Ejercicos para hallar el rango de de una matriz

1. 1. Si
𝑎 𝑏 𝑐
𝑝 𝑞 𝑟
𝑥 𝑦 𝑧
= 7, 𝑐𝑎𝑙𝑐𝑢𝑙𝑎 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒:
A =
𝑎 + 2𝑥 𝑏 + 2𝑦 𝑐 + 2𝑧
𝑝 𝑞 𝑟
2𝑥 + 𝑝 2𝑦 + 𝑞 2𝑧 + 𝑟
Solución:
𝑎 + 2𝑥 𝑏 + 2𝑦 𝑐 + 2𝑧
𝑝 𝑞 𝑟
2𝑥 + 𝑝 2𝑦 + 𝑞 2𝑧 + 𝑟
𝑎 + 2𝑥 𝑏 + 2𝑦 𝑐 + 2𝑧
𝑝 𝑞 𝑟
2𝑥 2𝑦 2𝑧
𝑎 𝑏 𝑐
𝑝 𝑞 𝑟
2𝑥 2𝑦 2𝑧
= 2
𝑎 𝑏 𝑐
𝑏 𝑞 𝑟
𝑐 𝑦 𝑧
= 2 7 = 14
→ det 𝐴 = 14
F3 – F2
F1 – F3
2. 𝐶𝑎𝑙𝑐𝑢𝑙𝑎 𝑙𝑎 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒 𝑑𝑒𝑙 𝑡𝑒𝑟𝑐𝑒𝑟 𝑜𝑟𝑑𝑒𝑛.
A =
3 4 −5
8 7 −2
2 −1 8
Solución:
det(A) = (3)(7)(8) + (4)(-2)(2) + (8)(-1)(-5) – (2)(7)(-5) –
(- 1)(-2)(3) – (8)(4)(8)
det(A) = 168 – 16 + 40 + 70 – 6 – 256
det(A) = 0

2. 3. 𝐷𝑎𝑑𝑎 𝐴 =
1 2 3
4 5 6
7 8 0
, 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑠𝑢𝑠 𝑐𝑜𝑓𝑎𝑐𝑡𝑜𝑟𝑒𝑒𝑠 𝑝𝑎𝑟𝑎 𝑜𝑏𝑡𝑒𝑛𝑒𝑟 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑎𝑑𝑗𝑢𝑛𝑡𝑎:
Solución:
𝑀11 = (−1)1+1 5 6
8 0
= −48 𝑀12 = −1 1+2 4 6
7 0
= 42 𝑀13 = −1 1+3 4 5
7 8
= −3
𝑀21 = (−1)2+1 2 3
8 0
= 24 𝑀22 = −1 2+2 1 3
7 0
= −21 𝑀23 = −1 2+3 1 2
7 8
= 6
𝑀31 = (−1)3+1 2 3
5 6
= −3 𝑀32 = −1 3+2 1 3
4 6
= 6 𝑀33 = −1 3+3 1 2
4 5
= −3
→ 𝑐𝑜𝑓 𝐴 =
−48 42 −3
24 −21 6
−3 6 −3
→ 𝑎𝑑𝑗 𝐴 =
−48 24 −3
42 −21 6
−3 6 −3

3. 4. 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑒𝑙 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒 𝑑𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑚𝑎𝑡𝑟𝑖𝑧.
𝐴 =
1 2 3 4
1 1 0 1
2 1 3 −1
3 1 3 0
Solución:
1 2 3 4
1 1 0 1
2 1 3 −1
3 1 3 0
1 2 3 4
0 −1 −3 −3
2 1 3 −1
3 1 3 0
1 2 3 4
0 −1 −3 −3
0 −3 −3 −9
3 1 3 0
1 2 3 4
0 −1 −3 −3
0 −3 −3 −9
0 −5 −6 −12
Podemos desarrollar el determinante por la primera columna.
det 𝐴 =
−1 −3 −3
−3 −3 −9
−5 −6 −12
= (−1)3
1 3 3
3 3 9
5 6 12
−
1 3 3
0 −6 0
5 6 12
−
1 3 3
0 −6 0
0 −9 −3
det 𝐴 = −18
F2 – F1 F3 – 2F1 F4 – 3F1
F2 – 3F1 F3 – 5F1

