Question

Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution? Discuss.

Short answer

No, it is not possible to find any vector in \({\rm{Nul }}A\) which spans \({\rm{Nul }}A\).

Step-by-step solution

Describe the given statement

It is given that a homogeneous system has ten linear equations with twelve unknowns. The twelve unknowns are multiples of one fixed nonzero solution. This implies that the maximum rank of the matrix formed from the homogeneous system is 10 as it has 10 pivot places.

Use the rank theorem

Consider a homogeneous system \(Ax = 0\), where \(A\) is \(10 \times 12\) matrix. The value of \(n\) of the unknown is 12 and \({\rm{rank}}\,A \le 10\). By the rank theorem, \({\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A = n\).

Put the values to get:

\(\begin{aligned} {\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A &= n\\{\rm{dim}}\,{\rm{Nul}}\,\,A &= n - {\rm{rank}}\,A\\{\rm{dim}}\,{\rm{Nul}}\,\,A &\ge 12 - 10\\{\rm{dim}}\,{\rm{Nul}}\,\,A &\ge 2\end{aligned}\)

Draw a conclusion

As the value of \({\rm{dim}}\,{\rm{Nul }}A\) is 2 or greater than 2, the number of non-pivot columns is at least 12. Thus, it is not possible to find any vector in \({\rm{Nul }}A\) which spans \({\rm{Nul }}A\).