ALGEBRA LINEAL

JACOB CASTILLO. C.I: 20920329

Usando algoritmo de Gauss-Jordan determina si la siguiente matriz es invertible. En caso de serlo calcule su inversa.

1. A= [−14

1 6

−5 −83

7

1 3 29]

Planteamos lo siguiente:

(−14

1 6

−5 −83

7

1 3 29|1 0 00 1 00 0 1)

f 3❑↔f 1: (

1 3 29

−5 −83

7

−14

1 6|0 0 10 1 01 0 0)

2da f : 5 f 1−f 2:

(1 3 2

9

0 533

−539

−14

1 6 |0 0 10 −1 51 0 0)

f 3: 14 . f 1 + f 3

(1 3 2

9

0 533

−539

0 74

10918

|0 0 10 −1 5

1 0 14 )

f 2: 3

53f 2:

(1 3 2

9

0 1 −13

0 74

10918

|0 0 1

0 −353

1553

1 0 14 )

f 3: 74f 2−f 3 :

(1 3 2

9

0 1 −13

0 0 −23936

| 0 0 1

0 −353

1553

−1 −21212

1353 )

f 1: −3 f 2+ f 1:

(1 0 11

9

0 1 −13

0 0 −23936

| 0 953

853

0 −353

1553

−1 −21212

1353

)f 3:

−36239

f 3 :

(1 0 119

0 1 −13

0 0 1 | 0 953

853

0 −353

1553

36239

18912667

−46812667

)f 2:

−13f 3+ f 2 :

(1 0 119

0 1 00 0 1 | 0 9

538

5312

239−34662671351

181737671351

36239

18912667

−46812667

)f 1:

−119f 3+f 1:

(1 0 00 1 00 0 1|

−44239

3052802014053

118486860421559

12239

−34662671351

181737671351

36239

18912667

−46812667

)La matriz inversa es:

A−1 =(−44239

3052802014053

118486860421559

12239

−34662671351

181737671351

36239

18912667

−46812667

)

II. Determinar la traza de la siguiente matriz:

3)

Tr(C) = -6 +0+ (-4) = -10.

III. Determine los valores de λ para los cuales existe la inversa de la matriz

5.)

|E| = |−3 −1 20 λ 3

−43

2 1|= |λ 32 1|= λ.1-2.3 = λ-6 ⇒ λ=6

Si λ≠ 6 E es invertible.

V. Encontrar la matriz A sabiendo que (4. A )−1 = [ 2 3−4 −4]

Aplicando la propiedad la inversa del producto de dos matrices, tenemos.

A−1. 4−1 = [ 2 3−4 −4]

Luego, A−1. 14 =[ 2 3

−4 −4]A−1= 4[ 2 3

−4 −4]A−1 =[ 8 12

−16 −16 ]Entonces A = [−16 −12

16 8 ]