4. 5. 𝑆𝑒𝑎 𝐴 𝑢𝑛𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑐𝑢𝑎𝑑𝑟𝑎𝑑𝑎 𝑑𝑒 𝑜𝑟𝑑𝑒𝑛 3.
𝐴 =
𝑥 1 1
𝑥 + 1 2 2
𝑥 2 − 𝑥 1
𝑎) 𝑆𝑖 𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑒𝑙 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒 𝑑𝑒 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 2𝐴 𝑒𝑠 2𝐴 = 8. ¿ 𝐶𝑢á𝑛𝑡𝑜 𝑣𝑎𝑙𝑒 𝑒𝑙 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒 𝑑𝑒 𝐴?
𝑏) 𝐶𝑎𝑙𝑐𝑢𝑙𝑎 𝑝𝑎𝑟𝑎 𝑞𝑢é 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑥 𝑠𝑒 𝑐𝑢𝑚𝑝𝑙𝑒 𝑞𝑢𝑒 2𝐴 = 8.
Solución:
a) Si en una matriz cuadrada multiplicamos por un mismo número todos los elementos de la matriz, su
determinante queda multiplicado por ese número elevado al orden de la matriz, en este caso, orden 3.
2𝐴 = 23
. 𝐴 = 8 → 8 𝐴 = 8 → 𝐴 = 1
b)
𝑥 1 1
𝑥 + 1 2 2
𝑥 2 − 𝑥 1
= 2𝑥 + 𝑥 + 1 2 − 𝑥 + 2𝑥 − 2𝑥 − 2𝑥 2 − 𝑥 − 𝑥 + 1
= 2𝑥 + 2𝑥 − 𝑥2
+ 2 − 𝑥 + 2𝑥 − 2𝑥 − 4𝑥 + 2𝑥2
− 𝑥 − 1 = 𝑥2
-2x+1
2𝐴 = 8 → 𝐴 = 1 → 𝑥2
−2x+1=1 → 𝑥2
−2x=0
→ 𝑥 𝑥 − 2 = 0 →
𝑥 = 0
𝑥 = 2

5. 6. 𝑆𝑒𝑎𝑛 𝑙𝑎𝑠 𝑚𝑎𝑡𝑟𝑖𝑐𝑒𝑠:
𝐴 =
0 1 2
0 2 1
3 1 1
𝐵 =
−2 −2 1
2 −1 −3
0 0 −3
¿ 𝑆𝑒 𝑐𝑢𝑚𝑝𝑙𝑒 𝑞𝑢𝑒 𝐴. 𝐵 = 𝐴 . 𝐵
Solución:
𝐴. 𝐵 =
0 1 2
0 2 1
3 1 1
.
−2 −2 1
2 −1 −3
0 0 −3
=
−2 −2 34
11 −16 231
9 −1 108
= 10
𝐴 =
0 1 2
0 2 1
3 1 1
= −2
𝐵 =
−2 −2 1
2 −1 −3
0 0 −3
= −5
𝐴 . 𝐵 =10
→ 𝐴. 𝐵 = 𝐴 . 𝐵

6. 7. 𝐶𝑎𝑙𝑐𝑢𝑙𝑎 𝑙𝑎 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑑𝑒 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧
A =
−1 1 4
0 3 −1
2 −1 5
Solución:
𝐴 =
−1 1 4
0 3 −1
2 −1 5
= −40 ≠ 0 → 𝐴 𝑒𝑠 𝑖𝑛𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒
𝑎𝑑𝑗 𝐴 =
30 −1
−1 5
−
0 −1
2 5
0 3
2 −1
−
1 4
−1 5
−1 4
2 5
−
−1 1
2 −1
1 4
3 −1
−
−1 4
0 −1
−1 1
0 3
=
14 −2 −6
−9 −13 1
−13 −1 −3
→ 𝑎𝑑𝑗(𝐴)𝑡
=
14 −9 −13
−2 −13 −1
−6 1 −3
𝐴−1 =
𝑎𝑑𝑗(𝐴)𝑡
𝐴
= −
1
40
.
14 −9 −13
−2 −13 −1
−6 1 −3
=
−
7
20
9
40
13
40
1
20
13
20
1
40
3
20
−
1
40
3
40

