Answer:
Let [tex]A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\a_{41}&a_{42}&a_{43}\end{array}\right][/tex] the matrix associated to the system and
[tex]b=\left[\begin{array}{ccc}\alpha_{1}\\\alpha_{2}\\\alpha_{3}\\\alpha_4\end{array}\right][/tex] the vector of constant values of the system.
The equation of the line joining p and q is
[tex]p+t(q-p), t\in\mathbb{R}[/tex]. Since p and q are solutions of the linear system then, [tex]Ap=b[/tex] and [tex]Aq=b[/tex]
Let [tex]w=q-p[/tex]. Observe that [tex]Aw=A(q-p)=Aq-Ap=b-b=0[/tex]. Then w is a solution of the homogeneous system Ax=0.
Now, let s=p+r(q-p) for some [tex]r\in\mathbb{R}[/tex] be a point in the line joining p and q. Observe that
[tex]As=A(p+r(q-p))=Ap+rA(q-p)=Ap+rAw=b+r*0=b[/tex]
Then As=b. This means that s is a solution of the linear system.
