Finite Math Examples

[\begin{array}{c} 1 & 4 & 5 & 0 \\ 0 & 2 & 1 & 3 \\ 2 & 5 & 4 & 1 \\ 1 & 5 & 0 & 2 \end{array}]

$[\begin{array}{c} 1 & 4 & 5 & 0 \\ 0 & 2 & 1 & 3 \\ 2 & 5 & 4 & 1 \\ 1 & 5 & 0 & 2 \end{array}]$

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} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} |

$| \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} |$

Step 1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

1 | \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} |

$1 | \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} |$

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

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

Step 1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

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} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} |

$| \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} |$

Step 1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} |

$5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} |$

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} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |

$| \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |$

Step 1.10

Multiply element

a_{14}

$a_{14}$ by its cofactor.

0 | \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |

$0 | \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |$

Step 1.11

Add the terms together.

1 | \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} | - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0 | \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |

$1 | \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} | - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0 | \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |$

1 | \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} | - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0 | \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |

Step 2

Multiply

0

$0$ by

| \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |

$| \begin{array}{c} 0 & 2 & 1 \\ 2 & 5 & 4 \\ 1 & 5 & 0 \end{array} |$ .

1 | \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} | - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

Step 3

Evaluate

| \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} |

$| \begin{array}{c} 2 & 1 & 3 \\ 5 & 4 & 1 \\ 5 & 0 & 2 \end{array} |$ .

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Step 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 column

2

$2$ by its cofactor and add.

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Step 3.1.1

Consider the corresponding sign chart.

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

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

Step 3.1.2

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

-

$-$ position on the sign chart.

Step 3.1.3

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

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

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

Step 3.1.4

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

Step 3.1.5

The minor for

a_{22}

$a_{22}$ is the determinant with row

2

$2$ and column

2

$2$ deleted.

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

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

Step 3.1.6

Multiply element

a_{22}

$a_{22}$ by its cofactor.

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

$4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} |$

Step 3.1.7

The minor for

a_{32}

$a_{32}$ is the determinant with row

3

$3$ and column

2

$2$ deleted.

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

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

Step 3.1.8

Multiply element

a_{32}

$a_{32}$ by its cofactor.

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

$0 | \begin{array}{c} 2 & 3 \\ 5 & 1 \end{array} |$

Step 3.1.9

Add the terms together.

1 (- 1 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 5 & 1 \end{array} |) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 5 & 1 \end{array} |) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 (- 1 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 5 & 1 \end{array} |) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

Step 3.2

Multiply

0

$0$ by

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

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

1 (- 1 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.3

Evaluate

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

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

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Step 3.3.1

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

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

1 (- 1 (5 \cdot 2 - 5 \cdot 1) + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 (5 \cdot 2 - 5 \cdot 1) + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.3.2

Simplify the determinant.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Multiply

5

$5$ by

2

$2$ .

1 (- 1 (10 - 5 \cdot 1) + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 (10 - 5 \cdot 1) + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.3.2.1.2

Multiply

- 5

$- 5$ by

1

$1$ .

1 (- 1 (10 - 5) + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 (10 - 5) + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 (- 1 (10 - 5) + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

Step 3.3.2.2

Subtract

5

$5$ from

10

$10$ .

1 (- 1 \cdot 5 + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 \cdot 5 + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 (- 1 \cdot 5 + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

1 (- 1 \cdot 5 + 4 | \begin{array}{c} 2 & 3 \\ 5 & 2 \end{array} | + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

Step 3.4

Evaluate

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

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

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Step 3.4.1

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

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

1 (- 1 \cdot 5 + 4 (2 \cdot 2 - 5 \cdot 3) + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 \cdot 5 + 4 (2 \cdot 2 - 5 \cdot 3) + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.4.2

Simplify the determinant.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Multiply

2

$2$ by

2

$2$ .