7. 8. Resolver:
𝐷 =
𝑎 𝑎2 𝑎3
𝑏 𝑏2
𝑏3
𝑐 𝑐2 𝑐3
Solución:
Sacando el factor común de las tres filas tenemos:
𝐷 = 𝑎𝑏𝑐
1 𝑎 𝑎2
1 𝑏 𝑏2
1 𝑐 𝑐2
Transformamos:
1 𝑎 𝑎2
1 𝑏 𝑏2
1 𝑐 𝑐2
1 𝑎 𝑎2
0 𝑏 − 𝑎 𝑏2 − 𝑎2
0 𝑐 − 𝑎 𝑐2 − 𝑎2
→ 𝐷 = 𝑎𝑏𝑐
1 𝑎 𝑎2
0 𝑏 − 𝑎 𝑏2 − 𝑎2
0 𝑐 − 𝑎 𝑐2 − 𝑎2
Desarrollamos el det. por la primera columna:
𝐷 = 𝑎𝑏𝑐 −1 1+1 1
𝑏 − 𝑎 𝑏 − 𝑎 𝑏 + 𝑎
𝑐 − 𝑎 𝑐 − 𝑎 𝑐 + 𝑎
𝐷 = 𝑎𝑏𝑐(𝑏 − 𝑎) 𝑐 − 𝑎
1 𝑏 + 𝑎
1 𝑐 + 𝑎
𝐷 = 𝑎𝑏𝑐(𝑏 − 𝑎) 𝑐 − 𝑎 (𝑐 − 𝑏)
F2 – F1
F3 – F1

8. 9. 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑛𝑑𝑜 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑑𝑒 𝐺𝑎𝑢𝑠𝑠, ℎ𝑎𝑙𝑙𝑎
𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒𝑙 𝑑𝑒𝑡𝑒𝑟𝑖𝑛𝑎𝑛𝑡𝑒 𝑑𝑒 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧:
𝐴 =
1 −2 1 −1
−2 2 −1 2
2 −3 1 −2
3 −2 1 −2
Solución:
𝐴 =
1 −2 1 −1
−2 2 −1 2
2 −3 1 −2
3 −2 1 −2
1 −2 1 −1
0 −2 1 0
0 0 −1/2 0
0 0 0 1
= 1
−2 1 0
0 −1/2 0
0 0 1
𝑑𝑒𝑡. 𝐴 = 1
F3 – 2F1– (-1/2)F2
F4 – 3F1 – (-2)F2
F2 – (-2)F1
10. 𝐻𝑎𝑙𝑙𝑎𝑟 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑑𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑚𝑎𝑡𝑟𝑖𝑧.
𝐴 =
cos 𝛼 −sin 𝛼 0
sin 𝛼 cos 𝛼 0
0 0 1
Solución:
= 1(−1)3+3 cos 𝛼 −sin𝛼
sinα cos 𝛼
=
cos 𝛼 −sin𝛼
sinα cos 𝛼
= 𝑐𝑜𝑠2
𝛼 + 𝑠𝑖𝑛2
𝛼 = 1
𝑅𝑎𝑛 𝐴 = 3

9. 11. 𝐷𝑎𝑑𝑎 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑀 =
1 2
1 1
𝑐𝑎𝑙𝑐𝑢𝑙𝑎 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 (𝑀−1
. 𝑀𝑡
)2
Solución:
det 𝑀 =
1 2
1 1
= 1 − 2 = −1
Hallamos la inversa de M
𝑀−1
=
1
det(𝑀)
. 𝑎𝑑𝑗𝑀𝑡
=
1
−1
1 2
1 1
=
−1 2
1 −1
La matriz traspuesta de M es:
𝑀𝑡 =
1 2
1 1
𝑡
=
1 1
2 1
→ 𝑀−1. 𝑀𝑡 =
−1 2
1 −1
.
1 1
2 1
=
3 1
−1 0
→ (𝑀−1
. 𝑀𝑡
)2
=
3 1
−1 0
.
3 1
−1 0
=
8 3
−3 −1

