Unless otherwise specified, assume that all matrices in these exercises are \(n \times n\). Determine which of the matrices in Exercises 1-10 are invertible. Use a few calculations as possible. Justify your answer.

2.\(\left( {\begin{aligned}{*{20}{c}}{ - 4}&6\\6&{ - 9}\end{aligned}} \right)\)

Short Answer

Expert verified

The matrix \(\left( {\begin{aligned}{*{20}{c}}{ - 4}&6\\6&{ - 9}\end{aligned}} \right)\) is not invertible.

Step by step solution

01

State the condition for an invertible matrix

Step 1: State the condition for an invertible matrix

02

Determine whether the matrix is invertible

It is not immediately apparent that the columns of the matrix \(\left( {\begin{aligned}{*{20}{c}}{ - 4}&6\\6&{ - 9}\end{aligned}} \right)\) are multiples. If the determinant is zero, the fastest check is likely to be determinant. The matrix is not invertible according to theorem 4 since its determinant is zero.

Thus, the matrix \(\left( {\begin{aligned}{*{20}{c}}{ - 4}&6\\6&{ - 9}\end{aligned}} \right)\) is not invertible.

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

Show that \({I_n}A = A\) when \(A\) is \(m \times n\) matrix. (Hint: Use the (column) definition of \({I_n}A\).)

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

Use the inverse found in Exercise 1 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)

Let Ube the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of Ulists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

16. a. If A and B are \({\bf{3}} \times {\bf{3}}\) and \(B = \left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}\end{aligned}} \right)\), then \(AB = \left( {A{{\bf{b}}_1} + A{{\bf{b}}_2} + A{{\bf{b}}_3}} \right)\).

b. The second row of ABis the second row of Amultiplied on the right by B.

c. \(\left( {AB} \right)C = \left( {AC} \right)B\)

d. \({\left( {AB} \right)^T} = {A^T}{B^T}\)

e. The transpose of a sum of matrices equals the sum of their transposes.

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