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Found in: Page 165

### 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

# 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 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

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