Answer
True
Work Step by Step
For every positive integer $n$, we have $(I_n)^{-1}=I_n \rightarrow det (I_n)=1$.
According to Theorem 3.3.18, we also have $(I_n)^{-1}=\frac{1}{det(I_n)}adj(I_n)$.
Hence, it is true to say that $adj(I_n)=I_n$
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 3 - Determinants - 3.3 Cofactor Expansions - True-False Review - Page 232: j
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