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Chapter 4: Q21E (page 191)

Suppose a nonhomogeneous system of nine linear equations in ten unknowns has a solution for all possible constants on the right sides of the equations. Is it possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other? Discuss.

Short Answer

Expert verified

No, it is not possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other.

Step by step solution

01

Describe the given statement

It is given that a nonhomogeneous system has nine linear equations with ten unknowns. A solution exists for all the possible constants on the right side of the equations. This implies that the maximum rank of matrix formed from the nonhomogeneous system is 9 as it has 9 pivot places.

02

Use the rank theorem

The value of \(n\) of the unknown is 10, and the rank is 9. By the rank theorem,\({\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A = n\).

Put the values as shown below:

\(\begin{aligned} {\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A &= n\\9 + {\rm{dim}}\,{\rm{Nul}}\,\,A &= 10\\{\rm{dim}}\,{\rm{Nul}}\,\,A &= 10 - 9\\{\rm{dim}}\,{\rm{Nul}}\,\,A &= 1\end{aligned}\)

03

Draw a conclusion

As the value of \({\rm{dim}}\,{\rm{Nul }}A\) is 1, the number of non-pivot columns is 1. Thus, it is not possible to find two nonzero solutions of the associated homogeneous system.

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Most popular questions from this chapter

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

Suppose a nonhomogeneous system of six linear equations in eight unknowns has a solution, with two free variables. Is it possible to change some constants on the equations’ right sides to make the new system inconsistent? Explain.

In statistical theory, a common requirement is that a matrix be of full rank. That is, the rank should be as large as possible. Explain why an m n matrix with more rows than columns has full rank if and only if its columns are linearly independent.

Let H be an n-dimensional subspace of an n-dimensional vector space V. Explain why \(H = V\).

In Exercise 3, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

3. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\\{\bf{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{2}}\\{ - {\bf{2}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{0}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{0}}\\{ - {\bf{1}}}\end{array}} \right)\)
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