Matemática discreta Ejemplos

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & 3 & 0 & 1 4 & 3 & 0 & 3 1 & 2 & 2 & 4 1 & 2 & 3 & 4 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

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

Paso 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.

Toca para ver más pasos...

Paso 1.1

Consider the corresponding sign chart.

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

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

Paso 1.2

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

-

$-$ position on the sign chart.

Paso 1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ ∣ \begin{matrix} 3 & 0 & 3 2 & 2 & 4 2 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

0 ∣ ∣ ∣ ∣ \begin{matrix} 3 & 0 & 3 2 & 2 & 4 2 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ ∣ \begin{matrix} 4 & 0 & 3 1 & 2 & 4 1 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

- 3 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 0 & 3 1 & 2 & 4 1 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 3 1 & 2 & 4 1 & 2 & 4 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

0 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 3 1 & 2 & 4 1 & 2 & 4 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 1.9

The minor for

a_{14}

$a_{14}$ is the determinant with row

1

$1$ and column

4

$4$ deleted.

∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 1.10

Multiply element

a_{14}

$a_{14}$ by its cofactor.

- 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 1.11

Add the terms together.

0 ∣ ∣ ∣ ∣ \begin{matrix} 3 & 0 & 3 2 & 2 & 4 2 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣ - 3 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 0 & 3 1 & 2 & 4 1 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣ + 0 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 3 1 & 2 & 4 1 & 2 & 4 \end{matrix} ∣ ∣ ∣ ∣ - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 ∣ ∣ ∣ ∣ \begin{matrix} 3 & 0 & 3 2 & 2 & 4 2 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣ - 3 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 0 & 3 1 & 2 & 4 1 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣ + 0 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 3 1 & 2 & 4 1 & 2 & 4 \end{matrix} ∣ ∣ ∣ ∣ - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

Paso 2

Multiplica

0

$0$ por

∣ ∣ ∣ ∣ \begin{matrix} 3 & 0 & 3 2 & 2 & 4 2 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 0 & 3 1 & 2 & 4 1 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣ + 0 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 3 1 & 2 & 4 1 & 2 & 4 \end{matrix} ∣ ∣ ∣ ∣ - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 3

Multiplica

0

$0$ por

∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 3 1 & 2 & 4 1 & 2 & 4 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 0 & 3 1 & 2 & 4 1 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣ + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4

Evalúa

∣ ∣ ∣ ∣ \begin{matrix} 4 & 0 & 3 1 & 2 & 4 1 & 3 & 4 \end{matrix} ∣ ∣ ∣ ∣

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

Toca para ver más pasos...

Paso 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.

Toca para ver más pasos...

Paso 4.1.1

Consider the corresponding sign chart.

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

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

Paso 4.1.2

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

-

$-$ position on the sign chart.

Paso 4.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

∣ ∣ ∣ \begin{matrix} 2 & 4 3 & 4 \end{matrix} ∣ ∣ ∣

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

Paso 4.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

4 ∣ ∣ ∣ \begin{matrix} 2 & 4 3 & 4 \end{matrix} ∣ ∣ ∣

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

Paso 4.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

∣ ∣ ∣ \begin{matrix} 1 & 4 1 & 4 \end{matrix} ∣ ∣ ∣

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

Paso 4.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 1 & 4 1 & 4 \end{matrix} ∣ ∣ ∣

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

Paso 4.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

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

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

Paso 4.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

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

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

Paso 4.1.9

Add the terms together.

0 - 3 (4 ∣ ∣ ∣ \begin{matrix} 2 & 4 3 & 4 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 4 1 & 4 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 (4 ∣ ∣ ∣ \begin{matrix} 2 & 4 3 & 4 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 4 1 & 4 \end{matrix} ∣ ∣ ∣ + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

Paso 4.2

Multiplica

0

$0$ por

∣ ∣ ∣ \begin{matrix} 1 & 4 1 & 4 \end{matrix} ∣ ∣ ∣

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

0 - 3 (4 ∣ ∣ ∣ \begin{matrix} 2 & 4 3 & 4 \end{matrix} ∣ ∣ ∣ + 0 + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.3

Evalúa

∣ ∣ ∣ \begin{matrix} 2 & 4 3 & 4 \end{matrix} ∣ ∣ ∣

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

Toca para ver más pasos...