1 (- 1 \cdot 5 + 4 (4 - 5 \cdot 3) + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 \cdot 5 + 4 (4 - 5 \cdot 3) + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.4.2.1.2

Multiply

- 5

$- 5$ by

3

$3$ .

1 (- 1 \cdot 5 + 4 (4 - 15) + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 \cdot 5 + 4 (4 - 15) + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 (- 1 \cdot 5 + 4 (4 - 15) + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 \cdot 5 + 4 (4 - 15) + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.4.2.2

Subtract

15

$15$ from

4

$4$ .

1 (- 1 \cdot 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 \cdot 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 (- 1 \cdot 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 \cdot 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 (- 1 \cdot 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 1 \cdot 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.5

Simplify the determinant.

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Step 3.5.1

Simplify each term.

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Step 3.5.1.1

Multiply

- 1

$- 1$ by

5

$5$ .

1 (- 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 5 + 4 \cdot - 11 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.5.1.2

Multiply

4

$4$ by

- 11

$- 11$ .

1 (- 5 - 44 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 5 - 44 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 (- 5 - 44 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 5 - 44 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.5.2

Subtract

44

$44$ from

- 5

$- 5$ .

1 (- 49 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 (- 49 + 0) - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 3.5.3

Add

- 49

$- 49$ and

0

$0$ .

1 \cdot - 49 - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 | \begin{array}{c} 0 & 1 & 3 \\ 2 & 4 & 1 \\ 1 & 0 & 2 \end{array} | + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4

Evaluate

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

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

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

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

Step 4.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

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

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

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

Step 4.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

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

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

Step 4.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

Step 4.1.9

Add the terms together.

1 \cdot - 49 - 4 (0 | \begin{array}{c} 4 & 1 \\ 0 & 2 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 | \begin{array}{c} 4 & 1 \\ 0 & 2 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 (0 | \begin{array}{c} 4 & 1 \\ 0 & 2 \end{array} | - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

Step 4.2

Multiply

0

$0$ by

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

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

1 \cdot - 49 - 4 (0 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.3

Evaluate

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

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

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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

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

1 \cdot - 49 - 4 (0 - 1 (2 \cdot 2 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 (2 \cdot 2 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

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

2

$2$ .

1 \cdot - 49 - 4 (0 - 1 (4 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 (4 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.3.2.1.2

Multiply

- 1

$- 1$ by

1

$1$ .

1 \cdot - 49 - 4 (0 - 1 (4 - 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 (4 - 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 (0 - 1 (4 - 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 (4 - 1) + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.3.2.2

Subtract

1

$1$ from

4

$4$ .

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 | \begin{array}{c} 2 & 4 \\ 1 & 0 \end{array} |) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.4

Evaluate

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

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

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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

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

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (2 \cdot 0 - 1 \cdot 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (2 \cdot 0 - 1 \cdot 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

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

2

$2$ by

0

$0$ .

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (0 - 1 \cdot 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (0 - 1 \cdot 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.4.2.1.2

Multiply

- 1

$- 1$ by

4

$4$ .

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (0 - 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (0 - 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (0 - 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 (0 - 4)) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.4.2.2

Subtract

4

$4$ from

0

$0$ .

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 1 \cdot 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

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

- 1

$- 1$ by

3

$3$ .

1 \cdot - 49 - 4 (0 - 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 3 + 3 \cdot - 4) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.5.1.2

Multiply

3

$3$ by

- 4

$- 4$ .

1 \cdot - 49 - 4 (0 - 3 - 12) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 3 - 12) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 (0 - 3 - 12) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (0 - 3 - 12) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.5.2

Subtract

3

$3$ from

0

$0$ .

1 \cdot - 49 - 4 (- 3 - 12) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 (- 3 - 12) + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 4.5.3

Subtract

12

$12$ from

- 3

$- 3$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 | \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} | + 0$

Step 5

Evaluate

| \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} |

$| \begin{array}{c} 0 & 2 & 3 \\ 2 & 5 & 1 \\ 1 & 5 & 2 \end{array} |$ .

