Suggested languages for you:

Americas

Europe

Q22E

Expert-verifiedFound in: Page 191

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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**.

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

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.

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:

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

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 \({\rm{Nul }}A\).

94% of StudySmarter users get better grades.

Sign up for free