Ecuaciones Resueltas

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MATERIA : MAT - 1130 NOMBRE: SUYO COPA ANEL PARALELO :”B”

1) Discutir : x1 + ( k + 1) x 2 + x3 = 0 x1 + x 2 + ( k + 1) x3 = 0

( k + 1) x1 + x2 + x3 = 0 solución: Hallando el determinante: 1 k +1 1 A= 1 1 k +1 = 0 k +1 1 1 1[1 − (k + 1)] − ( k + 1)[1 − (k + 1) 2 ] + 1[1 − (k + 1)] = 0 1[1 − k − 1] − ( k + 1)[1 − k 2 − 2k − 1] + (1 − k − 1) = 0 − k − ( k + 1)[ −k 2 − 2k ] − k = 0

− 2k − ( − k 3 − 2k 2 − k 2 − 2k ) = 0 k 3 + 3k 2 = 0 k 2 ( k + 3) = 0

Reemplazando k=0

k =0 k = −3 x1 + x 2 + x3 = 0

x1 + x 2 + x3 = 0

x2 = a ∀a, b ∈ R x3 = b  x1 = − a − b

x1 + x 2 + x3 = 0 x1 + x 2 + x3 = 0 Para k=0 ∃ infinitas Soluciones Reemplazando k=-3 x1 − 2 x 2 + x3 = 0

x1 − 2 x2 + x3 = 0

x1 + x 2 − 2 x3 = 0

3 x1 − 3 x3 = 0

− 2 x1 + x 2 + x3 = 0

− 3x1 + 3 x3 = 0

x1 − 2 x 2 + x3 = 0 3 x1 − 3 x3 = 0

Con k=-3 ∃ infinitas soluciones Con k≠0 , k≠-3 ∃ solución única. 2) Discutir : x1 + kx 2 + 2 x3 = 4 2 x1 + 2 x 2 + 4kx3 = 2 Solución: x1 + kx 2 + 2 x3 = 4 (2 − 2k ) x 2 + (4k − 4) x3 = −6 Reemplazando: 2(1 − k ) x 2 − 4(1 − k ) x3 = −6 (1 − k )[2 x 2 − 4 x3 = −6 1− k = 0 → k = 1 Reemplazando K=1 x1 + kx 2 + 2 x3 = 4 2 x1 + 2 x 2 + 4kx3 = 2 x1 + x 2 + 2 x3 = 4 0 = −6 Con K=1 No Existe solución Con K≠1 ∃ infinitas soluciones 3 )Si S ={(x,y,z) ∈ R 3 /x=y+z} T ={(x,y,z) ∈ R 3 /3x-3y=z} Hallar la Dim(T+S) Solución: Dim(T+S)=Dim(T)+Dim(S)-Dim(T ∩ S) Hallando la Dim(T) 3x − 3 y = z z=b ; y=a ∀a, b ∈ ℜ b 3 x − 3a = b → x = a + 3

b 1 (x,y,z)= (a+ , a, b )=a(1,1,0)+b( ,0,1) 3 3 1 Base(T)={(1,1,0),( ,0,1)} 3 Dim(T)= 2 Para S

x=y+z y=a ; z=b ∀a, b ∈ ℜ x=a+b (x,y,z)=(a+b,a,b)=a(1,1,0)+b(1,0,1) Base (S)={(1,1,0)(1,0,1)} Dim(S)= 2 Para Dim(T ∩ S) T : 3x − 3 y = z S:x= y+z

3x − 3 y − z = 0 x− y−z =0

x− y−z =0 2z = 0 → z = 0 y=a

; x-a-0=0 → x=a

(x,y,z)=(a,a,0)=a(1,1,0) Base(S ∩ T)= )={(1,1,0)} Dim(S ∩ T)=1 Dim (T+S)=2+2-1 =3 4) : ℜ 3 → ℜ 2 una T.L. Si N(T)={(0,1,-1)(2,1,3)} Hallar T(x,y,z)=? Solución: Expresando como una C.L. α 1 (0,1,−1) + α 2 ( 2,1,3) = ( x, y, z ) 2α 2 = x