10. 12. 𝐷𝑎𝑑𝑎 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴 =
1 0 −1
0 𝑎 3
4 1 −𝑎
, ℎ𝑎𝑙𝑙𝑎:
𝑎) 𝐿𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑎 𝑝𝑎𝑟𝑎 𝑙𝑜𝑠 𝑞𝑢𝑒 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴 𝑝𝑜𝑠𝑒𝑎 𝑖𝑛𝑣𝑒𝑟𝑠𝑎.
𝑏) 𝐿𝑎 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑑𝑒 𝐴 𝑝𝑎𝑟𝑎 𝑎 = 2.
Solución:
𝑎) 𝐿𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴 𝑝𝑜𝑠𝑒𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑐𝑢𝑎𝑛𝑑𝑜 𝑠𝑢 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒 𝑠𝑒𝑎 𝑑𝑖𝑠𝑡𝑖𝑛𝑡𝑜 𝑑𝑒 𝑐𝑒𝑟𝑜.
𝐴 =
1 0 −1
0 𝑎 3
4 1 −𝑎
= −𝑎2 + 4𝑎 − 3 = 0 → 𝑎 = 1, 𝑎 = 3
→ 𝐿𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴 𝑝𝑜𝑠𝑒𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑐𝑢𝑎𝑛𝑑𝑜 𝑎 ≠ 1 𝑦 𝑎 ≠ 3.
𝑏) 𝑃𝑎𝑟𝑎 𝑎 = 2, 𝐴 =
1 0 −1
0 2 3
4 1 −2
𝑦 𝐴 = 1
𝐿𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑣𝑖𝑒𝑛𝑒 𝑑𝑎𝑑𝑎 𝑝𝑜𝑟 𝐴−1
=
(𝐴𝑖𝑗)𝑡
𝐴
, 𝑠𝑖𝑒𝑛𝑑𝑜 𝐴𝑖𝑗 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑑𝑒 𝑙𝑜𝑠 𝑎𝑑𝑗𝑢𝑛𝑡𝑜𝑠 𝑑𝑒 𝐴.
𝐴𝑖𝑗 =
−7 12 −8
−1 2 −1
2 −3 2
→ 𝐴−1
=
1
1
−7 −1 2
12 2 −3
−8 −1 2
=
−7 −1 2
12 2 −3
−8 −1 2

11. 13. 𝐶𝑜𝑚𝑝𝑟𝑢𝑒𝑏𝑎 𝑞𝑢𝑒 𝑒𝑙 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑎𝑛𝑛𝑡𝑒 𝑒𝑠 𝑚ú𝑙𝑡𝑖𝑝𝑙𝑜 𝑑𝑒 42.
𝐴 =
6 −7 18
9 21 4
−3 14 10
Solución:
6 −7 18
9 21 4
−3 14 10
→ 3
2 −7 18
3 21 4
−1 14 10
→ 3.7
2 −1 18
3 3 4
−1 2 10
→ 3.7.2
2 −1 9
3 3 2
−1 2 5
→
6 −7 18
9 21 4
−3 14 10
= 42
2 −1 9
3 3 2
−1 2 5

12. 14. 𝐷𝑎𝑑𝑎 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑚𝑎𝑡𝑟𝑖𝑧:
𝐶 =
4 3 −2
0 1 4
5 1 2
𝐷𝑒𝑠𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑟 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐶 𝑒𝑛 𝑑𝑜𝑠 𝑚𝑎𝑡𝑟𝑖𝑐𝑒𝑠 𝐴 𝑦 𝐵 𝑑𝑒 𝑓𝑜𝑟𝑚𝑎 𝑞𝑢𝑒: 𝐶 = 𝐴 + 𝐵
Solución:
𝐶 =
4 3 −2
0 1 4
5 1 2
=
4 3 −2
0 1 4
2 + 3 1 + 0 1 + 1
→ 𝐴 =
4 3 −2
0 1 4
2 1 1
𝑦 𝐵 =
4 3 −2
0 1 4
3 0 1
𝐶 =
4 3 −2
0 1 4
5 1 2
= 62
𝐴 =
4 3 −2
0 1 4
2 1 1
= 16
𝐵 =
4 3 −2
0 1 4
3 0 1
= 46
→ 𝐴 + 𝐵 = 𝐶

