선형 대수 예제

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

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

단계 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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단계 1.1

Consider the corresponding sign chart.

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

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

단계 1.2

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

-

$-$ position on the sign chart.

단계 1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

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

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

단계 1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

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

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

단계 1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

단계 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 & 3 & 3 \\ 1 & 2 & 4 \\ 1 & 2 & 4 \end{array} |

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

단계 1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

단계 1.9

The minor for

a_{14}

$a_{14}$ is the determinant with row

1

$1$ and column

4

$4$ deleted.

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

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

단계 1.10

Multiply element

a_{14}

$a_{14}$ by its cofactor.

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

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

단계 1.11

Add the terms together.

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

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

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

단계 2

0

$0$ 에

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

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

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

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

단계 3

0

$0$ 에

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

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

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

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

단계 4

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

$| \begin{array}{c} 4 & 0 & 3 \\ 1 & 2 & 4 \\ 1 & 3 & 4 \end{array} |$ 의 값을 구합니다.

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단계 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

1

$1$ by its cofactor and add.

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단계 4.1.1

Consider the corresponding sign chart.

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

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

단계 4.1.2

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

-

$-$ position on the sign chart.

단계 4.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

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

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

단계 4.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

단계 4.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

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

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

단계 4.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

단계 4.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

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

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

단계 4.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

단계 4.1.9

Add the terms together.

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

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

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

단계 4.2

0

$0$ 에

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

$| \begin{array}{c} 1 & 4 \\ 1 & 4 \end{array} |$ 을 곱합니다.

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

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

단계 4.3

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

$| \begin{array}{c} 2 & 4 \\ 3 & 4 \end{array} |$ 의 값을 구합니다.

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단계 4.3.1

2 \times 2

$2 \times 2$ 행렬의 행렬식은

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

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 공식을 이용해 계산합니다.

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

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

단계 4.3.2

행렬식을 간단히 합니다.

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단계 4.3.2.1

각 항을 간단히 합니다.

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단계 4.3.2.1.1

2

$2$ 에

4

$4$ 을 곱합니다.

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

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

단계 4.3.2.1.2

- 3

$- 3$ 에

4

$4$ 을 곱합니다.

0 - 3 (4 (8 - 12) + 0 + 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} |) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |

$0 - 3 (4 (8 - 12) + 0 + 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} |) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$

0 - 3 (4 (8 - 12) + 0 + 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} |) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |

$0 - 3 (4 (8 - 12) + 0 + 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} |) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$

단계 4.3.2.2

8

$8$ 에서

12

$12$ 을 뺍니다.

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

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

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

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

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

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

단계 4.4

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

$| \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} |$ 의 값을 구합니다.

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단계 4.4.1

2 \times 2

$2 \times 2$ 행렬의 행렬식은

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

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 공식을 이용해 계산합니다.

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

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

단계 4.4.2

행렬식을 간단히 합니다.

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단계 4.4.2.1

각 항을 간단히 합니다.

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단계 4.4.2.1.1

3

$3$ 에

1

$1$ 을 곱합니다.

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

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

단계 4.4.2.1.2

- 1

$- 1$ 에

2

$2$ 을 곱합니다.

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

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

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

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

단계 4.4.2.2

3

$3$ 에서

2

$2$ 을 뺍니다.

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

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

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

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

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

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

단계 4.5

행렬식을 간단히 합니다.

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단계 4.5.1

각 항을 간단히 합니다.

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단계 4.5.1.1

4

$4$ 에

- 4

$- 4$ 을 곱합니다.

0 - 3 (- 16 + 0 + 3 \cdot 1) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |

$0 - 3 (- 16 + 0 + 3 \cdot 1) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$

단계 4.5.1.2

3

$3$ 에

1

$1$ 을 곱합니다.

0 - 3 (- 16 + 0 + 3) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |

$0 - 3 (- 16 + 0 + 3) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$

0 - 3 (- 16 + 0 + 3) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |

$0 - 3 (- 16 + 0 + 3) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$

단계 4.5.2

- 16

$- 16$ 를

0

$0$ 에 더합니다.

0 - 3 (- 16 + 3) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |

$0 - 3 (- 16 + 3) + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$

단계 4.5.3

- 16

$- 16$ 를

3

$3$ 에 더합니다.

0 - 3 \cdot - 13 + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |

$0 - 3 \cdot - 13 + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$

0 - 3 \cdot - 13 + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |

$0 - 3 \cdot - 13 + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$

0 - 3 \cdot - 13 + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |

$0 - 3 \cdot - 13 + 0 - 1 | \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$

단계 5

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

$| \begin{array}{c} 4 & 3 & 0 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} |$ 의 값을 구합니다.

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단계 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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단계 5.1.1

Consider the corresponding sign chart.

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

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

단계 5.1.2

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

-

$-$ position on the sign chart.

단계 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} 2 & 2 \\ 2 & 3 \end{array} |

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

단계 5.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

단계 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} 1 & 2 \\ 1 & 3 \end{array} |

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

단계 5.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

단계 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} 1 & 2 \\ 1 & 2 \end{array} |

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

단계 5.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

단계 5.1.9

Add the terms together.

