Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

[M] Repeat Exercise 37 for three random-integer-valued \(5 \times 7\) matrices \(A\) whose ranks are 5, 4, and 3. Make a conjecture about how \(CR\) is related to \(A\) for any matrix \(A\). Prove your conjecture.

Short Answer

Expert verified

The weights of \({r_j}\) are needed to construct the \({j^{th}}\) column of \(A\) from the columns of \(C\) and \(C{{\mathop{\rm r}\nolimits} _j} = {{\mathop{\rm a}\nolimits} _j}\).

Step by step solution

01

Write the matrix as in Exercise 37

Consider matrix\(A\)as\(A = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&{ - 4}&5&3&{ - 3}&{ - 7}\\{ - 4}&6&7&{ - 2}&{ - 6}&{ - 5}&5\\5&{ - 7}&{ - 6}&5&{ - 6}&2&8\\{ - 3}&5&8&{ - 1}&{ - 7}&{ - 4}&8\\6&{ - 8}&{ - 5}&4&4&9&3\end{array}} \right]\).

Matrix \(C\) is \(C = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&5&{ - 3}\\{ - 4}&6&{ - 2}&{ - 5}\\5&{ - 7}&5&2\\{ - 3}&5&{ - 1}&{ - 4}\\6&{ - 8}&4&9\end{array}} \right]\).

Matrix \(R\) is \(R = \left[ {\begin{array}{*{20}{c}}1&0&{\frac{{13}}{2}}&0&5&0&{ - 3}\\0&1&{\frac{{11}}{2}}&0&{\frac{1}{2}}&0&2\\0&0&0&1&{\frac{{ - 11}}{2}}&0&7\\0&0&0&0&0&1&1\end{array}} \right]\).

02

Make a conjecture about how \(CR\) is related to \(A\)

When \(A\) is nonzero, \(A = CR\). \(CR = \left[ {\begin{array}{*{20}{c}}{C{{\mathop{\rm r}\nolimits} _1}}&{C{{\mathop{\rm r}\nolimits} _2}}& \cdots &{C{{\mathop{\rm r}\nolimits} _n}}\end{array}} \right]\), where \({{\mathop{\rm r}\nolimits} _1},...,{{\mathop{\rm r}\nolimits} _n}\) represent the columns of \(R\). The columns of \(R\) may or may not have the pivot columns of \(R\). Let the first column of \(R\), \({e_i}\) be the \({i^{th}}\) pivot column of \(R\); the \({i^{th}}\) column in the identity matrix. Thus, \(C{e_i}\) is the \({i^{th}}\) pivot column of \(A\). Multiply \(C\) with the pivot column of \(R\) and the result is the corresponding pivot column of \(A\) in its pivot position because both \(A\) and \(R\) have the pivot columns in the same location.

Consider \({r_j}\) as a non-pivot column of \(R\). So, \({r_j}\) contains the weights needed to construct the \({j^{th}}\) column of \(A\) from the pivot columns of \(A\) as explained in examples 9 and 10 of section 4.3. Therefore, the weights of \({r_j}\) are used to construct the \({j^{th}}\) column of \(A\) from the columns of \(C\) and \(C{{\mathop{\rm r}\nolimits} _j} = {{\mathop{\rm a}\nolimits} _j}\).

Thus, \({r_j}\) contains the weights needed to construct the \({j^{th}}\) column of \(A\) from the columns of \(C\) and \(C{{\mathop{\rm r}\nolimits} _j} = {{\mathop{\rm a}\nolimits} _j}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

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 \(V\).)\(W\)

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.

Define by \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 1 \right)}\end{array}} \right)\). For instance, if \({\mathop{\rm p}\nolimits} \left( t \right) = 3 + 5t + 7{t^2}\), then \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}3\\{15}\end{array}} \right)\).

  1. Show that \(T\) is a linear transformation. (Hint: For arbitrary polynomials p, q in \({{\mathop{\rm P}\nolimits} _2}\), compute \(T\left( {{\mathop{\rm p}\nolimits} + {\mathop{\rm q}\nolimits} } \right)\) and \(T\left( {c{\mathop{\rm p}\nolimits} } \right)\).)
  2. Find a polynomial p in \({{\mathop{\rm P}\nolimits} _2}\) that spans the kernel of \(T\), and describe the range of \(T\).

Use Exercise 28 to explain why the equation\(Ax = b\)has a solution for all\({\rm{b}}\)in\({\mathbb{R}^m}\)if and only if the equation\({A^T}x = 0\)has only the trivial solution.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free