Download 1 5 3

UNIDAD 1 Álgebra de matrices
1. Ejercicios de refuerzo:
operaciones combinadas con matrices
Pág. 1 de 1
Soluciones
1
Dadas las matrices siguientes:
(
3 –5 1
A = 0 –3 7
–1 4 –1
0
0
1
( )
1 5 3
0 –1 0
B=
3 0 –2
0 7 1
)
(
4 4 –2
C = –1 0 0
0 –3 –1
)
Efectúa:
b) B · C – A t
a) A · B + C
c) B t · A t + C · C t
d) B · B t · A t + B
Resolución
)( )(
)( )(
)( ) (
(
(
3 –5 1
a) A · B + C = 0 –3 7
–1 4 –1
1 5 3
0
4 4 –2
0 –1 0
0 ·
+ –1 0 0 =
3 0 –2
1
0 –3 –1
0 7 1
6 20 7
4 4 –2
10 24 5
= 21 3 –14 + –1 0 0 = 20 3 –14
–4 –2 0
0 –3 –1
–4 –5 –1
)
)
)
(
( )( )( )
( )( )(
1 5 3
3 0 –1
4 4 –2
0
–1
0
–5
–3 4
· –1 0 0 –
=
b) B · C – A t =
3 0 –2
1 7 –1
0 –3 –1
0 7 1
0 0 1
–1 –5 –5
3 0 –1
–4 –5
1 0 0
–5 –3 4
6 3
=
–
=
12 18 –4
1 7 –1
11 11
–7 –3 –1
0 0 1
–7 –3
–4
–4
–3
–2
3 0 –1
1 0 3 0
4 4 –2
4 –1 0
–5
–3 4
+ –1 0 0 · 4 0 –3 =
c) B t · A t + C · C t = 5 –1 0 7 ·
1 7 –1
3 0 –2 1
0 –3 –1
–2 0 –1
0 0 1
(
)(
)(
)(
)
)(
)
6 21 –4
36 –4 –10
42 17 –15
= 20 3 –2 + –4 1 0 = 16 4 –2
7 –14 0
–10 0 10
–3 –14 10
(
(
(
)( )
)( ) ( )
)( )( )
)(
1 5 3
3 0 –1
1 5 3
1 0 3 0
0
–1
0
–5
–3
4
0 –1 0
d) B · B t · A t + B =
· 5 –1 0 7 ·
+
=
3 0 –2
1 7 –1
3 0 –2
3 0 –2 1
0 7 1
0 0 1
0 7 1
35 –5 –3 38
3 0 –1
1 5 3
–5 1 0 –7
–5 –3 4
0 –1 0
=
·
+
=
–3 0 13 –2
1 7 –1
3 0 –2
38 –7 –2 50
0 0 1
0 7 1
127
–20
=
4
147
–6
–3
91
7
–14
1 5 3
128
2
0 –1 0
–20
+
=
–12
3 0 –2
7
–14
0 7 1
147
–1
–4
91
14
–11
2
–14
–13