Matemáticas

MATRICES INVERSAS PROBLEMA 1
A=

2 4 3

4 6 5 6 1 -2

2 4 3

4 6 2 4 5 6 4 5 1 -2 3 1

A= -20+72+24-90-12+32 = 6 5 1 6 -2

A11=

(5 · -2)- (1 · 6) = -10-6 = 16

A12= -

4 3

6 -2- [(4 · -2)- (3 · 6)] = 8 + 18 = 26

A13=

4 3

5 1

(4 · 1)- (3 · 5)= 4 - 15 = -11

A21= -

4 1 2 3 2 3

6 -2 6 -2 4 1

- [(4 · -2)- (1 · 6)] = 8 + 6 = 14

A22=

(2 · -2)- (3· 6) = -4 - 18 = -22

A23 = -

- [(2 · 1)- (3 · 4)] = -2 + 12 = 10

A31=

4 5

6 6

(4 · 6)- (5 · 6) = 24 - 30 = -6

A32 = -

2 4

6 6

- [(2 · 6)- (4 · 6)] = -12 + 24 = 12

A-1=

-16 26 -11

14 -6 -22 12 10 -6

2 4 3

4 6 5 6 1 -2
26 (2) -22 (4) +12 (3) = 0 26 (4) -22 (5) +12 (1) = 6 26 (6) -22 (6) +12 (-2)= 0

-16 (2) + 14 (4) -6 (3) = 6 -16 (4) + 14 (5) -6 (1)= 0 -16 (6) + 14 (6) -6 (-2)= 0

-11 (2) + 10 (4) -6 (3) = 0 -11 (4) + 10 (5) -6 (1) = 0 -11 (6) + 10 (6) -6 (-2)= 6

A-1 =

6 0 0

0 0 6 0 0 6

A-1 =

1 0 0

0 0 1 0 0 1

PROBLEMA 2A=

3 1 1

1 0 -1 2 1 1

3 1 1

1 0 3 1 -1 2 1 -1 1 1 1 1

A= -3 + 2 - 6 - 1 = -8

A11 =

-1 1 1 1

2 1 2 1

(-1 · 1)- (1 · 2) = -1-2 = -3

A12 = -

- [(1 · 1)- (1 · 2)] = -1 + 2= 1

A13 =

1 1

2 1

(-1 · 1)- (1 · 2) = -1-2 = -3

A21= -

1 1

0 1

- [(1 · 1)- (1 · 0)] = -1 + 0 = -1

A22=

3 1 3 1

0 1 1 1

(3 · 1)- (1 · 0) = 3 - 0 = 3

A23= -

-[(3 · 1)- (1 · 1)] = -3 + 1 = -2

A31=

1 -1

0 2

(1 · 2)- (-1 · 0) = 2 - 0 = 2

A32= -

3 1

0 2

- [(3 · 2)- (1 · 0)] = -6 - 0 = -6

A33=

3 1

1 -1

(3 · -1)- (1 · 1) = -3- 1 = -4

A =

-1

-3 1 2

-1 2 3 -6 -2 -4

3 1 1

1 0 -1 2 1 1

-3(3) -1 (1) + 2(1) = -8 -3(1) - 1(-1) + 2(1) = 0 -3(0) - 1(2) + 2(1)= 0

1(3) + 3(1) - 6(1) = 0 1(1) + 3(-1) - 6(1) =-8 1(0) + 3(2) - 6 (1)= 0

2(3) - 2(1) - 4(1) = 0 2(1) - 2(-1) - 4(1) = 0 2(0) - 2(2) - 4(1)= -8

A-1 =

-8 0 0

0 0 -8 0 0 -8

A =

-1

-1 0 0

0 0 -1 0 0 -1

PROBLEMA 3
A=...