Examples

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

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

Step 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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Step 1.1

Consider the corresponding sign chart.

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

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

Step 1.2

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

-

$-$ position on the sign chart.

Step 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} |

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

Step 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} |

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

Step 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} |

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

Step 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} |

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

Step 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} |

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

Step 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} |

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

Step 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} |

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

Step 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} |

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

Step 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} |

Step 2

Multiply

0

$0$ by

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

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

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} |

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

Step 3

Multiply

0

$0$ by

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

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

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} |

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

Step 4

Evaluate

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

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

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Step 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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Step 4.1.1

Consider the corresponding sign chart.

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

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

Step 4.1.2

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

-

$-$ position on the sign chart.

Step 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} |

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

Step 4.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

Step 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} |

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

Step 4.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

Step 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} |

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

Step 4.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

Step 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} |

Step 4.2

Multiply

0

$0$ by

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

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

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} |

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

Step 4.3

Evaluate

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

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

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Step 4.3.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

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

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

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} |

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

Step 4.3.2

Simplify the determinant.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

Multiply

2

$2$ by

4

$4$ .

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} |

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

Step 4.3.2.1.2

Multiply

- 3

$- 3$ by

4

$4$ .

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} |

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

Step 4.3.2.2

Subtract

12

$12$ from

8

$8$ .

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} |

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

Step 4.4

Evaluate

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

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

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Step 4.4.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

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

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

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} |

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

Step 4.4.2

Simplify the determinant.

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Step 4.4.2.1

Simplify each term.

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Step 4.4.2.1.1

Multiply

3

$3$ by

1

$1$ .

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} |

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

Step 4.4.2.1.2

Multiply

- 1

$- 1$ by

2

$2$ .

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} |

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

Step 4.4.2.2

Subtract

2

$2$ from

3

$3$ .

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} |

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

Step 4.5

Simplify the determinant.

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Step 4.5.1

Simplify each term.

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Step 4.5.1.1

Multiply

4

$4$ by

- 4

$- 4$ .

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

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

Step 4.5.1.2

Multiply

3

$3$ by

1

$1$ .

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} |

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

Step 4.5.2

Add

- 16

$- 16$ and

0

$0$ .

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

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

Step 4.5.3

Add

- 16

$- 16$ and

3

$3$ .

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} |

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

Step 5

Evaluate

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

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

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Step 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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Step 5.1.1

Consider the corresponding sign chart.

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

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

Step 5.1.2

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

-

$-$ position on the sign chart.

Step 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} |

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

Step 5.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

Step 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} |

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

Step 5.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

Step 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} |

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

Step 5.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

Step 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} |)

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

Step 5.2

Multiply

0

$0$ by

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

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

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)

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

Step 5.3

Evaluate

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

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

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Step 5.3.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

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

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

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

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

Step 5.3.2

Simplify the determinant.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Multiply

2

$2$ by

3

$3$ .

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

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

Step 5.3.2.1.2

Multiply

- 2

$- 2$ by

2

$2$ .

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)

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

Step 5.3.2.2

Subtract

4

$4$ from

6

$6$ .

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)

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

Step 5.4

Evaluate

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

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

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Step 5.4.1

The determinant of a

2 \times 2

$2 \times 2$ matrix can be found using the formula

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

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

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

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

Step 5.4.2

Simplify the determinant.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Multiply

3

$3$ by

1

$1$ .

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

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

Step 5.4.2.1.2

Multiply

- 1

$- 1$ by

2

$2$ .

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)

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

Step 5.4.2.2

Subtract

2

$2$ from

3

$3$ .

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)

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

Step 5.5

Simplify the determinant.

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Step 5.5.1

Simplify each term.

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Step 5.5.1.1

Multiply

4

$4$ by

2

$2$ .

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

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

Step 5.5.1.2

Multiply

- 3

$- 3$ by

1

$1$ .

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)

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

Step 5.5.2

Subtract

3

$3$ from

8

$8$ .

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

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

Step 5.5.3

Add

5

$5$ and

0

$0$ .

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

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

Step 6

Simplify the determinant.

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Step 6.1

Simplify each term.

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Step 6.1.1

Multiply

- 3

$- 3$ by

- 13

$- 13$ .

0 + 39 + 0 - 1 \cdot 5

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

Step 6.1.2

Multiply

- 1

$- 1$ by

5

$5$ .

0 + 39 + 0 - 5

$0 + 39 + 0 - 5$

0 + 39 + 0 - 5

$0 + 39 + 0 - 5$

Step 6.2

Add

0

$0$ and

39

$39$ .

39 + 0 - 5

$39 + 0 - 5$

Step 6.3

Add

39

$39$ and

0

$0$ .

39 - 5

$39 - 5$

Step 6.4

Subtract

5

$5$ from

39

$39$ .

34

$34$

34

$34$

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