¡Descarga Mateemáticas Empresariales II: Capítulo 1. Ecuaciones Lineales. Matrices - Prof. Pérez y más Apuntes en PDF de Matemáticas solo en Docsity!
MATEM¡TICAS EMPRESARIALES II
HOMEWORK ASSIGNMENT 1
CHAPTER 1. SYSTEMS OF LINEAR EQUATIONS. MATRICES.
- Solve the following systems by Gaussian elimination. Work directly with the equations and check your results by using the table.
(a)
x + 2y = 3 3 x + 4y = 5 (b)
2 x y z = 4 x + y 2 z = 2 (c)
x + 2y 3 z = 0 2 x + 4y 2 z = 2 3 x + 6y 5 z = 3
(d)
4 x + y = 2 x + 2y = 5 6 x + 3y = 12
(e)
x + 2y 3 z + 4u = 2 2 x + 5y 2 z + u = 1 5 x + 12y 7 z + 9u = 7
(f)
x + y 2 t = 1 x + 2y z = 1 y + z 2 t = 0 2 x 2 z + 8t = m
- Solve graphically
x y = 2 x + y = 3 2 x y = 0
- Obtain AB and BA for:
(a) A =
A ; B =
A (^) (b) A =
; B =
A
(c) A =
BB
CC
A ; B^ =^
(d) A =
; B =
A
- Obtain the n-th powers of the following matrices: (a)
a 1 0 a
; (b)
A :
- Find the determinants of : a)
; b)
A (^) ; c)
a 0 b c
; d)
a b c 0 d e 0 0 f
A :
- Compute
by making zeros in the Örst column.
- Find
by Gaussian reduction.
- Find the inverses, in case they exist, of the following matrices:
(a)
2 a 1 1
(b)
a 0 0 b
(c)
a b c d
(d)
cos sin sin cos
(e)
A (^) (f)
A (^) (g)
A (^) (h)
A
Explain in each case for which values of the variables (if there are any) is the matrix invertible.
- Find the inverses of the following matrices by Gaussís method:
(a)
A (^) (b)
A (^) (c)
1 b c d
- Find the inverse of
A (^) (if it exists) by writing A ^1 =
a b c d e f g h i
A (^) and solving the
appropriate 9 9 system of equations.
- Given A =
A
1 Önd the third column of its inverse matrix. Check your result without
Önding the full inverse matrix A ^1. (This is the only interesting part of this exercise. The answer is (2; 1 ; 1)T^ ).
- Find the ranks of the following matrices:
(a)
(b)
A
- Find the rank of A =
BB
CC
A by Örst removing redundant lines and then applying
Gaussian reduction.
- Find the ranks of the following matrices by Gaussian elimination (changing the order of the rows 1 and 2 in (b)):
(a)
BB
CC
A ;^ (b)
BB
CC
A (c)
BB
CC
A
- Solve the following systems: (i) by Cramerís rule; (ii) by the inverse matrix method.
(a)
x + y = 2 y + z = 4 x + z = 0
(b)
x y 4 z = 2 x + 2z = 1 2 x y z = 1
SELECTED ANSWERS
- (a)
x + 2y = 3 3 x + 4y = 5 =)
x + 2y = 3 2 y = 4 =)
x = 1 y = 2 1 2 3 3 4 5
(b)
x = 2 + z y = z
(c)
x + 2y 3 z = 0 2 x + 4y 2 z = 2 3 x + 6y 5 z = 3
x + 2y 3 z = 0 4 z = 2 4 z = 3
x + 2y 3 z = 0 4 z = 2 0 = 1 0 @
A !
A !
A
(d)
x = 1 y = 2
A
(f)
x + y 2 t = 1 x + 2y z = 1 y + z 2 t = 0 2 x 2 z + 8t = m
x + y 2 t = 1 y z + 2t = 0 y + z 2 t = 0 2 y 2 z + 4t = m + 2
x + y 2 t = 1 y z + 2t = 0 2 y 2 z + 4t = m + 2
x + y 2 t = 1 y z + 2t = 0 0 = m + 2
m 6 = 2 : Incompatible m = 2 :
x = 1 z + 4t y = z 2 t
BB
2 0 2 8 m
CC
A!
BB
0 2 2 4 m + 2
CC
A!
BB
0 0 0 0 m + 2
CC
A
If m + 2 = 0 :
- a)
A (^) and
A (^) ; b)
and
A :
c)
BB
CC
A and^72 :
- b)
A
n = 3n ^1
A :
- a) 17 ; b) 121.
- b)
1 =a 0 0 1 =b
if a 6 = 0 and b 6 = 0; (c)
ad bc
d b c a
; (d)
cos sin sin cos
g)
A (^) ; (h)
A
1
^2 + 1
^2 + 1 0 0
A
- (a)
A !
A
(b)
A !
A
(c)
1 b 1 0 c d 0 1
1 b 1 0 0 d bc c 1
1 b 1 0 0 1 c d bc
d bc
A !
B
1 0 b c d bc
b d bc 0 1
c d bc
d bc
C
A =
B
d d bc
b d bc 0 1
c d bc
d bc
C
A :
- (a) rank = 2 if 2 6 = 2 ; rank = 1 if 2 = 2 :
(b)
A !
A =)
r = 3 if 6 = 2 and 6 = 0 r = 2 if = 2 or = 0 (but not both) r = 1 if = 2 and = 0
- R 1 = R 2 + R 3 ; C 4 = C 1 + C 2 : Removing R 1 and C 4 ;
rank
BB
CC
A = rank
A (^) : Since
= 0 but
= 0; the
rank is 2 :
If 6 = 1; determinate, solution: fx = 1; y = 3; z = 0g :
If = 1; indeterminate, solution: fx = 1 z; y = 3 3 zg or fx = 1 t; y = 3 3 t; z = tg
By Cramer: =
1 6 = 0 =) Determinate 1 = 0 =) Donít know yet
a) 6 = 1 =) x =
= 1; y =
z = x =
b) = 1 :
x y = 2 + z y = 3 3 z
x = 1 2 z y = 3 3 z z = arbitrary
general solution.
- a) Determinate if a 6 = 1; solution fx 1 = a; x 2 = 0g :
Compatible indeterminate if a = 1 : solution x 1 = 1 x 2 :
b) Incompatible if a 6 = 3: Compatible indeterminate if a = 3: Solution fx = 1 y zg.
c)
x 1 + 2x 2 x 3 + x 4 = 1 2 x 1 3 x 2 + x 3 x 4 = 1 x 1 + x 2 + ax 3 + 2x 4 = 4
x 1 + 2x 2 + x 4 x 3 = 1 2 x 1 3 x 2 x 4 + x 3 = 1 x 1 + x 2 + 2x 4 + ax 3 = 4
Since
= 2 6 = 0 the system
x 1 + 2x 2 + x 4 = 1 + x 3 2 x 1 3 x 2 x 4 = 1 x 3 x 1 + x 2 + 2x 4 = 4 ax 3
is Cramer.
x 1 =
1 + x 3 2 1 1 x 3 3 1 4 ax 3 1 2 1 2 1 2 3 1 1 1 2
= 2 (1 + a) x 3 ; x 2 =
1 1 + x 3 1 2 1 x 3 1 1 4 ax 3 2 1 2 1 2 3 1 1 1 2
a
x 3 ;
x 4 =
1 2 1 + x 3 2 3 1 x 3 1 1 4 ax 3 1 2 1 2 3 1 1 1 2
ax 3 :
