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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

Let be a linear transformation from a vector space \(V\) \(T:V \to W\)in to vector space \(W\). Prove that the range of T is a subspace of . (Hint: Typical elements of the range have the form \(T\left( {\mathop{\rm x}\nolimits} \right)\) and \(T\left( {\mathop{\rm w}\nolimits} \right)\) for some \({\mathop{\rm x}\nolimits} ,\,{\mathop{\rm w}\nolimits} \)in \(V\).)\(W\)

In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.

\({\left( {{\bf{2}} - t} \right)^{\bf{3}}}\), \({\left( {{\bf{3}} - t} \right)^2}\), \({\bf{1}} + {\bf{6}}t - {\bf{5}}{t^{\bf{2}}} + {t^{\bf{3}}}\)

(calculus required) Define \(T:C\left( {0,1} \right) \to C\left( {0,1} \right)\) as follows: For f in \(C\left( {0,1} \right)\), let \(T\left( t \right)\) be the antiderivative \({\mathop{\rm F}\nolimits} \) of \({\mathop{\rm f}\nolimits} \) such that \({\mathop{\rm F}\nolimits} \left( 0 \right) = 0\). Show that \(T\) is a linear transformation, and describe the kernel of \(T\). (See the notation in Exercise 20 of Section 4.1.)

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)

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