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GO Classes DPP | GATE CS, DA | Linear Algebra | Determinant

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If $\lambda$ is an eigenvalue of $A$, then $\lambda^{100}$ is an eigenvalue of $A^{100}$.

So the eigenvalues of $A^{100}$ are :

$(-1)^{100}=1$, $1^{100}=1$, and $0^{100}=0$.

Hence the eigenvalues of $A^{100}+I$ are :

$1+1=2$, $1+1=2$, and $0+1=1$.

Therefore,

$|A^{100}+I|=(2)(2)(1)=4$
 

Answer : $\boxed{|A^{100}+I|=4}$

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