/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q23Q Suppose \(CA = {I_n}\)(the \(n \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose \(CA = {I_n}\)(the \(n \times n\) identity matrix). Show that the equation \(Ax = 0\) has only the trivial solution. Explain why Acannot have more columns than rows.

Short Answer

Expert verified

Amust have at least as many rows as columns because each pivot is in a different row.

Step by step solution

01

The equation \(Ax = 0\) has the trivial solution

When \(x\) satisfies the equation \(Ax = 0\), then \(CA{\mathop{\rm x}\nolimits} = C \times 0 = 0\). So, \({I_n}{\mathop{\rm x}\nolimits} = 0\) and \({\mathop{\rm x}\nolimits} = 0\). This demonstrates that there are no free variables in the equation \(Ax = 0\).

02

Explanation for A cannot have more columns than rows

Therefore, every variable is a basic variable since the equation \(Ax = 0\) has no free variables. Each column of Ais a pivot column. Thus, Amust have at least as many rows as columns because each pivot is in a different row.

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 91Ó°ÊÓ!

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

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.

3. \[\left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\I&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}W&X\\Y&Z\end{array}} \right]\]

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

37. Construct a random \({\bf{4}} \times {\bf{4}}\) matrix Aand test whether \(\left( {A + I} \right)\left( {A - I} \right) = {A^2} - I\). The best way to do this is to compute \(\left( {A + I} \right)\left( {A - I} \right) - \left( {{A^2} - I} \right)\) and verify that this difference is the zero matrix. Do this for three random matrices. Then test \(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^{\bf{2}}}\) the same way for three pairs of random \({\bf{4}} \times {\bf{4}}\) matrices. Report your conclusions.

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).

[M] Suppose memory or size restrictions prevent your matrix program from working with matrices having more than 32 rows and 32 columns, and suppose some project involves \(50 \times 50\) matrices A and B. Describe the commands or operations of your program that accomplish the following tasks.

a. Compute \(A + B\)

b. Compute \(AB\)

c. Solve \(Ax = b\) for some vector b in \({\mathbb{R}^{50}}\), assuming that \(A\) can be partitioned into a \(2 \times 2\) block matrix \(\left[ {{A_{ij}}} \right]\), with \({A_{11}}\) an invertible \(20 \times 20\) matrix, \({A_{22}}\) an invertible \(30 \times 30\) matrix, and \({A_{12}}\) a zero matrix. [Hint: Describe appropriate smaller systems to solve, without using any matrix inverse.]

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.