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If A=[0100]A=\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}, then A1999A^{1999} is equal to

Solution

✅ Correct Option: 3

Direct multiplication gives A2=OA^2=O, the null matrix.

Hence every higher power is null, and A1999=A2⋅A1997=OA^{1999}=A^{2}\cdot A^{1997}=O.

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