Solucionario De Algebra Lineal 2

1. Determinar una matriz ortogonal simétrica P cuya primera fila sea (1/3, 2/3, 2/3). SLp286e6.93

Luego:

2. Determinar la matriz P que diagonaliza por similaridad a: A = SLp365e8.46b

3.
En el espacio vectorial real V, T:VV es un operador lineal, A = {u1, u2} y B = {w1, w2} son bases. Si w1 = u1 + u2, w2 = 2u1 + 3u2 y T(u1) = 3u1 – 2u2 y T(u2) = u1 +4u2. Hallar la matriz de T relativa a la base B. SLp433e10.43
w1 = u1 + u2 ; w2 = 2u1 + 3u2
T(u1) = 3u1 – 2u2 ; T(u2) = u1 + 4u2
Pero:
Además:
Entonces:
Luego:
4. Determinar los valores de k para que el siguiente sistema de ecuaciones lineales admita: a) solución única; b) ninguna solución; y, c) infinitas soluciones.ARp213e6.32
2x ky + z = 2k + 5
x + y kz = 1
4x + y kz = k
Reordenado las ecuaciones se tiene: (Tratando de evitar tener un pívot función de k)
x | y | z | B | Observaciones |
1 | 1 | –k | –1 | L1 |
4 | 1 | –k | k | L2 |
2 | –k | 1 | –2k + 5 | L3 |
1 | 1 | –k | –1 | L1’ = L1 |
0 | –3 | 3k | k + 4 | L2’ = L2 – 4L1 |
0 | –k – 2 | 2k + 1 | –2k + 7 | L3’ = L4 – 2L1 |1 | 1 | –k | –1 | L1” = L1’ |
0 | –3 | 3k | k + 4 | L2” = L2’ |
0 | 0 | –3(k – 1)(k + 1) | (k – 1)(k + 13) | L3” = 3L3’ – (k + 2)L2’ |

El sistema admite a) Solución única si k 1 k –1; b) Ninguna solución si k = –1; c) Infinitas soluciones si k = 1
5. Hallar una matriz no singular P tal B = PtAP sea diagonal. Así mismo determinar B y la signatura de ASLp161e4.118
C1 | C2 | C3 | C4 | | | | | Observaciones |
1 | 1 | 2 | 3 | 1 | 0 | 0 | 0 | L1 |
1 | 2 | 5 | 1 | 0 | 1 | 0 | 0 | L1 |
2 | 5 | 6 | 9 | 0 | 0 | 1 | 0 | L3 |
3 | 1 | 9 | 11 | 0 | 0 | 0 | 1 | L4 |
1 | 1 | 2 | 3 | 1 | 0 | 0 | 0 | L1’ = L1 |
0 | 1 | 3 | 2 | 1 | 1 | 0 | 0 | L2’ = L2 – L1 |
0 | 3 | 2 | 3 | 2 | 0 | 1 | 0 | L3’ = L3 + 2L1 |
0 | 2 | 3 | 2 | 3 | 0 | 0 | 1| L4’ = L4 + 3L1 |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | C1’ = C1 |
0 | 1 | 3 | 2 | 1 | 1 | 0 | 0 | C2’ = C2 – C1 |
0 | 3 | 2 | 3 | 2 | 0 | 1 | 0 | C3’ = C3 + 2C1 |
0 | 2 | 3 | 2 | 3 | 0 | 0 | 1 | C4’ = C4 + 3C1 |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | L1” = L1’ |
0 | 1 | 3 | 2 | 1 | 1 | 0 | 0 | L2” = L2’ |
0 | 0 | 7 | 9 | 1 | 3 | 1 | 0 | L3” = L3’ + 3L2’ |
0 | 0 | 9 | 2 | 5 | 2 | 0 | 1 | L4”= L4’ – 2L2’ |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | C1” = C1’ |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | C2” = C2’ |
0 | 0 | 7 | 9 | 1 | 3 | 1 | 0 | C3” = C3’ + 3C2’ |
0 | 0 | 9 | 2 | 5 | 2 | 0 | 1 | C4” = C4’ – 2C2’ |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | L1”’ = L1” |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | L2”’ = L2” |
0 | 0 | 7 | 9 | 1 | 3 | 1 | 0 | L3”’ = L3” |
0 | 0 | 0 | 67 | 26 | 13 | 9 | 7 | L4”’ =–7L4” 9L3” |
1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | C1”’ = C1” |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | C2”’ = C2” |
0 | 0 | 7 | 0 | 1 | 3 | 1 | 0 | C3”’ = C3” |
0 | 0 | 0 | 469 | 26 | 13 | 9 | 7 | C4”’ = –7C4” 9C3” |
Entonces:


6. Siendo F:R5R3 la aplicación lineal definida según: SLp388e9.16
F(x, y, z, s, t) = (x + 2y + z –3s + 4t, 2x + 5y + 4z – 5s + 5t, x + 4y + 5z – s – 2t)
Hallaruna base y la dimensión para a) Ker F; y, b) Im F
a) Base para el núcleo de F. La imagen debe ser el vector (0, 0, 0), entonces:
x | y | z | s | t | Observaciones |
1 | 2 | 1 | –3 | 4 | L1 |
2 | 5 | 4 | –5 | 5 | L2 |
1 | 4 | 5 | –1 | –2 | L3 |
1 | 2 | 1 | –3 | 4 | L1’ = L1 |
0 | 1 | 2 | 1 | –3 | L2’ = L2 – 2L1 |
0 | 2 | 4 | 2 | –6 | L3’ = L3 – L1 |
1 | 2 | 1 | –3| 4 | L1” = L1’ |
0 | 1 | 2 | 1 | –3 | L2” = L2’ |
0 | 0 | 0 | 0 | 0 | L3” = L3’ – 2L2’ |
1 | 0 | –3 | –5 | 10 | L1”’ = L1” – 2L2’ |
0 | 1 | 2 | 1 | –3 | L2”’ = L2” |
0 | 0 | 0 | 0 | 0 | L3”’ = L3” |
Con z = 1; s = 0; t = 0; se obtienen: x = 3; y = –2
Con z = 0; s = 1; t = 0; se obtienen: x = 5; y = –1
Con z = 0; s = 0; t = 1; se obtienen: x = –10; y = 3...