13. 15. 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑑𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑚𝑎𝑡𝑟𝑖𝑧 𝑒𝑛 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑑𝑒𝑙 𝑝𝑎𝑟á𝑚𝑒𝑡𝑟𝑜.
𝐴 =
𝑘 3 0
3 2 𝑘
3 𝑘 0
Solución:
𝐴 =
𝑘 3 0
3 2 𝑘
3 𝑘 0
= 𝑘 𝑘2 − 9 → 𝐴 ≠ 0 𝑐𝑢𝑎𝑛𝑑𝑜 𝑘 ≠ 0, −3 𝑦 3; 𝐴 = 0 si k = 0, −3 o 3.
* 𝑆𝑖 𝑘 ≠ 0, −3 𝑦 3, 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑑𝑒 𝐴 𝑠𝑒𝑟á 3. 𝑃𝑎𝑟𝑎 𝑘 = 0, −3 𝑜 3 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑠𝑒𝑟𝑎 𝑚𝑒𝑛𝑜𝑟 𝑞𝑢𝑒 3.
* 𝑆𝑖 𝑘 = 0, 𝐴 =
0 3 0
3 2 0
3 0 0
→ 𝑠𝑢 𝑟𝑎𝑛𝑔𝑜 𝑒𝑠 2. 𝐸𝑙 𝑚𝑒𝑛𝑜𝑟 𝐴1 =
0 3
3 0
= −9
* 𝑆𝑖 𝑘 = −3, 𝐴 =
−3 3 0
3 2 −3
3 −3 0
→ 𝑠𝑢 𝑟𝑎𝑛𝑔𝑜 𝑒𝑠 2. 𝐸𝑙 𝑚𝑒𝑛𝑜𝑟 𝐴2 =
−3 3
3 2
= −15
* 𝑆𝑖 𝑘 = 3, 𝐴 =
3 3 0
3 2 3
3 3 0
→ 𝑠𝑢 𝑟𝑎𝑛𝑔𝑜 𝑒𝑠 2. 𝐸𝑙 𝑚𝑒𝑛𝑜𝑟 𝐴3 =
3 3
3 2
= −3

14. 16. 𝑆𝑒𝑎 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧
1 2 1
0 𝑡 𝑡
−1 𝑡 −1
𝑎) 𝐻𝑎𝑙𝑙𝑎 𝑠𝑢 𝑟𝑎𝑛𝑔𝑜 𝑒𝑛 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑑𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑡.
𝑏) 𝐶𝑎𝑙𝑐𝑢𝑙𝑎 𝑠𝑢 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑝𝑎𝑟𝑎 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑜 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑡 𝑝𝑎𝑟𝑎 𝑙𝑜𝑠 𝑞𝑢𝑒 𝑒𝑙 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒 𝑑𝑒 𝑒𝑠𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑣𝑎𝑙𝑒 1.
Solución:
𝑎)
1 2 1
0 𝑡 𝑡
−1 𝑡 −1
= −𝑡 − 𝑡2 − 𝑡 = −𝑡 𝑡 + 2 → 𝑆𝑖 𝑡 ≠ 0 𝑦 − 2, 𝑒𝑙 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒 𝑒𝑠 𝑑𝑖𝑠𝑡𝑖𝑛𝑡𝑜 𝑑𝑒 0.
* 𝑆𝑖 𝑡 ≠ 0 𝑦 𝑡 ≠ −2 → 𝑠𝑢 𝑟𝑎𝑛𝑔𝑜 𝑒𝑠 3.
* 𝑆𝑖 𝑡 = 0 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑒𝑠
1 2 1
0 0 0
−1 0 −1
→ 𝑠𝑢 𝑟𝑎𝑛𝑔𝑜 𝑒𝑠 2.
* 𝑆𝑖 𝑡 = −2 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑒𝑠
1 2 1
0 2 2
−1 −2 −1
. 𝑇𝑎𝑚𝑏𝑖é𝑛 𝑐𝑜𝑛 𝑟𝑎𝑛𝑔𝑜 2.
b) 𝐸𝑙 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒 𝑣𝑎𝑙𝑒 1 𝑐𝑢𝑎𝑛𝑑𝑜 − 𝑡2 − 2𝑡 = 1 → 𝑡2 + 2𝑡 + 1 = 0 → 𝑡 + 1 2 = 0 → t = −1
𝑃𝑎𝑟𝑎 𝑡 = −1, 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑒𝑠 𝐴 =
1 2 1
0 −1 −1
−1 −1 −1
. 𝑆𝑢 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑠𝑒𝑟á: 𝐴−1 =
0 1 −1
1 0 1
−1 −1 −1

