Ejercicios De Pau De Matrices
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Ejercicio 2.1. Murcia. Septiembre 2008. Dada la matriz A=
1 2 2 1
, encontrar una matriz B tal que A•B =
0 3 3 0
Solución: A•B =
0 3 3 0
→ B = A-1•
0 3 3 0
Utilizamos el método de Gauss-Jordan para calcular la inversa de A:
A =
1 2 2 1
1 0 0 1
→
1 2 0 -3
F2 – 2 F1
→
1 0 0 1
- 1/3 2 / 3 2 / 3 - 1/3
1 0 -2 1
→
3 0 -1 2 0 -3 -2 1
→
3F1 + 2 F2
= A-1
F1 /3 F2 /–3
B = A-1•
0 3 3 0
=
- 1/3 2 / 3 • 2 / 3 - 1/3
0 3 3 0
2 -1 -1 2
=
Ejercicio 2.2. Castilla La Mancha. Junio 2008. (1) Despeja la matriz X en la ecuación 2•X – B = A•X
1 0 1 (2) Halla la matriz X de la ecuación anterior sabiendo que A = 2 1 0 -1 3 1 1 -2 B = -3 3 . 4 -3 Solución: (1) 2•X – B = A•X → 2•X – A•X = B → (2•I – A)•X = B → X =(2•I – A)-1•B
1 0 0 (2) 2•I – A = 2• 0 1 0 0 0 1
1 0 1 2 0 0 – 2 1 0 =0 2 0 -1 3 1 0 0 2
1 0 1 – 2 1 0 = -1 3 1
y
1 0 -1 = -2 1 0 1 -3 1
.
1 0 -1 1 0 0 (2•I – A) = - 2 1 0 0 1 0 1 -3 1 0 0 1
1 0 -1 1 0 0 0 1 -2 2 1 0 → 1 -3 1 0 0 1
→ F2 + 2•F1
1 0 -1 1 0 0 0 1 -2 2 1 0 0 - 3 2 -1 0 1
→ F3 – F1
4•F1 – F3
→
→
1 0 0 2 1 0 5 3 1
→
−1 − 3 -1 -1 −1 -1 5 3 1
→
F3 + 3•F2
4 0 0 0 1 -2 0 0 -4
→
1 0 -1 0 1 -2 0 0 -4
−1 − 3 -1 2 1 0 5 3 1
4 0 0 0 2 0 0 0 -4
→ 2•F2 – F3
4 0 0 0 2 0 0 0 -4
− 1 / 4 − 3 / 4 - 1/4 - 1/2 − 1 / 2 - 1/2 - 5/4 - 3/4 − 1 / 4
= (2•I – A)-1
F1/4, F2/2, F3/(–4)
− 1 / 4 − 3 / 4 - 1/4 1 -2 X = (2•I – A) •B = - 1/2 − 1 / 2 - 1/2 • - 3 3 = - 5/4 - 3/4 − 1 / 4 4 -3 -1
(-1/4) • 1+ (-3/4) • (-3) + (-1/4) • 4 (-1/4) • (-2) + (-3/4) • 3 + (-1/4) • (-3) = (-1/2) • 1+ (-1/2) • (-3) + (-1/2) • 4 (-1/2) • (-2) + (-1/2) • 3 + (-1/2) • (-3) = (-5/4) • 1+ (-3/4) • (-3) + (-1/4) • 4 (-5/4) • (-2) + (-3/4) • 3 + (-1/4) • (-3)
- 1/4 + 9/4 - 4/4 2/4 - 9/4 + 3/4 1 -1 = - 1/2 + 3/2 - 4/2 2/2 - 3/2 + 3/2 = - 1 1 - 5/4 + 9/4 - 4/4 10/4 - 9/4 + 3/4 0 1
Ejercicio 2.3. Valencia. Junio 2008. Determina la matriz X que verifica la ecuación AX + I = ABt, siendo I la matriz identidad, A =
1 1 2 1 ,B= -1 1 -1 1
Bt la traspuesta de la matriz B.
y
Solución: AX + I = ABt → X = A-1•(ABt – I) 2
1
2 -1
t 1 1 →B= 1 1
B=
1 ABt =
1
2 -1 1• 2 + 1• 1 1• (-1) + 1• 1 3 0 1 1 • 1 1 = (-1) • 2 + 1• 1 (-1) • (-1) + 1• 1 = - 1 2
ABt – I =
3 0 1 0 2 0 = -1 2 0 1 -1 1 F2 + F1
A=
1 1 1 0 -1 1 0 1
→
2F1 - F1
1 1 0 2
1 0 1 1
→
F1 /2
→
1 0 0 1
1/ 2 - 1/2 1/ 2 1/ 2
= A-1
F2 /2 X = A-1•(ABt – I) =
=
1/ 2 - 1/2 2 0 • 1/ 2 1/ 2 -1 1
1/ 2 • 2 + (-1/2) • 1 1/2 • 0 + (-1/2) • 1 1/ 2 • 2 + 1/ 2 • (-1) 1/ 2 • 0 + 1/ 2 • 1
=
=
1/ 2 - 1/2 1/ 2 1/2
2 0 0 2
1 -1 1 1
Ejercicio 2.4. Valencia. Septiembre 2008. 1 3
Dada la matriz A =
.
