Suggested languages for you:

Americas

Europe

Q22E

Expert-verified
Found in: Page 191

Linear Algebra and its Applications

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 $${\rm{Nul }}A$$.

See the step by step solution

Step 1: 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.

Step 2: 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}

Step 3: 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 $${\rm{Nul }}A$$.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.