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