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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} 5 & 1 \\ 5 & 2 \end{array} |

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

Step 5.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

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

$0 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} |$

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

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

Step 5.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

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} 2 & 5 \\ 1 & 5 \end{array} |

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

Step 5.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

Step 5.1.9

Add the terms together.

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 | \begin{array}{c} 5 & 1 \\ 5 & 2 \end{array} | - 2 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.2

Multiply

0

$0$ by

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

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

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 | \begin{array}{c} 2 & 1 \\ 1 & 2 \end{array} | + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.3

Evaluate

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

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

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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

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

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (2 \cdot 2 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (2 \cdot 2 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

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

2

$2$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (4 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (4 - 1 \cdot 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.3.2.1.2

Multiply

- 1

$- 1$ by

1

$1$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (4 - 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (4 - 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (4 - 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 (4 - 1) + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.3.2.2

Subtract

1

$1$ from

4

$4$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 | \begin{array}{c} 2 & 5 \\ 1 & 5 \end{array} |) + 0$

Step 5.4

Evaluate

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

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

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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

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

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (2 \cdot 5 - 1 \cdot 5)) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (2 \cdot 5 - 1 \cdot 5)) + 0$

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

2

$2$ by

5

$5$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (10 - 1 \cdot 5)) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (10 - 1 \cdot 5)) + 0$

Step 5.4.2.1.2

Multiply

- 1

$- 1$ by

5

$5$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (10 - 5)) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (10 - 5)) + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (10 - 5)) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 (10 - 5)) + 0$

Step 5.4.2.2

Subtract

5

$5$ from

10

$10$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 \cdot 5) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 \cdot 5) + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 \cdot 5) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 \cdot 5) + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 \cdot 5) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 2 \cdot 3 + 3 \cdot 5) + 0$

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

- 2

$- 2$ by

3

$3$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 6 + 3 \cdot 5) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 6 + 3 \cdot 5) + 0$

Step 5.5.1.2

Multiply

3

$3$ by

5

$5$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 6 + 15) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 6 + 15) + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 6 + 15) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (0 - 6 + 15) + 0$

Step 5.5.2

Subtract

6

$6$ from

0

$0$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 (- 6 + 15) + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 (- 6 + 15) + 0$

Step 5.5.3

Add

- 6

$- 6$ and

15

$15$ .

1 \cdot - 49 - 4 \cdot - 15 + 5 \cdot 9 + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 \cdot 9 + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 \cdot 9 + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 \cdot 9 + 0$

1 \cdot - 49 - 4 \cdot - 15 + 5 \cdot 9 + 0

$1 \cdot - 49 - 4 \cdot - 15 + 5 \cdot 9 + 0$

Step 6

Simplify the determinant.

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Step 6.1

Simplify each term.

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Step 6.1.1

Multiply

- 49

$- 49$ by

1

$1$ .

- 49 - 4 \cdot - 15 + 5 \cdot 9 + 0

$- 49 - 4 \cdot - 15 + 5 \cdot 9 + 0$

Step 6.1.2

Multiply

- 4

$- 4$ by

- 15

$- 15$ .

- 49 + 60 + 5 \cdot 9 + 0

$- 49 + 60 + 5 \cdot 9 + 0$

Step 6.1.3

Multiply

5

$5$ by

9

$9$ .

- 49 + 60 + 45 + 0

$- 49 + 60 + 45 + 0$

- 49 + 60 + 45 + 0

$- 49 + 60 + 45 + 0$

Step 6.2

Add

- 49

$- 49$ and

60

$60$ .

11 + 45 + 0

$11 + 45 + 0$

Step 6.3

Add

11

$11$ and

45

$45$ .

56 + 0

$56 + 0$

Step 6.4

Add

56

$56$ and

0

$0$ .

56

$56$

56

$56$

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