Ans: (c) Yes
-
$\begin{array}{rcl}
A^2 +A+I & = & 0 \\
(A-I)(A^2+A+I) & = & (A-I) 0 \\
A^3+A^2+A-A^2-A-I & = & 0 \\
A^3 & = & I
\end{array}$ - By the result of (a), we have
$\begin{array}{rcl}
A^3 & = & I \\
A(A^2) & = & I \\
(A^2) A & = & I
\end{array}$Therefore, $A^{-1}=A^2$.
Hence $A$ is non-singular.
- By the result of (a) and (b), we have $A^3=I$ and $A^{-1}=A^2$. Hence, we have
$\begin{array}{cl}
& \left(A^{1000}+(A^{-1})^{2000}\right)^{-1} \\
= & \left(A^{1000}+(A^2)^{2000}\right)^{-1} \\
= & (A^{1000}+A^{4000})^{-1} \\
= & (A^{999} A +A^{3999}A)^{-1} \\
= & [(A^3)^{333}A+(A^3)^{1333}A]^{-1} \\
= & (I^{333}A+I^{1333}A)^{-1} \\
= & (2A)^{-1} \\
= & \dfrac{1}{2} A^{-1} \\
= & \dfrac{1}{2} A^2 \\
= & \dfrac{1}{2}(-A-I) \text{ , } \because A^2+A+I=0\\
= & \dfrac{-1}{2}I-\dfrac{1}{2}A
\end{array}$Take $\alpha=\beta=\dfrac{-1}{2}$, which are real numbers.
Therefore, the claim is correct.