4 2
(a) Halla su inversa. 6
(b) Resuelve la ecuación XA² + 5A =
8
.
10 − 20
Solución: (a) F2 - 4F1 A=
1 3
1 0
4 2
0 1
F2/(-10)
→
10F1 + 3F2 1
→
3
1
0 − 10
1 0
−2
0 1
2 / 5 − 1 / 10
3
0
1
→
−4 1
−2 3
0
0 − 10
−4 1
→
= A-1
(b) XA² + 5A =
→X= =
=
6
10 − 20
1
8
10 − 25
=
6
. → XA² =
• A-2 =
1
8
10 − 20
8
•
10 − 25
1 ⋅ 26 / 5 + 8 ⋅ (−42 / 50)
- 5A → XA² =
26 / 5
- 63/10
− 42 / 50 121 / 100
1 ⋅ (−63 / 10) + 8 ⋅ (121 / 100)
26 / 5 − 336 / 50 − 63 / 10 + 968 / 100 260 / 5 + 21
− 63 / 10 − 121 / 4
−2
3
2 / 5 − 1 / 10
( − 2) ⋅ ( − 2) + 3 ⋅ 2 / 5
•
−2
= 3
2 / 5 − 1 / 10
− 76 / 50
338 / 100
365 / 5
− 1462 / 40
- 63/10
− 42 / 50 121 / 100
8
→
10 − 25
=
=
− 38 / 25
169 / 50
73
− 731 / 20
=
(-2) ⋅ 3 + 3 ⋅ (-1/10)
2 / 5 ⋅ (−2) + (−1 / 10) ⋅ 2 / 5 2 / 5 ⋅ 3 + (−1 / 10) ⋅ (−1 / 10) 26 / 5
1
=
10 ⋅ (26 / 5) + (−25) ⋅ (−42 / 50) 10 ⋅ (−63 / 10) + (−25) ⋅ (121 / 100)
A-2 = (A-1)2 =
=
8
=
4 + 6/5
- 6 - 3/10
− 4 / 5 − 2 / 50 6 / 5 + 1 / 100
=
Ejercicio 2.5. Castilla La Mancha. Septiembre 2008. (1) Despeja la matriz X en la ecuación: X•A – X = B
(2)Halla la matriz X de la ecuación anterior sabiendo que 1 −1 2 0 −1 8 A= 0 1 3 y B = − 1 2 − 10 − 1 1 − 1 Solución: (1) X•A – X = B
→ X•(A – I) = B → X = B•(A – I)-1
(2)
1 −1 2 1 0 0 A–I= 0 1 3 – 0 1 0 −1 1 −1 0 0 1 F3
0 −1 2 1 0 0 A–I= 0 0 3 0 1 0 −1 1 − 2 0 0 1 F3
F2
→ 3F2 – 2F3
→ F1/(-3) F2/(-3) F3/3
→
-1
X = B•(A – I) =
=
=
−1 1 − 2 0 0 3 0 −1 2
3F1 + 2F3
→
−3 3 0 0 −1 2 0 0 3
0 0 1 0 1 0 1 0 0
→
0 2 3 1 0 0 0 1 0
→
F1 + F2
0 2 3 3 −2 0 0 1 0
−1 0 −1 −1 2 / 3 0 0 1/ 3 0
1 0 0 0 1 0 0 0 1
F1
→
−1 1 − 2 0 0 1 0 −1 2 1 0 0 0 0 3 0 1 0 −3 3 0 0 −3 0 0 0 3
0 −1 2 = 0 0 3 −1 1 − 2
0
−1
−1
2
(−1) ⋅ (−1)
−3 0 0 → 0 −3 0 0 0 3
= (A – I)-1
−1 0 −1 • −1 2 / 3 0 − 10 0 1/ 3 0 8
(-1) ⋅ 2/3 + 8 ⋅ 1/3
(−1) ⋅ (−1) + 2 ⋅ (−1)
2 ⋅ 2 / 3 + (−10) ⋅ 1 / 3
1
2
0
−1
−2
1
3 0 3 3 −2 0 0 1 0
=
0 (−1) ⋅ (−1)
=
→
1 2 2 1
, encontrar una matriz B tal que A•B =
0 3 3 0
Solución: A•B =
0 3 3 0
→ B = A-1•
0 3 3 0
Utilizamos el método de Gauss-Jordan para calcular la inversa de A:
A =
1 2 2 1
1 0 0 1
→
1 2 0 -3
F2 – 2 F1
→
1 0 0 1
- 1/3 2 / 3 2 / 3 - 1/3
1 0 -2 1
→
3 0 -1 2 0 -3 -2 1
→
3F1 + 2 F2
= A-1
F1 /3 F2 /–3
