Algebra lineal
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Transcript of Algebra lineal
![Page 1: Algebra lineal](https://reader036.fdocuments.es/reader036/viewer/2022072002/563db9b2550346aa9a9f0d0d/html5/thumbnails/1.jpg)
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 |
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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
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¿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
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