Express every solution of the given system as the sum of a specific solution plus a solution of the associated homogeneous system:

2x_1+x_2-x_3-x_4=-1

3x_1+x_2+x_3-2x_4=-2

-x_1-x_2+2x_3+x_4=2

-2x_1-x_2+2x_4=3

Solution: This question essentially asks you to use Theorem 3.4. Theorem 3.4 states that to solve the linear system of equations;

A\mathbf{X}=\mathbf{b}      (*)

it is sufficient to find some/ any (among all the solutions – if one exists) solution \mathbf{X}_1, find the solution to the homogeneous system, \mathbf{X}_0:

A\mathbf{X}=\mathbf{0},

and that the general solution to (*) will be \mathbf{X}_1+\mathbf{X_0}.

Realistically you wouldn’t use this method to solve this problem (Q.4 (ii)) – we are more seeing how this theorem works as ye will be using it later in solving linear differential equations.

Namely if

F(y(x),y'(x),y''(x),...,y^{(n)}(x))=f(x)

is a suitably linear differential equation, then the general solution may be found by finding any particular solution, y_p(x) (by trial and error as much as anything) and adding it to homogeneous solution, y_H(x) (which we know how to find).

Now back to Q. 4 (ii). I don’t know any fool-proof strategy for looking for a particular solution but in this example you might notice that if we add all the equations together, then the second and forth columns all cancel and we are left with:

2x_1+2x_3=2

\Rightarrow x_1+x_3=1

Now a reasonable guess to make is that x_1=0 for some solution giving us, in this case, that x_3=1. Throwing this into the third equation:

-x_2+2+x_4=2

\Rightarrow x_2=x_4

Throwing this into equation 4, using all these assumptions:

-x_2+2x_2=3

\Rightarrow x_2=3=x_4

Hence (0,3,1,3) is a potential particular solution. Substituting into the four equations shows that it is indeed a solution.

Now to solve the homogeneous system. Writing in augmented matrix form (with the constant terms all zero):

\left(\begin{array}{cccc}2&1&-1&-1\\ 3&1&1&-2\\ -1&-1&2&1\\ -2&-1&0&2\end{array}\right)

Apply elementary row operations to get:

\left(\begin{array}{cccc}1&0&0&1\\ 0&1&0&-4\\ 0&0&1&-1\\ 0&0&0&0\end{array}\right)

Now reading off the solutions (note that #parameters=#variables – rank – so here one parameter. We have the choice because no row is saying that x_i must be something in particular).

Let x_4=t. Hence x_3-x_4=0\Rightarrow x_3=t. Also x_2-4x_4=0\Rightarrow x_2=4t. Finally x_1+x_4=0\Rightarrow x_1=-t.

Hence using Theorem 3.4, the solution is the sum of the particular and homogeneous solutions:

(-t,3+4t,1+t,3+t).

 

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