Questão resolvida - Determine o valor do determinante da matriz A descrita abaixo - Álgebra Linear I - IFRN

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Tiago Lima - Instagram: @professor_disciplinas_exatas
 
• Determine o valor do determinante da matriz A descrita abaixo:
 
A =
5 0 0 0 2 1
2 0 0 0 3 4
3 1 5 7 6 11
6 0 6 1 8 9
7 0 1 0 9 10
1 0 0 0 1 1
Resolução:
 
Vamos usar o médoto do escalonamento ou eliminação gaussiana para obter a matriz 
triangular; o deterinante da matriz é o produto dos elementos da diagonal principal;
 
 
 ∼ 
5 0 0 0 2 1
2 0 0 0 3 4
3 1 5 7 6 11
6 0 6 1 8 9
7 0 1 0 9 10
1 0 0 0 1 1
5 0 0 0 2 1
2 - ⋅ 5
2
5
0 0 0 3 - ⋅ 2
2
5
4 - ⋅ 1
2
5
3 1 5 7 6 11
6 0 6 1 8 9
7 0 1 0 9 10
1 0 0 0 1 1
L L - L2 → 2
2
5
1
∼ ∼ 
5 0 0 0 2 1
0 0 0 0
11
5
18
5
3 1 5 7 6 11
6 0 6 1 8 9
7 0 1 0 9 10
1 0 0 0 1 1
5 0 0 0 2 1
0 0 0 0
11
5
18
5
3 - ⋅ 5
3
5
1 5 7 6 - ⋅ 2
3
5
11 - ⋅ 1
3
5
6 0 6 1 8 9
7 0 1 0 9 10
1 0 0 0 1 1
 
 
L L - L3 → 3
3
5
1
∼ ∼ 
5 0 0 0 2 1
0 0 0 0
11
5
18
5
0 1 5 7
24
5
52
5
6 0 6 1 8 9
7 0 1 0 9 10
1 0 0 0 1 1
5 0 0 0 2 1
0 0 0 0
11
5
18
5
0 1 5 7
24
5
52
5
6- ⋅5
6
5
0 6 1 8- ⋅2
6
5
9- ⋅1
6
5
7 0 1 0 9 10
1 0 0 0 1 1
L L - L4 → 4
6
5
1
 
 
 
 
 
∼ ∼ 
5 0 0 0 2 1
0 0 0 0
11
5
18
5
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
7 0 1 0 9 10
1 0 0 0 1 1
5 0 0 0 2 1
0 0 0 0
11
5
18
5
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
7 - ⋅ 5
7
5
0 1 0 9 - ⋅ 2
7
5
10 - ⋅
7
5
1 0 0 0 1 1
L L - L5 → 5
7
5
1
∼ ∼ 
5 0 0 0 2 1
0 0 0 0
11
5
18
5
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 1 0
31
5
43
5
1 0 0 0 1 1
5 0 0 0 2 1
0 0 0 0
11
5
18
5
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 1 0
31
5
43
5
1 - ⋅ 5
1
5
0 0 0 1 - ⋅ 1
1
5
1 - ⋅
1
5
 
∼ ∼ 
5 0 0 0 2 1
0 0 0 0
11
5
18
5
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 1 0
31
5
43
5
0 0 0 0
3
5
4
5
5 0 0 0 2 1
0 1 5 7
24
5
52
5
0 0 0 0 -
11
5
-
18
5
0 0 6 1
28
5
39
5
0 0 1 0
31
5
43
5
0 0 0 0
3
5
4
5
L L - L6 → 5
1
5
1
L ⟺ - L3 2
 
 
 
 
 
 ∼ 
5 0 0 0 2 1
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 0 0
11
5
18
5
0 0 1 0
31
5
43
5
0 0 0 0
3
5
4
5
 
 ∼ 
5 0 0 0 2 1
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 0 0
11
5
18
5
0 0 1 - ⋅ 6
1
6
0 - ⋅ 1
1
6
- ⋅
31
5
1
6
28
5
- ⋅
43
5
1
6
39
5
0 0 0 0
3
5
4
5
 
 
 
∼ ∼ ∼
5 0 0 0 2 1
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 0 0
11
5
18
5
0 0 0 -
1
6
79
15
73
10
0 0 0 0
3
5
4
5
5 0 0 0 2 1
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 0 -
1
6
79
15
73
10
0 0 0 0 -
11
5
-
18
5
0 0 0 0
3
5
4
5
L L - L5 → 5
1
6
3
L ⟺ -L4 3
L ⟺ - L5 4 L L - -6 → 6
3
11
 
 
 
 
 ∼
5 0 0 0 2 1
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 0 -
1
6
79
15
73
10
0 0 0 0 -
11
5
-
18
5
0 0 0 0 - - ⋅ -
3
5
3
11
11
5
- - ⋅
4
5
3
11
4
5
 
 
Finalmente, a matriz triangular é :
 
 
5 0 0 0 2 1
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 0 -
1
6
79
15
73
10
0 0 0 0 -
11
5
-
18
5
0 0 0 0 0 -
2
11
 
 
 
 
 
 
 
 
 
 
 
Assim, o determinante da matriz A, produto dos termos da diagonal da matriz triangular
 
 
 em vermelho , é dado por :( )
 
 
 detA = = 5 ⋅ 1 ⋅ 6 ⋅ - ⋅ - ⋅ - = - 2
5 0 0 0 2 1
0 1 5 7
24
5
52
5
0 0 6 1
28
5
39
5
0 0 0 -
1
6
79
15
73
10
0 0 0 0 -
11
5
-
18
5
0 0 0 0 0 -
2
11
1
6
11
5
2
11