If the characteristic polynomial of a matrix A is p(λ) = λ2+ 1, then A is invertible
P(λ\lambdaλ )=λ\lambdaλ 2+1 ⟹ \implies⟹ ∃\exist∃ A-1
det(A-λ\lambdaλ E)=∣−λ1−1−λ∣\begin{vmatrix} - \lambda& 1 \\ -1 & - \lambda \end{vmatrix}∣∣−λ−11−λ∣∣=P(λ\lambdaλ )
A=(01−10)A=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}A=(0−110)
detA=1 don`t equal 0 ⟹ ∃\implies\exist⟹∃ A-1
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