Ejercicio

1. ALGEBRA LINEAL
Dada 𝑓: 𝑅3 → 𝑅3, una transformación lineal invertible tal que,
𝑓
𝑥
𝑦
𝑧
=
𝑥 − 2𝑦 + 𝑧
𝑦 − 3𝑧
𝑥 + 2𝑧
𝐵1 =
1
1
0
,
0
1
1
,
1
0
1
y, 𝐵2 =
1
1
1
,
0
1
1
,
0
0
1
bases de 𝑅3
.
a) Probar que 𝑓 es invertible
b) Hallar 𝐴 = 𝑓 𝐵2
𝐵1
c) Si 𝑓(𝑣) 𝐵2
=
1
−3
1
hallar 𝑣. Sugerencia 𝑣 𝐵1
= 𝐴−1
𝑓(𝑣) 𝐵2

2. 𝑢 𝑓(𝑢)
𝑓
𝑅3
𝑅3
𝑓−1
a) 𝑃. 𝐷. 𝑓 𝑒𝑠 𝑖𝑛𝑦𝑒𝑐𝑡𝑖𝑣𝑎 → 𝑓 𝑒𝑠 𝑖𝑛𝑣𝑒𝑟𝑡𝑖𝑏𝑙𝑒
𝑑𝑖𝑚𝑅3
= 𝑑𝑖𝑚𝑅3
3 = 3
𝑁 𝑓 𝑓 𝑢 = 0 𝑅3
𝑥 − 2𝑦 + 𝑧
𝑦 − 3𝑧
𝑥 + 2𝑧
=
0
0
0
𝑥 − 2𝑦 + 𝑧 = 0
𝑦 − 3𝑧 = 0
𝑥 + 2𝑧 = 0
𝐸𝑥𝑖𝑠𝑡𝑒 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 ú𝑛𝑖𝑐𝑎 𝑡𝑟𝑖𝑣𝑖𝑎𝑙 𝑠𝑖:
1 −2 1
0 1 −3
1 0 2
≠ 0
1 −2 1
0 1 −3
1 0 2 𝐹3←𝐹3−𝐹1
=
1 −2 1
0 1 −3
0 2 1 𝐹3←𝐹3−2𝐹2
=
1 −2 1
0 1 −3
0 0 7
= 7
7 ≠ 0 𝑁 𝑓 = 0 𝑅3
∴ 𝑓 𝑒𝑠 𝑖𝑛𝑦𝑒𝑐𝑡𝑖𝑣𝑎 → 𝑓 𝑒𝑠 𝑖𝑛𝑣𝑒𝑟𝑡𝑖𝑏𝑙𝑒

3. b) 𝑓
1
1
0
=
−1
1
1
𝑓
0
1
1
=
−1
−2
2
𝑓
1
0
1
=
2
−3
3
−1
1
1
= 𝛼
1
1
1
+ 𝛽
0
1
1
+ 𝛾
0
0
1
𝛼 = −1
𝛼 + 𝛽 = 1
𝛼 + 𝛽 + 𝛾 = 1
1 0 0
1 1 0
1 1 1
−1
1
1
−1
−2
2
2
−3
3 𝐹2←𝐹2−𝐹1
𝐹3←𝐹3−𝐹1
~
1 0 0
0 1 0
0 1 1
−1
2
2
−1
−1
3
2
−5
1 𝐹3←𝐹3−𝐹2
~
1 0 0
0 1 0
0 0 1
−1
2
0
−1
−1
4
2
−5
6
𝐴 =
−1 −1 2
2 −1 −5
0 4 6

4. c) 𝐴−1
=
1
𝐴
𝐶𝑜𝑓(𝐴) 𝑡
𝐴 =
−1 −1 2
2 −1 −5
0 4 6 𝐹2←𝐹2+2𝐹1
=
−1 −1 2
0 −3 −1
0 4 6 𝐹2←𝐹2+𝐹3
=
−1 −1 2
0 1 5
0 4 6 𝐹3←𝐹3−4𝐹2
=
−1 −1 2
0 1 5
0 0 −14
= 14
𝐶𝑜𝑓 𝐴 =
+
−1 −5
4 6
−
2 −5
0 6
+
2 −1
0 4
−
−1 2
4 6
+
−1 2
0 6
−
−1 −1
0 4
+
−1 2
−1 −5
−
−1 2
2 −5
+
−1 −1
2 −1
→ 𝐶𝑜𝑓 𝐴 =
14 −12 8
14 −6 4
7 −1 3
𝐶𝑜𝑓(𝐴) 𝑡
=
14 14 7
−12 −6 −1
8 4 3
𝐴−1
=
1
14
14 14 7
−12 −6 −1
8 4 3
→ 𝐴−1
=
1 1
1
2
−
6
7
−
3
7
−
1
14
4
7
2
7
3
14

5. 𝑣 𝐵1
= 𝐴−1
𝑓(𝑣) 𝐵2
𝑣 𝐵1
=
1 1
1
2
−
6
7
−
3
7
−
1
14
4
7
2
7
3
14
1
−3
1
→ 𝑣 𝐵1
=
−
3
2
5
14
−
1
14
𝑣 = −
3
2
1
1
0
+
5
14
0
1
1
+ −
1
14
1
0
1
𝑣 =
−
3
2
−
3
2
0
+
0
5
14
5
14
+
−
1
14
0
−
1
14
→ 𝒗 =
−
11
7
−
8
7
2
7