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Prove the part of the theorem which lets w be any solution of Ax = b, and defines vh = w - p. Show that v_h is a solution of Ax = 0. This shows that every solution of Ax = b has the form w = p + v_h, with p a particular solution of Ax = b and v_h a solution of Ax = 0. Let w and p be solutions of Ax = b. Substitute for v_h from the equation w = p+ v_h.

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Answer:

It is proved that [tex]V_h[/tex] is a solution of Ax=b.

Step-by-step explanation:

As given let w and p be a solution of Ax=b. Then,

[tex]Aw=0[/tex] and [tex]Ap=0[/tex]

Defining,

[tex]V_h=w-p[/tex]

To show [tex]V_h[/tex] is a solution of Ax=b, we have to show [tex]A(V_h)= 0[/tex]. Then,

[tex]A(V_h)[/tex]  

[tex]=A(w-p)[/tex]        (By putting value of [tex]V_h[/tex])

[tex]=Aw-Ap[/tex]       (By distributive law)

[tex]=0[/tex]                     (Since w and p are solution of Ax=b)

Hence [tex]V_h[/tex] become a solution of the homogenous system Ax=b.

Since w and p are solution of Ax=b, and any linear combination of the solutions again a solution of Ax=b, this shows that every solution of Ax=b is of the form [tex]w=p+V_h[/tex]  where [tex]V_h[/tex] is a solution of Ax=0.

The proofing of the theorem illustrates that Vh = solution of Ax = 0 and p = particular solution of Ax = b.

Proving the theorem.

Let w be the solution of Ax = b.

  • Aw = b ........ equation i

p is a particular solution of Ax = b.

  • Ap = b ....... equation ii

Subtract equation ii from i

Aw - Ap = 0

A(w - p) = 0

(w - p) is a solution of Ax = 0

Let vh = w - p

This is a homogeneous equation Ax = 0. A

Any solution w of system Ax = b can be written into form Vh = w - p; w = Vh + p.

where,

Vh = solution of Ax = 0.

p = particular solution of Ax = b.

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