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Chapter 2: Q25Q (page 93)

Verify the boxed statement preceding Example 1 i.e., Let A and B be square matrices. If \[AB = I\], then Aand B are both invertible, with \[B = {A^{ - {\bf{1}}}}\] and \[A = {B^{ - {\bf{1}}}}\].

Short Answer

Expert verified

Square matrices A and B are both invertible with \[B = {A^{ - 1}}\] and \[A = {B^{ - 1}}\].

Step by step solution

01

Use the inverse matrix theorem

Given that \[AB = I\]. Then A is invertible based on statements (a) and (k) of the inverse matrix theorem.

02

Use the fact of the invertible matrix

\[{A^{ - 1}}\]exists.

Multiply by \[{A^{ - 1}}\] on both sides of \[AB = I\].

\[\begin{array}{c}{A^{ - 1}}AB = {A^{ - 1}}I\\IB = {A^{ - 1}}\\B = {A^{ - 1}}\end{array}\]

This implies B is invertible. The inverse of B is

\[\begin{array}{c}{B^{ - 1}} = {\left( {{A^{ - 1}}} \right)^{ - 1}}\\ = A.\end{array}\]

03

Draw a conclusion

Hence, A and B are both invertible with\[B = {A^{ - 1}}\]and\[A = {B^{ - 1}}\].

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Most popular questions from this chapter

Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

Use the inverse found in Exercise 3 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{5}}{{\bf{x}}_{\bf{2}}} = - {\bf{9}}\\ - {\bf{7}}{{\bf{x}}_{\bf{1}}} - {\bf{5}}{{\bf{x}}_{\bf{2}}} = {\bf{11}}\end{aligned}\)

Suppose A, B,and Care invertible \(n \times n\) matrices. Show that ABCis also invertible by producing a matrix Dsuch that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation. Explain why T is both one-to-one and onto \({\mathbb{R}^n}\). Use equations (1) and (2). Then give a second explanation using one or more theorems.
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