Is it possible for a nonhomogeneous system of seven equations in six unknowns to have a unique solution for some right-hand side of constants? Is it possible for such a system to have a unique solution for every right-hand side? Explain.

Short Answer

Expert verified

Yes,it possible for a nonhomogeneous system of seven equations in six unknowns to have a unique solution for some right-hand side of constants.

No, it is not possible for such a system to have a unique solution for every right-hand side.

Step by step solution

01

Describe the given statement

It is given that a nonhomogeneous system has seven linear equations with six unknowns. The system has unique solutions for some right-hand side of constants. This implies that the system has at most six pivot positions.

02

Use the rank theorem

Consider the nonhomogeneous system \(Ax = b\), where \(A\) is \(7 \times 6\) matrix. As the system hasat most six pivot positions, \({\rm{rank}}\,A \le 6\), and the value of unknown’s \(n\) is 6 . By the rank theorem, \({\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A = n\).

Put the values as shown:

\(\begin{aligned} {\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A &= n\\{\rm{dim}}\,{\rm{Nul}}\,\,A &= n - {\rm{rank}}\,A\\{\rm{dim}}\,{\rm{Nul}}\,\,A &\ge 6 - 6\\{\rm{dim}}\,{\rm{Nul}}\,\,A &\ge 0\end{aligned}\)

03

Draw a conclusion

If \({\rm{dim}}\,{\rm{Nul}}\,\,A = 0\), the system \(Ax = b\) has no free variable and its solution is unique. The value of \({\rm{dimcol}}\,A\) is also 6. Moreover, \({\rm{col}}\,A\) is a subspace of \({\mathbb{R}^7}\) as \({\rm{rank}}\,A \le 6\). So, a value of \(b\) must exist in \({\mathbb{R}^7}\)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

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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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.)

In Exercise 2, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

2. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{5}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{7}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{5}}}\end{array}} \right)\)

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