StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q13SE

Expert-verifiedFound in: Page 165

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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 \({\left( {{\rm{adj}}\,A} \right)^{ - 1}} = \frac{1}{{\det A}}A\).

Given, *A* is **invertible**. Therefore,

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

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

By **inverse formula**,

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

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

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

94% of StudySmarter users get better grades.

Sign up for free