (3 \cdot - 2 + 12 \cdot 0 - 2 (- 12 + 10))$

解题步骤 1.5.4.2.2

将

$- 12$ 和

$10$ 相加。

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

解题步骤 1.5.5

化简行列式。

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解题步骤 1.5.5.1

化简每一项。

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解题步骤 1.5.5.1.1

将

$3$ 乘以

$- 2$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (- 6 + 12 \cdot 0 - 2 \cdot - 2)$

解题步骤 1.5.5.1.2

将

$12$ 乘以

$0$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (- 6 + 0 - 2 \cdot - 2)$

解题步骤 1.5.5.1.3

将

$- 2$ 乘以

$- 2$ 。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (- 6 + 0 + 4)$

解题步骤 1.5.5.2

将

$- 6$ 和

$0$ 相加。

$1 \cdot 0 + 3 \cdot 1 + 0 + 2 (- 6 + 4)$

解题步骤 1.5.5.3

将

$- 6$ 和

$4$ 相加。

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

解题步骤 1.6

化简行列式。

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解题步骤 1.6.1

化简每一项。

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解题步骤 1.6.1.1

将

$0$ 乘以

$1$ 。

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

解题步骤 1.6.1.2

将

$3$ 乘以

$1$ 。

$0 + 3 + 0 + 2 \cdot - 2$

解题步骤 1.6.1.3

将

$2$ 乘以

$- 2$ 。

$0 + 3 + 0 - 4$

解题步骤 1.6.2

将

$0$ 和

$3$ 相加。

$3 + 0 - 4$

解题步骤 1.6.3

将

$3$ 和

$0$ 相加。

$3 - 4$

解题步骤 1.6.4

从

$3$ 中减去

$4$ 。

$- 1$

解题步骤 2

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

解题步骤 3

Set up a

$4 \times 8$ matrix where the left half is the original matrix and the right half is its identity matrix.

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 3 & - 12 & - 2 & - 6 & 0 & 1 & 0 & 0 \\ - 2 & 10 & 2 & 5 & 0 & 0 & 1 & 0 \\ - 1 & 6 & 1 & 3 & 0 & 0 & 0 & 1 \end{array}]$

解题步骤 4

求行简化阶梯形矩阵。

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解题步骤 4.1

Perform the row operation

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

$2, 1$ a

$0$ .

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解题步骤 4.1.1

Perform the row operation

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

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 3 - 3 \cdot 1 & - 12 - 3 \cdot - 3 & - 2 - 3 \cdot 0 & - 6 - 3 \cdot - 2 & 0 - 3 \cdot 1 & 1 - 3 \cdot 0 & 0 - 3 \cdot 0 & 0 - 3 \cdot 0 \\ - 2 & 10 & 2 & 5 & 0 & 0 & 1 & 0 \\ - 1 & 6 & 1 & 3 & 0 & 0 & 0 & 1 \end{array}]$

解题步骤 4.1.2

化简

$R_{2}$ 。

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 0 & - 3 & - 2 & 0 & - 3 & 1 & 0 & 0 \\ - 2 & 10 & 2 & 5 & 0 & 0 & 1 & 0 \\ - 1 & 6 & 1 & 3 & 0 & 0 & 0 & 1 \end{array}]$

解题步骤 4.2

Perform the row operation

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

$3, 1$ a

$0$ .

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解题步骤 4.2.1

Perform the row operation

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

$3, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 0 & - 3 & - 2 & 0 & - 3 & 1 & 0 & 0 \\ - 2 + 2 \cdot 1 & 10 + 2 \cdot - 3 & 2 + 2 \cdot 0 & 5 + 2 \cdot - 2 & 0 + 2 \cdot 1 & 0 + 2 \cdot 0 & 1 + 2 \cdot 0 & 0 + 2 \cdot 0 \\ - 1 & 6 & 1 & 3 & 0 & 0 & 0 & 1 \end{array}]$

解题步骤 4.2.2

化简

$R_{3}$ 。

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 0 & - 3 & - 2 & 0 & - 3 & 1 & 0 & 0 \\ 0 & 4 & 2 & 1 & 2 & 0 & 1 & 0 \\ - 1 & 6 & 1 & 3 & 0 & 0 & 0 & 1 \end{array}]$

解题步骤 4.3

Perform the row operation

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

$4, 1$ a

$0$ .

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解题步骤 4.3.1

Perform the row operation

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

$4, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 3 & 0 & - 2 & 1 & 0 & 0 & 0 \\ 0 & - 3 & - 2 & 0 & - 3 & 1 & 0 & 0 \\ 0 & 4 & 2 & 1 & 2 & 0 & 1 & 0 \\ - 1 + 1 \cdot 1 & 6 - 3 & 1 + 0 & 3 - 2 & 0 + 1 \cdot 1 & 0 + 0 & 0 + 0 & 1 + 0 \end{array}]$

解题步骤 4.3.2

化简

$R_{4}$ 。

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

解题步骤 4.4

Multiply each element of

$R_{2}$ by

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

$2, 2$ a

$1$ .

