A homogeneous system of twelve linear equations in eight unknowns has two fixed solutions that are not multiples of each other, and all other solutions are linear combinations of these two solutions. Can the set of all solutions be described with fewer than twelve homogeneous linear equations? If so, how many? Discuss.

It is given that a homogeneous system has twelve linear equations with eight unknowns. The system has two fixed solutions that are not multiples of each other, and all other solutions are linear combinations of these two solutions. This implies that the number of non-pivot columns is 2, that is, the value of \({\rm{dim}}\,{\rm{Nul }}A\) is 2.

Consider the homogeneous system \(Ax = 0\), where \(A\) is \(12 \times 8\) matrix. The value of \(n\) of the unknown is 8, and the linear equations (rows) are 12. By the rank theorem, \({\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A = n\).

Put the values as shown below:

\(\begin{aligned} {\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A &= n\\{\rm{rank}}\,A &= n - {\rm{dim}}\,{\rm{Nul}}\,\,A\\{\rm{rank}}\,A &= 8 - 2\\{\rm{rank}}\,A &= 6\end{aligned}\)

As the value of \({\rm{rank }}A\) is 6, the value of \({\rm{dimcol}}\,A\) is also 6. It means the echelon form of matrix \(A\) has six non-zero rows. Thus, the set of all solutions can be described with six homogeneous linear equations.