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Linear Algebra and its Applications
Found in: Page 165
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Question: 13. Show that if A is invertible, then adj A is invertible, and \({\left( {adj\,A} \right)^{ - {\bf{1}}}} = \frac{{\bf{1}}}{{detA}}A\).

Hence, adj A is invertible and \({\left( {{\rm{adj}}\,A} \right)^{ - 1}} = \frac{1}{{\det A}}A\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Use the definition of invertible

Given, A is invertible. Therefore,

\(A{A^{ - 1}} = {A^{ - 1}}A = I\).

Also, \({A^{ - 1}}\) is invertible and \({\left( {{A^{ - 1}}} \right)^{ - 1}} = A\).

Step 2: Use the inverse formula

By inverse formula,

\({A^{ - 1}} = \frac{1}{{\det A}}{\rm{adj}}\,A\)

This implies \({\rm{adj}}\,A\) is also invertible.

Step 3:  Perform the substitution

\(\begin{array}{c}{\left( {{A^{ - 1}}} \right)^{ - 1}} = A\\{\left( {\frac{1}{{\det A}}{\rm{adj}}\,A} \right)^{ - 1}} = A\\\det A{\left( {{\rm{adj}}\,A} \right)^{ - 1}} = A\\{\left( {{\rm{adj}}\,A} \right)^{ - 1}} = \frac{1}{{\det A}}A\end{array}\)

Most popular questions for Math Textbooks

Question: In Exercise 12, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.

12. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{3}}\\{ - {\bf{2}}}&{\bf{2}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}\end{array}} \right)\)

Find the determinants in Exercises 5-10 by row reduction to echelon form.

\(\left| {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{3}}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{5}}&{\bf{4}}\\{ - {\bf{1}}}&{\bf{0}}&{\bf{5}}&{\bf{3}}\\{\bf{3}}&{ - {\bf{3}}}&{ - {\bf{2}}}&{\bf{3}}\end{array}} \right|\)

Question: 12. Use the concept of area of a parallelogram to write a statement about a \(2 \times 2\) matrix A that is true if and only if A is invertible.

In Exercise 33-36, verify that \(\det EA = \left( {\det E} \right)\left( {\det A} \right)\)where E is the elementary matrix shown and \(A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\).

35. \(\left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]\)

Question: Use Cramer’s rule to compute the solutions of the systems in Exercises1-6.

2. \(\begin{array}{l}4{x_1} + {x_2} = 6\\3{x_1} + 2{x_2} = 7\end{array}\)
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