Paso 4.3.1

El determinante de una matriz

2 \times 2

$2 \times 2$ puede obtenerse usando la fórmula

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

0 - 3 (4 (2 \cdot 4 - 3 \cdot 4) + 0 + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.3.2

Simplifica el determinante.

Toca para ver más pasos...

Paso 4.3.2.1

Simplifica cada término.

Toca para ver más pasos...

Paso 4.3.2.1.1

Multiplica

2

$2$ por

4

$4$ .

0 - 3 (4 (8 - 3 \cdot 4) + 0 + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.3.2.1.2

Multiplica

- 3

$- 3$ por

4

$4$ .

0 - 3 (4 (8 - 12) + 0 + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 (4 (8 - 12) + 0 + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.3.2.2

Resta

12

$12$ de

8

$8$ .

0 - 3 (4 \cdot - 4 + 0 + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 (4 \cdot - 4 + 0 + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 (4 \cdot - 4 + 0 + 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.4

Evalúa

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

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

Toca para ver más pasos...

Paso 4.4.1

El determinante de una matriz

2 \times 2

$2 \times 2$ puede obtenerse usando la fórmula

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

0 - 3 (4 \cdot - 4 + 0 + 3 (1 \cdot 3 - 1 \cdot 2)) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.4.2

Simplifica el determinante.

Toca para ver más pasos...

Paso 4.4.2.1

Simplifica cada término.

Toca para ver más pasos...

Paso 4.4.2.1.1

Multiplica

3

$3$ por

1

$1$ .

0 - 3 (4 \cdot - 4 + 0 + 3 (3 - 1 \cdot 2)) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.4.2.1.2

Multiplica

- 1

$- 1$ por

2

$2$ .

0 - 3 (4 \cdot - 4 + 0 + 3 (3 - 2)) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 (4 \cdot - 4 + 0 + 3 (3 - 2)) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.4.2.2

Resta

2

$2$ de

3

$3$ .

0 - 3 (4 \cdot - 4 + 0 + 3 \cdot 1) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 (4 \cdot - 4 + 0 + 3 \cdot 1) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 (4 \cdot - 4 + 0 + 3 \cdot 1) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.5

Simplifica el determinante.

Toca para ver más pasos...

Paso 4.5.1

Simplifica cada término.

Toca para ver más pasos...

Paso 4.5.1.1

Multiplica

4

$4$ por

- 4

$- 4$ .

0 - 3 (- 16 + 0 + 3 \cdot 1) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.5.1.2

Multiplica

3

$3$ por

1

$1$ .

0 - 3 (- 16 + 0 + 3) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 (- 16 + 0 + 3) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.5.2

Suma

- 16

$- 16$ y

0

$0$ .

0 - 3 (- 16 + 3) + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 4.5.3

Suma

- 16

$- 16$ y

3

$3$ .

0 - 3 \cdot - 13 + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 \cdot - 13 + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

0 - 3 \cdot - 13 + 0 - 1 ∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Paso 5

Evalúa

∣ ∣ ∣ ∣ \begin{matrix} 4 & 3 & 0 1 & 2 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

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

Toca para ver más pasos...

Paso 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.

Toca para ver más pasos...

Paso 5.1.1

Consider the corresponding sign chart.

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

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

Paso 5.1.2

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

-

$-$ position on the sign chart.

Paso 5.1.3

The minor for

a_{11}

$a_{11}$ is the determinant with row

1

$1$ and column

1

$1$ deleted.

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

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

Paso 5.1.4

Multiply element

a_{11}

$a_{11}$ by its cofactor.

4 ∣ ∣ ∣ \begin{matrix} 2 & 2 2 & 3 \end{matrix} ∣ ∣ ∣

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

Paso 5.1.5

The minor for

a_{12}

$a_{12}$ is the determinant with row

1

$1$ and column

2

$2$ deleted.

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

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

Paso 5.1.6

Multiply element

a_{12}

$a_{12}$ by its cofactor.

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

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

Paso 5.1.7

The minor for

a_{13}

$a_{13}$ is the determinant with row

1

$1$ and column

3

$3$ deleted.

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

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

Paso 5.1.8

Multiply element

a_{13}

$a_{13}$ by its cofactor.

