5. Calcule los siguientes determinantes (Sugerencia: emplee algunas propiedades para intentar transformarla en una matriz triangular)

A=[−12 3 104 6 50 2 1 ]

det A=[−12 3 104 6 50 2 1 ]=+ (−12 )|6 5

2 1|−3|4 50 1|+10|4 6

0 2|

det A=+(−12 ) ( (6∗1 )− (2∗5 ) )−3 ( (4∗1 )−(0∗5 ) )+10(( 4∗2 )−(0∗6))det A=−12 ( (6 )−(10 ) )−3(( 4 )−(0))+10 ((8 )−(0))det A=−12 (−4 )−3 (4 )+10 (8)det A=48−12+80det A=116

B=[−2 01 2

0 7−1 0

5 04 2

−1 53 2

]det B=[−2 0

1 20 7

−1 05 04 2

−1 53 2

]¿+(−2 )|2 −1 0

0 −1 52 3 2|−0|1 −1 0

5 −3 54 3 2|+0|1 2 0

5 0 54 2 2|−7|1 2 −1

5 0 −14 2 3 |

−2|2 −1 00 −1 52 3 2|=2|−1 5

3 2|−(−1 )|0 52 2|+0|0 −1

2 3 |

det A=(−2 ) (2 ( (−1∗2 )−(3∗5 ) ) )−(−1 ) ( (0∗2 )−(2∗5 ) )det A=(−2 ) (2 ( (−2 )−(15 ) )−(−1)( (0 )−(10 ) ) )det A=(−2 ) (2 (−17 )−(−1) (−10 ) ) det A=(−2)(−34−10)det A=(−2)(−44 )det A=88

−7|1 2 −15 0 −14 2 3 |=1|0 −1

2 3 |−2|5 −14 3 |+(−1)|5 0

4 2|det A=(−7 ) ( (1¿ (0∗3 )−(2∗−1 ) ) )−(2 ) ( (5∗3 )−(4∗−1 )+ (−1 ) (5∗2 )−(4∗0))det A=(−7 ) (1 )( (0 )−(−2))− (2 )((15 )−(−4 ))+(−1 )((10 )−(0))det A=(−7 ) ( (1 ) (2 ) )−( (2 ) (19 ) )+((−1)(10))det A=(−7 )((2 )−(38 )+(−10))det A=(−7)(−46)det A=322

det B=[−2 01 2

0 7−1 0

5 04 2

−1 53 2

] (88 )+ (322 )=410

C=[8 −13 1

0 4 1−1 2 0

2 20 03 2

−2 5 14 −1 66 −1 1

]det C=[

8 −13 1

0 4 1−1 2 0

2 20 03 2

−2 5 14 −1 66 −1 1

]1

¿8|1 −12 −2

2 05 1

0 42 6

−1 6−1 1

|−(−1 )|3 −12 −1

2 05 1

0 43 6

−1 6−1 1

|+0|3 12 2

2 05 1

0 03 2

−1 6−1 1

|−4|3 12 2

−1 0−2 1

0 03 2

4 66 1

|+1|3 1

2 2−1 2−2 5

0 03 2

4 −16 −1

|8|1 −1

2 −22 05 1

0 42 6

−1 6−1 1

|=256

−(−1 )|3 −12 −1

2 05 1

0 43 6

−1 6−1 1

|=335

−4|3 12 2

−1 0−2 1

0 03 2

4 66 1

|=752

1|3 12 2

−1 2−2 5

0 03 2

4 −16 −1

|=−60

det C=[8 −13 1

0 4 1−1 2 0

2 20 03 2

−2 5 14 −1 66 −1 1

]256+335+752+(−60)=1283

2