Respuesta :
Answer:
The following are the solution to the given points:
Step-by-step explanation:
for point A:
[tex]\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)[/tex]
[tex]=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)[/tex]
The set A is not part of the subspace [tex]R^3[/tex]
for point B:
[tex]\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)[/tex]
[tex]=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0[/tex]
[tex]\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B[/tex]
The set B is part of the subspace [tex]R^3[/tex]
for point C: [tex]\to C={(x,y,z)|x<y<z}[/tex]
In this, the scalar multiplication can't behold
[tex]\to for (-2,-1,2) \varepsilon C[/tex]
[tex]\to -1(-2,-1,2)= (2,1,-1)[/tex] ∉ C
this inequality is not hold
The set C is not a part of the subspace [tex]R^3[/tex]
for point D:
[tex]\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)[/tex]
The scalar multiplication s is not to hold
[tex]\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)[/tex] ∉ D
this is an inequality, which is not hold
The set D is not part of the subspace [tex]R^3[/tex]
For point E:
[tex]\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\[/tex]
The [tex]x_1, x_2[/tex] is the arbitrary, in which [tex]ax_1+bx_2[/tex]is arbitrary
[tex]\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E[/tex]
The set E is the part of the subspace [tex]R^3[/tex]
For point F:
[tex]\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))[/tex]
The [tex]x_1, x_2[/tex] arbitrary so, they have [tex]ax_1+bx_2[/tex] as the arbitrary [tex]\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F[/tex]
The set F is the subspace of [tex]R^3[/tex]