B = A-1•
0 3 3 0
=
- 1/3 2 / 3 • 2 / 3 - 1/3
0 3 3 0
2 -1 -1 2
=
Ejercicio 2.2. Castilla La Mancha. Junio 2008. (1) Despeja la matriz X en la ecuación 2•X – B = A•X
1 0 1 (2) Halla la matriz X de la ecuación anterior sabiendo que A = 2 1 0 -1 3 1 1 -2 B = -3 3 . 4 -3 Solución: (1) 2•X – B = A•X → 2•X – A•X = B → (2•I – A)•X = B → X =(2•I – A)-1•B
1 0 0 (2) 2•I – A = 2• 0 1 0 0 0 1
1 0 1 2 0 0 – 2 1 0 =0 2 0 -1 3 1 0 0 2
1 0 1 – 2 1 0 = -1 3 1
y
1 0 -1 = -2 1 0 1 -3 1
.
1 0 -1 1 0 0 (2•I – A) = - 2 1 0 0 1 0 1 -3 1 0 0 1
1 0 -1 1 0 0 0 1 -2 2 1 0 → 1 -3 1 0 0 1
→ F2 + 2•F1
1 0 -1 1 0 0 0 1 -2 2 1 0 0 - 3 2 -1 0 1
→ F3 – F1
4•F1 – F3
→
→
1 0 0 2 1 0 5 3 1
→
−1 − 3 -1 -1 −1 -1 5 3 1
→
F3 + 3•F2
4 0 0 0 1 -2 0 0 -4
→
1 0 -1 0 1 -2 0 0 -4
−1 − 3 -1 2 1 0 5 3 1
4 0 0 0 2 0 0 0 -4
→ 2•F2 – F3
4 0 0 0 2 0 0 0 -4
− 1 / 4 − 3 / 4 - 1/4 - 1/2 − 1 / 2 - 1/2 - 5/4 - 3/4 − 1 / 4
= (2•I – A)-1
F1/4, F2/2, F3/(–4)
− 1 / 4 − 3 / 4 - 1/4 1 -2 X = (2•I – A) •B = - 1/2 − 1 / 2 - 1/2 • - 3 3 = - 5/4 - 3/4 − 1 / 4 4 -3 -1
(-1/4) • 1+ (-3/4) • (-3) + (-1/4) • 4 (-1/4) • (-2) + (-3/4) • 3 + (-1/4) • (-3) = (-1/2) • 1+ (-1/2) • (-3) + (-1/2) • 4 (-1/2) • (-2) + (-1/2) • 3 + (-1/2) • (-3) = (-5/4) • 1+ (-3/4) • (-3) + (-1/4) • 4 (-5/4) • (-2) + (-3/4) • 3 + (-1/4) • (-3)
- 1/4 + 9/4 - 4/4 2/4 - 9/4 + 3/4 1 -1 = - 1/2 + 3/2 - 4/2 2/2 - 3/2 + 3/2 = - 1 1 - 5/4 + 9/4 - 4/4 10/4 - 9/4 + 3/4 0 1
Ejercicio 2.3. Valencia. Junio 2008. Determina la matriz X que verifica la ecuación AX + I = ABt, siendo I la matriz identidad, A =
1 1 2 1 ,B= -1 1 -1 1
Bt la traspuesta de la matriz B.
y
Solución: AX + I = ABt → X = A-1•(ABt – I) 2
1
2 -1
t 1 1 →B= 1 1
B=
1 ABt =
1
2 -1 1• 2 + 1• 1 1• (-1) + 1• 1 3 0 1 1 • 1 1 = (-1) • 2 + 1• 1 (-1) • (-1) + 1• 1 = - 1 2
ABt – I =
3 0 1 0 2 0 = -1 2 0 1 -1 1 F2 + F1
A=
1 1 1 0 -1 1 0 1
→
2F1 - F1
1 1 0 2
1 0 1 1
→
F1 /2
→
1 0 0 1
1/ 2 - 1/2 1/ 2 1/ 2
= A-1
F2 /2 X = A-1•(ABt – I) =
=
1/ 2 - 1/2 2 0 • 1/ 2 1/ 2 -1 1
1/ 2 • 2 + (-1/2) • 1 1/2 • 0 + (-1/2) • 1 1/ 2 • 2 + 1/ 2 • (-1) 1/ 2 • 0 + 1/ 2 • 1
=
=
1/ 2 - 1/2 1/ 2 1/2
2 0 0 2
1 -1 1 1
Ejercicio 2.4. Valencia. Septiembre 2008. 1 3
Dada la matriz A =
.