0 - 3 \cdot - 13 + 0 - 1 (4 | \begin{array}{c} 2 & 2 \\ 2 & 3 \end{array} | - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 1 & 2 \end{array} |)

$0 - 3 \cdot - 13 + 0 - 1 (4 | \begin{array}{c} 2 & 2 \\ 2 & 3 \end{array} | - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 1 & 2 \end{array} |)$

0 - 3 \cdot - 13 + 0 - 1 (4 | \begin{array}{c} 2 & 2 \\ 2 & 3 \end{array} | - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 1 & 2 \end{array} |)

$0 - 3 \cdot - 13 + 0 - 1 (4 | \begin{array}{c} 2 & 2 \\ 2 & 3 \end{array} | - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 1 & 2 \end{array} |)$

단계 5.2

0

$0$ 에

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

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

0 - 3 \cdot - 13 + 0 - 1 (4 | \begin{array}{c} 2 & 2 \\ 2 & 3 \end{array} | - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 | \begin{array}{c} 2 & 2 \\ 2 & 3 \end{array} | - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)$

단계 5.3

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

$| \begin{array}{c} 2 & 2 \\ 2 & 3 \end{array} |$ 의 값을 구합니다.

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단계 5.3.1

2 \times 2

$2 \times 2$ 행렬의 행렬식은

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

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 공식을 이용해 계산합니다.

0 - 3 \cdot - 13 + 0 - 1 (4 (2 \cdot 3 - 2 \cdot 2) - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 (2 \cdot 3 - 2 \cdot 2) - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)$

단계 5.3.2

행렬식을 간단히 합니다.

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단계 5.3.2.1

각 항을 간단히 합니다.

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단계 5.3.2.1.1

2

$2$ 에

3

$3$ 을 곱합니다.

0 - 3 \cdot - 13 + 0 - 1 (4 (6 - 2 \cdot 2) - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 (6 - 2 \cdot 2) - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)$

단계 5.3.2.1.2

- 2

$- 2$ 에

2

$2$ 을 곱합니다.

0 - 3 \cdot - 13 + 0 - 1 (4 (6 - 4) - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 (6 - 4) - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)$

0 - 3 \cdot - 13 + 0 - 1 (4 (6 - 4) - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 (6 - 4) - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)$

단계 5.3.2.2

6

$6$ 에서

4

$4$ 을 뺍니다.

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)$

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)$

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | + 0)$

단계 5.4

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

$| \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} |$ 의 값을 구합니다.

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단계 5.4.1

2 \times 2

$2 \times 2$ 행렬의 행렬식은

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

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ 공식을 이용해 계산합니다.

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 (1 \cdot 3 - 1 \cdot 2) + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 (1 \cdot 3 - 1 \cdot 2) + 0)$

단계 5.4.2

행렬식을 간단히 합니다.

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단계 5.4.2.1

각 항을 간단히 합니다.

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단계 5.4.2.1.1

3

$3$ 에

1

$1$ 을 곱합니다.

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 (3 - 1 \cdot 2) + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 (3 - 1 \cdot 2) + 0)$

단계 5.4.2.1.2

- 1

$- 1$ 에

2

$2$ 을 곱합니다.

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 (3 - 2) + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 (3 - 2) + 0)$

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 (3 - 2) + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 (3 - 2) + 0)$

단계 5.4.2.2

3

$3$ 에서

2

$2$ 을 뺍니다.

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 \cdot 1 + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 \cdot 1 + 0)$

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 \cdot 1 + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 \cdot 1 + 0)$

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 \cdot 1 + 0)

$0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 \cdot 1 + 0)$

단계 5.5

행렬식을 간단히 합니다.

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단계 5.5.1

각 항을 간단히 합니다.

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단계 5.5.1.1

4

$4$ 에

2

$2$ 을 곱합니다.

0 - 3 \cdot - 13 + 0 - 1 (8 - 3 \cdot 1 + 0)

$0 - 3 \cdot - 13 + 0 - 1 (8 - 3 \cdot 1 + 0)$

단계 5.5.1.2

- 3

$- 3$ 에

1

$1$ 을 곱합니다.

0 - 3 \cdot - 13 + 0 - 1 (8 - 3 + 0)

$0 - 3 \cdot - 13 + 0 - 1 (8 - 3 + 0)$

0 - 3 \cdot - 13 + 0 - 1 (8 - 3 + 0)

$0 - 3 \cdot - 13 + 0 - 1 (8 - 3 + 0)$

단계 5.5.2

8

$8$ 에서

3

$3$ 을 뺍니다.

0 - 3 \cdot - 13 + 0 - 1 (5 + 0)

$0 - 3 \cdot - 13 + 0 - 1 (5 + 0)$

단계 5.5.3

5

$5$ 를

0

$0$ 에 더합니다.

0 - 3 \cdot - 13 + 0 - 1 \cdot 5

$0 - 3 \cdot - 13 + 0 - 1 \cdot 5$

0 - 3 \cdot - 13 + 0 - 1 \cdot 5

$0 - 3 \cdot - 13 + 0 - 1 \cdot 5$

0 - 3 \cdot - 13 + 0 - 1 \cdot 5

$0 - 3 \cdot - 13 + 0 - 1 \cdot 5$

단계 6

행렬식을 간단히 합니다.

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단계 6.1

각 항을 간단히 합니다.

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단계 6.1.1

- 3

$- 3$ 에

- 13

$- 13$ 을 곱합니다.

0 + 39 + 0 - 1 \cdot 5

$0 + 39 + 0 - 1 \cdot 5$

단계 6.1.2

- 1

$- 1$ 에

5

$5$ 을 곱합니다.

0 + 39 + 0 - 5

$0 + 39 + 0 - 5$

0 + 39 + 0 - 5

$0 + 39 + 0 - 5$

단계 6.2

0

$0$ 를

39

$39$ 에 더합니다.

39 + 0 - 5

$39 + 0 - 5$

단계 6.3

39

$39$ 를

0

$0$ 에 더합니다.

39 - 5

$39 - 5$

단계 6.4

39

$39$ 에서

5

$5$ 을 뺍니다.

34

$34$

34

$34$

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