Finite Math Examples

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

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

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} 4 & 4 \\ 4 & 3 \end{array} |

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

Step 1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

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

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

Step 1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

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} 2 & 4 \\ 4 & 4 \end{array} |

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

Step 1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

Step 1.9

Add the terms together.

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

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

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

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

Step 2

Multiply

0

$0$ by

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

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

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

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

Step 3

Evaluate

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

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

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Step 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

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

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

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

Step 3.2

Simplify the determinant.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply

4

$4$ by

3

$3$ .

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

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

Step 3.2.1.2

Multiply

- 4

$- 4$ by

4

$4$ .

3 (12 - 16) + 0 + 1 | \begin{array}{c} 2 & 4 \\ 4 & 4 \end{array} |

$3 (12 - 16) + 0 + 1 | \begin{array}{c} 2 & 4 \\ 4 & 4 \end{array} |$

3 (12 - 16) + 0 + 1 | \begin{array}{c} 2 & 4 \\ 4 & 4 \end{array} |

$3 (12 - 16) + 0 + 1 | \begin{array}{c} 2 & 4 \\ 4 & 4 \end{array} |$

Step 3.2.2

Subtract

16

$16$ from

12

$12$ .

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

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

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

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

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

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

Step 4

Evaluate

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

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

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Step 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

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

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

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

Step 4.2

Simplify the determinant.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply

2

$2$ by

4

$4$ .

3 \cdot - 4 + 0 + 1 (8 - 4 \cdot 4)

$3 \cdot - 4 + 0 + 1 (8 - 4 \cdot 4)$

Step 4.2.1.2

Multiply

- 4

$- 4$ by

4

$4$ .

3 \cdot - 4 + 0 + 1 (8 - 16)

$3 \cdot - 4 + 0 + 1 (8 - 16)$

3 \cdot - 4 + 0 + 1 (8 - 16)

$3 \cdot - 4 + 0 + 1 (8 - 16)$

Step 4.2.2

Subtract

16

$16$ from

8

$8$ .

3 \cdot - 4 + 0 + 1 \cdot - 8

$3 \cdot - 4 + 0 + 1 \cdot - 8$

3 \cdot - 4 + 0 + 1 \cdot - 8

$3 \cdot - 4 + 0 + 1 \cdot - 8$

3 \cdot - 4 + 0 + 1 \cdot - 8

$3 \cdot - 4 + 0 + 1 \cdot - 8$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply

3

$3$ by

- 4

$- 4$ .

- 12 + 0 + 1 \cdot - 8

$- 12 + 0 + 1 \cdot - 8$

Step 5.1.2

Multiply

- 8

$- 8$ by

1

$1$ .

- 12 + 0 - 8

$- 12 + 0 - 8$

- 12 + 0 - 8

$- 12 + 0 - 8$

Step 5.2

Add

- 12

$- 12$ and

0

$0$ .

- 12 - 8

$- 12 - 8$

Step 5.3

Subtract

8

$8$ from

- 12

$- 12$ .

- 20

$- 20$

- 20

$- 20$

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