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 $adj(I_n)=I_n$
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