0 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 2 \end{matrix} ∣ ∣ ∣

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

Paso 5.1.9

Add the terms together.

0 - 3 \cdot - 13 + 0 - 1 (4 ∣ ∣ ∣ \begin{matrix} 2 & 2 2 & 3 \end{matrix} ∣ ∣ ∣ - 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 2 \end{matrix} ∣ ∣ ∣)

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

0 - 3 \cdot - 13 + 0 - 1 (4 ∣ ∣ ∣ \begin{matrix} 2 & 2 2 & 3 \end{matrix} ∣ ∣ ∣ - 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 0 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 2 \end{matrix} ∣ ∣ ∣)

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

Paso 5.2

Multiplica

0

$0$ por

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

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

0 - 3 \cdot - 13 + 0 - 1 (4 ∣ ∣ ∣ \begin{matrix} 2 & 2 2 & 3 \end{matrix} ∣ ∣ ∣ - 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 0)

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

Paso 5.3

Evalúa

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

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

Toca para ver más pasos...

Paso 5.3.1

El determinante de una matriz

2 \times 2

$2 \times 2$ puede obtenerse usando la fórmula

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

0 - 3 \cdot - 13 + 0 - 1 (4 (2 \cdot 3 - 2 \cdot 2) - 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 0)

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

Paso 5.3.2

Simplifica el determinante.

Toca para ver más pasos...

Paso 5.3.2.1

Simplifica cada término.

Toca para ver más pasos...

Paso 5.3.2.1.1

Multiplica

2

$2$ por

3

$3$ .

0 - 3 \cdot - 13 + 0 - 1 (4 (6 - 2 \cdot 2) - 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 0)

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

Paso 5.3.2.1.2

Multiplica

- 2

$- 2$ por

2

$2$ .

0 - 3 \cdot - 13 + 0 - 1 (4 (6 - 4) - 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 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{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 0)

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

Paso 5.3.2.2

Resta

4

$4$ de

6

$6$ .

0 - 3 \cdot - 13 + 0 - 1 (4 \cdot 2 - 3 ∣ ∣ ∣ \begin{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 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{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 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{matrix} 1 & 2 1 & 3 \end{matrix} ∣ ∣ ∣ + 0)

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

Paso 5.4

Evalúa

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

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

Toca para ver más pasos...

Paso 5.4.1

El determinante de una matriz

2 \times 2

$2 \times 2$ puede obtenerse usando la fórmula

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

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

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

Paso 5.4.2

Simplifica el determinante.

Toca para ver más pasos...

Paso 5.4.2.1

Simplifica cada término.

Toca para ver más pasos...

Paso 5.4.2.1.1

Multiplica

3

$3$ por

1

$1$ .

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

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

Paso 5.4.2.1.2

Multiplica

- 1

$- 1$ por

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)

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

Paso 5.4.2.2

Resta

2

$2$ de

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)

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

Paso 5.5

Simplifica el determinante.

Toca para ver más pasos...

Paso 5.5.1

Simplifica cada término.

Toca para ver más pasos...

Paso 5.5.1.1

Multiplica

4

$4$ por

2

$2$ .

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

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

Paso 5.5.1.2

Multiplica

- 3

$- 3$ por

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)

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

Paso 5.5.2

Resta

3

$3$ de

8

$8$ .

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

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

Paso 5.5.3

Suma

5

$5$ y

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

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

Paso 6

Simplifica el determinante.

Toca para ver más pasos...

Paso 6.1

Simplifica cada término.

Toca para ver más pasos...

Paso 6.1.1

Multiplica

- 3

$- 3$ por

- 13

$- 13$ .

0 + 39 + 0 - 1 \cdot 5

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

Paso 6.1.2

Multiplica

- 1

$- 1$ por

5

$5$ .

0 + 39 + 0 - 5

$0 + 39 + 0 - 5$

0 + 39 + 0 - 5

$0 + 39 + 0 - 5$

Paso 6.2

Suma

0

$0$ y

39

$39$ .

39 + 0 - 5

$39 + 0 - 5$

Paso 6.3

Suma

39

$39$ y

0

$0$ .

39 - 5

$39 - 5$

Paso 6.4

Resta

5

$5$ de

39

$39$ .

34

$34$

34

$34$

Ingresa TU problema