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A scientist solves a nonhomogeneous system of ten linear equations in twelve unknowns and finds that three of the unknowns are free variables. Can the scientist be certain that, if the right sides of the equations are changed, the new nonhomogeneous system will have a solution? Discuss.

Short Answer

Expert verified

No, the new nonhomogeneous system will not have any solution.

Step by step solution

01

Describe the given statement

It is given that a nonhomogeneous system has ten linear equations with twelve unknowns. There are three free variables in the system. It implies that the system has three non-pivot columns.

02

Use the rank theorem

Consider the nonhomogeneous system \(Ax = b\), where \(A\) is \(10 \times 12\) matrix. As the system hasthree non-pivot columns, \({\rm{dimNul}}\,A = 3\), and the value of unknown’s \(n\) is 12 . By the rank theorem, \({\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A = n\).

Put the values as shown:

\(\begin{aligned} {\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A &= n\\{\rm{rank}}\,A &= n - {\rm{dim}}\,{\rm{Nul}}\,\,A\\{\rm{rank}}\,A &= 12 - 3\\{\rm{rank}}\,A &= 9\end{aligned}\)

03

Draw a conclusion

As \({\rm{rank}}\,A\) is 9, \({\rm{col}}\,A\) must be a subspace of \({\mathbb{R}^{10}}\). It means a value of \(b\) exists in \({\mathbb{R}^{10}}\)at which the nonhomogeneous system \(Ax = b\) is inconsistent. Thus, the system \(Ax = b\) may not have a unique solution for all \(b\).

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Most popular questions from this chapter

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.

Determine the dimensions of \({\mathop{\rm Nul}\nolimits} A\) and \({\mathop{\rm Col}\nolimits} A\) for the matrices shown in Exercise 13-18

16. \(A = \left( {\begin{array}{*{20}{c}}3&4\\{ - 6}&{10}\end{array}} \right)\)

In Exercise 4, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)

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.

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