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解题步骤 4.4.1

Multiply each element of

$R_{2}$ by

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

$2, 2$ a

$1$ .

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

解题步骤 4.4.2

化简

$R_{2}$ 。

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

解题步骤 4.5

Perform the row operation

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

$3, 2$ a

$0$ .

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解题步骤 4.5.1

Perform the row operation

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

$3, 2$ a

$0$ .

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

解题步骤 4.5.2

化简

$R_{3}$ 。

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

解题步骤 4.6

Perform the row operation

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

$4, 2$ a

$0$ .

点击获取更多步骤...

解题步骤 4.6.1

Perform the row operation

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

$4, 2$ a

$0$ .

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

解题步骤 4.6.2

化简

$R_{4}$ 。

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

解题步骤 4.7

Multiply each element of

$R_{3}$ by

$- \frac{3}{2}$ to make the entry at

$3, 3$ a

$1$ .

点击获取更多步骤...

解题步骤 4.7.1

Multiply each element of

$R_{3}$ by

$- \frac{3}{2}$ to make the entry at

$3, 3$ a

$1$ .

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

解题步骤 4.7.2

化简

$R_{3}$ 。

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

解题步骤 4.8

Perform the row operation

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

$4, 3$ a

$0$ .

点击获取更多步骤...

解题步骤 4.8.1

Perform the row operation

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

$4, 3$ a

$0$ .

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

解题步骤 4.8.2

化简

$R_{4}$ 。

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

解题步骤 4.9

Multiply each element of

$R_{4}$ by

$- 2$ to make the entry at

$4, 4$ a

$1$ .

点击获取更多步骤...

解题步骤 4.9.1

Multiply each element of

$R_{4}$ by

$- 2$ to make the entry at

$4, 4$ a

$1$ .

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

解题步骤 4.9.2

化简

$R_{4}$ 。

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

解题步骤 4.10

Perform the row operation

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

$3, 4$ a

$0$ .

点击获取更多步骤...

解题步骤 4.10.1

Perform the row operation

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

$3, 4$ a

$0$ .

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

解题步骤 4.10.2

化简

$R_{3}$ 。

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

解题步骤 4.11

Perform the row operation

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

$1, 4$ a

$0$ .

点击获取更多步骤...

解题步骤 4.11.1

Perform the row operation

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

$1, 4$ a

$0$ .

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

解题步骤 4.11.2

化简

$R_{1}$ 。

$[\begin{array}{c} 1 & - 3 & 0 & 0 & - 3 & 4 & 6 & - 4 \\ 0 & 1 & \frac{2}{3} & 0 & 1 & - \frac{1}{3} & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & 3 & - 3 \\ 0 & 0 & 0 & 1 & - 2 & 2 & 3 & - 2 \end{array}]$

解题步骤 4.12

Perform the row operation

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

$2, 3$ a

$0$ .

点击获取更多步骤...

解题步骤 4.12.1

Perform the row operation

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

$2, 3$ a

$0$ .

$[\begin{array}{c} 1 & - 3 & 0 & 0 & - 3 & 4 & 6 & - 4 \\ 0 - \frac{2}{3} \cdot 0 & 1 - \frac{2}{3} \cdot 0 & \frac{2}{3} - \frac{2}{3} \cdot 1 & 0 - \frac{2}{3} \cdot 0 & 1 - \frac{2}{3} \cdot 0 & - \frac{1}{3} - \frac{2}{3} \cdot 1 & 0 - \frac{2}{3} \cdot 3 & 0 - \frac{2}{3} \cdot - 3 \\ 0 & 0 & 1 & 0 & 0 & 1 & 3 & - 3 \\ 0 & 0 & 0 & 1 & - 2 & 2 & 3 & - 2 \end{array}]$

解题步骤 4.12.2

化简

$R_{2}$ 。

$[\begin{array}{c} 1 & - 3 & 0 & 0 & - 3 & 4 & 6 & - 4 \\ 0 & 1 & 0 & 0 & 1 & - 1 & - 2 & 2 \\ 0 & 0 & 1 & 0 & 0 & 1 & 3 & - 3 \\ 0 & 0 & 0 & 1 & - 2 & 2 & 3 & - 2 \end{array}]$

解题步骤 4.13

Perform the row operation

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

$1, 2$ a

$0$ .

点击获取更多步骤...

解题步骤 4.13.1

Perform the row operation

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

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 0 + 3 \cdot 0 & 0 + 3 \cdot 0 & - 3 + 3 \cdot 1 & 4 + 3 \cdot - 1 & 6 + 3 \cdot - 2 & - 4 + 3 \cdot 2 \\ 0 & 1 & 0 & 0 & 1 & - 1 & - 2 & 2 \\ 0 & 0 & 1 & 0 & 0 & 1 & 3 & - 3 \\ 0 & 0 & 0 & 1 & - 2 & 2 & 3 & - 2 \end{array}]$

解题步骤 4.13.2

化简

$R_{1}$ 。

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

解题步骤 5

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

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

线性代数 示例

线性代数示例