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Chapter 4: Q30E (page 191)

Suppose \(A\) is \(m \times n\)and \(b\) is in \({\mathbb{R}^m}\). What has to be true about the two numbers rank \(\left[ {A\,\,\,{\rm{b}}} \right]\) and \({\rm{rank}}\,A\) in order for the equation \(Ax = b\) to be consistent?

Short Answer

Expert verified

The condition \({\rm{rank}}\,\left[ {A\,\,\,{\rm{b}}} \right] = {\rm{rank}}\,A\) must be true about the two numbers rank \(\left[ {A\,\,\,{\rm{b}}} \right]\) and \({\rm{rank}}\,A\) in order for the equation \(Ax = b\) to be consistent.

Step by step solution

01

Assume an arbitrary system and relate the given statement with it

Consider the nonhomogeneous system \(Ax = b\) with the joint matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\). If \(b\) is not a pivot column of the matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\), then the rank of matrix \(A\) and the joint matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\) are the same, that is, \({\rm{rank}}\,\left[ {A\,\,\,{\rm{b}}} \right] = {\rm{rank}}\,A\).

02

Use the existence and uniqueness theorem

By the existence and uniqueness theorem, a linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot. So, for the system to be consistent, \(b\) must not be a pivot column of the matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\).

03

Draw a conclusion

If \(b\) is not a pivot column of the matrix \(\left[ {A\,\,\,{\rm{b}}} \right]\), then \({\rm{rank}}\,\left[ {A\,\,\,{\rm{b}}} \right] = {\rm{rank}}\,A\). So, the condition \({\rm{rank}}\,\left[ {A\,\,\,{\rm{b}}} \right] = {\rm{rank}}\,A\) is true about the two numbers rank \(\left[ {A\,\,\,{\rm{b}}} \right]\) and \({\rm{rank}}\,A\) in order for the equation \(Ax = b\) to be consistent.

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Most popular questions from this chapter

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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.)
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