15. 17. 𝐷𝑎𝑑𝑜𝑠 𝑙𝑜𝑠 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑟𝑒𝑎𝑙𝑒𝑠 𝑎 𝑦 𝑏 𝑠𝑒 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴 =
𝑎 + 𝑏 𝑎 𝑎
𝑎 𝑎 + 𝑏 𝑎
𝑎 𝑎 𝑎 + 𝑏
.
𝑎) 𝑂𝑏𝑡𝑒𝑛𝑔𝑎 𝑒𝑙 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒 𝑑𝑒 𝐴.
𝑏) 𝐸𝑠𝑡𝑢𝑑𝑖𝑒 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑑𝑒 𝐴 𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑒𝑛𝑑𝑜 𝑑𝑒 𝑙𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑎 𝑦 𝑏.
Solución:
𝑎) 𝐴 =
𝑎 + 𝑏 𝑎 𝑎
𝑎 𝑎 + 𝑏 𝑎
𝑎 𝑎 𝑎 + 𝑏
=
3𝑎 + 𝑏 𝑎 𝑎
3𝑎 + 𝑏 𝑎 + 𝑏 𝑎
3𝑎 + 𝑏 𝑎 𝑎 + 𝑏
= 3𝑎 + 𝑏
1 𝑎 𝑎
1 𝑎 + 𝑏 𝑎
1 𝑎 𝑎 + 𝑏
= 3𝑎 + 𝑏
1 𝑎 𝑎
1 𝑎 + 𝑏 𝑎
1 𝑎 𝑎 + 𝑏
3𝑎 + 𝑏
1 𝑎 𝑎
0 𝑏 0
0 0 𝑏
= 3𝑎 + 𝑏 𝑏2
𝑏) 𝐶𝑜𝑚𝑜 3𝑎 + 𝑏 𝑏2
= 0 𝑐𝑢𝑎𝑛𝑑𝑜 𝑏 = 0 𝑜 − 3𝑎 𝑠𝑒 𝑡𝑒𝑛𝑑𝑟á:
* 𝑆𝑖 𝑏 ≠ 0 𝑦 𝑏 ≠ 3𝑎 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑑𝑒 𝐴 𝑠𝑒𝑟á 3.
* 𝑆𝑖 𝑏 = 0, 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑒𝑠 𝐴 =
𝑎 𝑎 𝑎
𝑎 𝑎 𝑎
𝑎 𝑎 𝑎
, 𝑐𝑢𝑦𝑜 𝑟𝑎𝑛𝑔𝑜 𝑒𝑠 1 𝑠𝑖 𝑎 ≠ 0; 𝑦 0 𝑠𝑖 𝑎 = 0.
* 𝑆𝑖 𝑏 ≠ 0 𝑝𝑒𝑟𝑜 𝑏 = −3𝑎, 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑒𝑠 𝐴 =
−2𝑎 𝑎 𝑎
𝑎 −2𝑎 𝑎
𝑎 𝑎 −2𝑎
→
0 𝑎 𝑎
0 −2𝑎 𝑎
0 𝑎 −2𝑎
→ 𝑆𝑢𝑠 𝑚𝑒𝑛𝑜𝑟𝑒𝑠 𝑑𝑒 𝑜𝑟𝑑𝑒𝑛 𝑑𝑜𝑠 𝑛𝑜 𝑛𝑢𝑙𝑜𝑠 𝑣𝑎𝑙𝑒𝑠 ± 3𝑎2
. 𝑃𝑜𝑟 𝑡𝑎𝑛𝑡𝑜, 𝑠𝑖 𝑏 = −3𝑎 𝑐𝑜𝑛 𝑎 ≠ 0, 𝑒𝑙 𝑟𝑎𝑛𝑔𝑜 𝑣𝑎𝑙𝑑𝑟á 2.
F2 – F1
F3 – F1

