Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Q15Q (page 93)

Can a square matrix with two identical columns be invertible? Why or why not?

Short Answer

Expert verified

It is not invertible as the identical column vectors are linearly dependent.

Step by step solution

01

State the condition of an invertible matrix

If a matrix contains two identical columns, then the columns are linearly dependent on each other.

02

Consider the matrix

For the matrix \(\left[ {\begin{aligned}{*{20}{c}}{ - 2}&3&3\\1&{ - 1}&{ - 1}\\{ - 1}&{ - 2}&{ - 2}\end{aligned}} \right]\),column 2 and column 3 are identical.

The equation \({x_1}\left[ {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 2}\end{aligned}} \right] + {x_2}\left[ {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 2}\end{aligned}} \right] = 0\) has one solution. Therefore, the vectors are linearly dependent.

So, the square matrix with two identical columns cannot be inverted as the columns are linearly dependent.

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.

Start your free trial

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

Show that if the columns of Bare linearly dependent, then so are the columns of AB.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

8. \[\left[ {\begin{array}{*{20}{c}}A&B\\{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}X&Y&Z\\{\bf{0}}&{\bf{0}}&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&I\end{array}} \right]\]

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

If a matrix \(A\) is \({\bf{5}} \times {\bf{3}}\) and the product \(AB\)is \({\bf{5}} \times {\bf{7}}\), what is the size of \(B\)?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation
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