Notasdeaula-GAAL-Determinantes (1)

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𝑛
𝑀
𝑛
𝑀
 
1
 
 
𝐴 det 𝐴 |𝐴|
2
|
1 0
0 1
| = 1
|
𝑐 𝑑
𝑎 𝑏
| = − |
𝑎 𝑏
𝑐 𝑑
|
|
𝑘𝑎 𝑘𝑏
𝑐 𝑑
| = 𝑘 |
𝑎 𝑏
𝑐 𝑑
| ;
|𝑎 + 𝑎′ 𝑏 + 𝑏′
𝑐 𝑑
| = |
𝑎 𝑏
𝑐 𝑑
| + |𝑎′ 𝑏′
𝑐 𝑑
|.
• 
𝐿𝑖 ⟷ 𝐿𝑗
• 
𝐿𝑖 ⟵ 𝑘𝐿𝑖
𝑘
𝐿𝑖 ⟵ 𝐿𝑖 + 𝑘𝐿𝑗
2
(
𝑎 𝑏
𝑐 𝑑
) 𝐿1 ⟵ 𝐿1 + 𝑘𝐿2 (
𝑎 + 𝑘𝑐 𝑏 + 𝑘𝑑
𝑐 𝑑
)
|
𝑎 + 𝑘𝑐 𝑏 + 𝑘𝑑
𝑐 𝑑
| = |
𝑎 𝑏
𝑐 𝑑
| + |
𝑘𝑐 𝑘𝑑
𝑐 𝑑
| =
= |
𝑎 𝑏
𝑐 𝑑
| + 𝑘 ⋅ |
𝑐 𝑑
𝑐 𝑑
| = |
𝑎 𝑏
𝑐 𝑑
| + 𝑘 ⋅ 0 
2
|
0 0
𝑐 𝑑
| = 0 ⋅ |
𝑎 𝑏
𝑐 𝑑
| = 0.
2
|
𝑎 0
0 𝑑
| = 𝑎 ⋅ |
1 0
0 𝑑
| = 𝑎 ⋅ 𝑑 ⋅ |
1 0
0 1
|.
|
1 0
0 1
| = 1
𝐿𝑖 ⟵ 𝐿𝑖 + 𝑘𝐿𝑗
𝐼𝑛
det 𝐼𝑛 = 1
𝑘
|𝐴|
𝛼 ⋅ |𝐴| = 1,
𝛼
|𝐴| ≠ 0
𝐴
𝑛
• 
𝐴
• 
𝑥
• 
𝑥
|
1 −2
5 −9
3 1
6 3
−1 2
2 8
−6 −2
6 1
|.
|
0 1
1/2 1/2
1 1
1 1/2
1/3 1/3
−1/3 2/3
1/3 0
0 0
|.
𝐴 𝐵
𝐴𝐵
𝐴 𝐵
det(𝐴𝐵) = det 𝐴 ⋅ det 𝐵.
𝐴
det 𝐴−1 =
1
det 𝐴
𝐴−1 ⋅ 𝐴 = 𝐼𝑛
det(𝐴−1 ⋅ 𝐴) = det 𝐼𝑛
det(𝐴−1 ⋅ 𝐴) = det 𝐴−1 ⋅ det 𝐴
det 𝐼𝑛 = 1
𝐴
𝐴𝑇
𝐴
det 𝐴𝑇 = det 𝐴
|
−3 0 7
2 5 1
−1 0 5
|.
|
1 0
3 1
0 −1
2 2
1 0
2 0
−2 1
0 1
|.
|
3 5
1 2
−2 6
−1 1
2 4
3 7
1 5
5 3
|.
|
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
| = −6,
|
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
𝑎 𝑏 𝑐
|
|
3𝑎 3𝑏 3𝑐
−𝑑 −𝑒 −𝑓
4𝑔 4ℎ 4𝑖
|
|
𝑎 + 𝑔 𝑏 + ℎ 𝑐 + 𝑖
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
|
|
−3𝑎 −3𝑏 −3𝑐
𝑑 𝑒 𝑓
𝑔 − 4𝑑 ℎ − 4𝑒 𝑖 − 4𝑓
|
𝐴
3 det 𝐴 = 5
det(2𝐴)
det(𝐴3)
𝐴 𝐵
2 det(𝐴 + 𝐵) ≠ det 𝐴 + det 𝐵
 −40
 −6
−18
−6 72 −6 18
40 125
𝐴 = (
1 0
0 0
) 𝐵 = (
0 0
0 1
)
det(𝐴 + 𝐵) = 1 det 𝐴 =
= det 𝐵 = 0.