α1 + α 2 = y − α 1 + 3α 2 = z

→ α2 = x / 2 → α1 + x / 2 = y → α1 = y − x / 2

x x − ( y − ) + 3( ) = z 2 2 x 3 −y+ + x= z 2 2 2x − y − z = 0 ∴ T ( x, y, z ) = (2 x − y − z ,0,0) 5) Calcular: A=

a 2 2ab b 2 b2 a 2 2ab 2ab b 2 a2

Solución: a 2 2ab b 2 A = b2 a 2 2ab 2ab b 2 a2 a 2 [a 4 − 2ab 3 ] − 2ab[ a 2 b 2 − 4a 2 b 2 ] + b 2 [b 4 − 2a 3b] a 6 − 2a 3b 3 − 2a 3 b 3 − 8a 3b 3 + b 6 − 2a 3b 3 a 6 + 2a 3 b 3 + b 6 Factorizando : (a 3 + b 3 ) 2 Reemplazando: A= (a 3 + b 3 ) 2 = a 3 + b 3 6) Hallar “a” y ”b” B={(1,1,0)(0,3,2)} Sea Base Si S ={(x,y,z) ∈ R 3 /x+ay-bz=0} Solución: Encontrando el Sub Espacio Generado de B α 1 (1,1,0) + α 2 (0,3,2) = ( x, y , z )

α1 = x α 1 + 3α 2 = y 2α 2 = z Reemplazando:

→ α1 = x → α2 = z / 2

X+3/2(z)=y

;

x-y-3/2z=0

B ={(x,y,z) ∈ R 3 / x-y-3/2z=0} S ={(x,y,z) ∈ R 3 /x+ay-bz=0} el Sub Espacio Generado por B debe ser el mismo que “S” 3 z=0 2 x + ay − bz = 0 x− y−

Comparando coeficientes a = −1 b=−

3 2

7) Hallar la Base y Dim [( S1 ∩ S 2 )+( S 3 + S 4 )] S1 ={(x,y,z) ∈ R 3 / 2y-3z= -4x} S 2 ={(1,2,0)(0,3,1)} S 3 ={(x,y,z) ∈ R 3 / x+5z=4y} S 4 ={(3,0,1)(-3,1,0)} Solución: Para S1 ∩ S 2 Para S 2

α 1 (1,2,0) + α 2 (0,3,1) = ( x, y, z ) α1 = x 2α 1 + 3α 2 = y α2 = z

→ 2 x + 3z = y → 2 x − y + 3 z = 0

S 2 : 2 x − y + 3z = 0 S1 : 4 x + 2 y − 3 z = 0 4 x + 2 y − 3z = 0 9 − 2y − z = 0 2

z=a

9 9 − 2y − a = 0 → y = a 2 4 9 4 x + 2( ) − 3a = 0 4 3 3 4x = − a → x = − a 2 8  3 9  (x,y,z)=  − a, a, a   8 4   3 9  = a − , ,1  8 4   3 9  Base ( S1 ∩ S 2 )={  − , ,1 }  8 4  Para( S 3 + S 4 ) Para S 3 x+5z=4y y=a ; z=b ; x+5b=4ª x=4ª-5b (x,y,z)=(4a -5b,a,b) ={a(4,1,0)+b(-5,0,1)} Base S 3 ={(4,1,0)(-5,0,1} S3 + S 4

 4  − 5  3  − 3 

1 0 0 1

4 1 0  0 5   4 1  3 =  1 0 − 4  7 0   0 4 

0  4   1 0   = 1 0     0 0  

1 5 4 0 0

0   1   8  5  28  −  20 

Base ( S 3 + S 4 ) ={(4,1,0)(0,5/4,1)(0,0,8/5)} Para ( S1 ∩ S 2 + S 3 + S 4 )

  3 9   3 9 1  1  −  − 1   8 4    8 4 32  32     0 25 4 25 0     3  3 = =  0   8 5   0 0 −  0   0 15  4 8   0 −   8   8  0 0 5  0   0  5   5   Base { ( S1 ∩ S 2 )+( S 3 + S 4 )}={(-3/8,9/4,1)(0,25,32/3)(0,0,-8/15)} Dim { ( S1 ∩ S 2 )+( S 3 + S 4 ) =3  3 −  8  4  0    0 

9 4 1 5 4

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