16. 18. 𝑆𝑒 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴 =
1 0 0 𝑥
0 𝑥 0 𝑥
1 0 𝑥 0
0 1 𝑥 𝑥
𝑎) 𝑅𝑒𝑠𝑢𝑒𝑙𝑣𝑒 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 det 𝐴 = 0
𝑏) ¿ 𝐸𝑛 𝑞𝑢é 𝑐𝑎𝑠𝑜𝑠 𝑎𝑑𝑚𝑖𝑡𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴?
Solución:
𝑎) 𝐴 =
1 0 0 𝑥
0 𝑥 0 𝑥
1 0 𝑥 0
0 1 𝑥 𝑥
1 0 0 𝑥
−1 𝑥 0 0
1 0 𝑥 0
−1 1 𝑥 0
= −𝑥
−1 𝑥 0
1 0 𝑥
−1 1 𝑥
= −𝑥 𝑥 − 𝑥. 2𝑥
𝐴 = 0 → 𝑥2 2𝑥 − 1 = 0 → 𝑥 = 0 𝑜 𝑥 =
1
2
𝑏) 𝐿𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴 𝑎𝑑𝑚𝑖𝑡𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑠𝑖𝑒𝑚𝑝𝑟𝑒 𝑞𝑢𝑒 𝑥 ≠ 0 𝑦 𝑥 ≠
1
2
F2 – F1
F4 – F1

17. 19. 𝑆𝑒 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴 =
1 2 𝜆
1 −1 −1
𝑦 𝑏 =
1 3
𝜆 0
0 2
, 𝑑𝑜𝑛𝑑𝑒 𝜆 𝑒𝑠 𝑢𝑛 𝑛ú𝑚𝑒𝑟𝑜 𝑟𝑒𝑎𝑙.
𝐸𝑛𝑐𝑢𝑒𝑛𝑡𝑟𝑎 𝑙𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝜆 𝑝𝑎𝑟𝑎 𝑙𝑜𝑠 𝑞𝑢𝑒 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴𝐵 𝑡𝑖𝑒𝑛𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑎.
Solución:
𝐴. 𝐵 =
1 2 𝜆
1 −1 −1
.
1 3
𝜆 0
0 2
=
1 + 2𝜆 3 + 2𝜆
1 − 𝜆 1
Para que tenga inversa es necesario que 𝐴𝐵 ≠ 0
𝐶𝑜𝑚𝑜
1 + 2𝜆 3 + 2𝜆
1 − 𝜆 1
= 2𝜆2
+ 3𝜆 − 2 = 0 si 𝜆 =
1
2
𝑜 − 2
→ 𝐿𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝐴𝐵 𝑡𝑒𝑛𝑑𝑟á 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑐𝑢𝑎𝑛𝑑𝑜 𝜆 ≠ −2 𝑦 𝜆 =
1
2

18. 20. 𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝑙𝑎 𝑓ó𝑟𝑚𝑢𝑙𝑎 𝐴−1 =
1
𝐴
(𝐴𝑖𝑗) 𝑡
𝑐𝑎𝑙𝑐𝑢𝑙𝑎 𝑙𝑎 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑑𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑚𝑎𝑡𝑟𝑖𝑧, 𝑠𝑖 𝑒𝑥𝑖𝑠𝑡𝑒.
𝐴 =
1 1 1
0 1 1
1 1 2
Solución:
𝐴 =
1 1 1
0 1 1
1 1 2
= 1
𝑎𝑑𝑗 𝐴𝑖𝑗 =
1 1 −1
−1 1 0
0 −1 1
→ 𝐴−1 = (𝐴𝑖𝑗) 𝑡
=
1 −1 0
1 1 −1
−1 0 1
𝑃𝑢𝑒𝑑𝑒 𝑐𝑜𝑚𝑝𝑟𝑜𝑏𝑎𝑟𝑠𝑒 𝑞𝑢𝑒 𝐴. 𝐴−1
= 𝐼
𝐸𝑛 𝑒𝑓𝑒𝑐𝑡𝑜: 𝐴. 𝐴−1
=
1 1 1
0 1 1
1 1 2
.
1 −1 0
1 1 −1
−1 0 1
=
1 + 1 − 1 −1 + 1 −1 + 1
1 − 1 1 −1 + 1
1 + 1 − 2 −1 + 1 −1 + 2
=
1 0 0
0 1 0
0 0 1