4 2
(a) Halla su inversa. 6
(b) Resuelve la ecuación XA² + 5A =
8
.
10 − 20
Solución: (a) F2 - 4F1 A=
1 3
1 0
4 2
0 1
F2/(-10)
→
10F1 + 3F2 1
→
3
1
0 − 10
1 0
−2
0 1
2 / 5 − 1 / 10
3
0
1
→
−4 1
−2 3
0
0 − 10
−4 1
→
= A-1
(b) XA² + 5A =
→X= =
=
6
10 − 20
1
8
10 − 25
=
6
. → XA² =
• A-2 =
1
8
10 − 20
8
•
10 − 25
1 ⋅ 26 / 5 + 8 ⋅ (−42 / 50)
- 5A → XA² =
26 / 5
- 63/10
− 42 / 50 121 / 100
1 ⋅ (−63 / 10) + 8 ⋅ (121 / 100)
26 / 5 − 336 / 50 − 63 / 10 + 968 / 100 260 / 5 + 21
− 63 / 10 − 121 / 4
−2
3
2 / 5 − 1 / 10
( − 2) ⋅ ( − 2) + 3 ⋅ 2 / 5
•
−2
= 3
2 / 5 − 1 / 10
− 76 / 50
338 / 100
365 / 5
− 1462 / 40
- 63/10
− 42 / 50 121 / 100
8
→
10 − 25
=
=
− 38 / 25
169 / 50
73
− 731 / 20
=
(-2) ⋅ 3 + 3 ⋅ (-1/10)
2 / 5 ⋅ (−2) + (−1 / 10) ⋅ 2 / 5 2 / 5 ⋅ 3 + (−1 / 10) ⋅ (−1 / 10) 26 / 5
1
=
10 ⋅ (26 / 5) + (−25) ⋅ (−42 / 50) 10 ⋅ (−63 / 10) + (−25) ⋅ (121 / 100)
A-2 = (A-1)2 =
=
8
=
4 + 6/5
- 6 - 3/10
− 4 / 5 − 2 / 50 6 / 5 + 1 / 100
=
Ejercicio 2.5. Castilla La Mancha. Septiembre 2008. (1) Despeja la matriz X en la ecuación: X•A – X = B
(2)Halla la matriz X de la ecuación anterior sabiendo que 1 −1 2 0 −1 8 A= 0 1 3 y B = − 1 2 − 10 − 1 1 − 1 Solución: (1) X•A – X = B
→ X•(A – I) = B → X = B•(A – I)-1
(2)
1 −1 2 1 0 0 A–I= 0 1 3 – 0 1 0 −1 1 −1 0 0 1 F3
0 −1 2 1 0 0 A–I= 0 0 3 0 1 0 −1 1 − 2 0 0 1 F3
F2
→ 3F2 – 2F3
→ F1/(-3) F2/(-3) F3/3
→
-1
X = B•(A – I) =
=
=
−1 1 − 2 0 0 3 0 −1 2
3F1 + 2F3
→
−3 3 0 0 −1 2 0 0 3
0 0 1 0 1 0 1 0 0
→
0 2 3 1 0 0 0 1 0
→
F1 + F2
0 2 3 3 −2 0 0 1 0
−1 0 −1 −1 2 / 3 0 0 1/ 3 0
1 0 0 0 1 0 0 0 1
F1
→
−1 1 − 2 0 0 1 0 −1 2 1 0 0 0 0 3 0 1 0 −3 3 0 0 −3 0 0 0 3
0 −1 2 = 0 0 3 −1 1 − 2
0
−1
−1
2
(−1) ⋅ (−1)
−3 0 0 → 0 −3 0 0 0 3
= (A – I)-1
−1 0 −1 • −1 2 / 3 0 − 10 0 1/ 3 0 8
(-1) ⋅ 2/3 + 8 ⋅ 1/3
(−1) ⋅ (−1) + 2 ⋅ (−1)
2 ⋅ 2 / 3 + (−10) ⋅ 1 / 3
1
2
0
−1
−2
1
3 0 3 3 −2 0 0 1 0
=
0 (−1) ⋅ (−1)
=
→
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