If \( A=\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix} \), then \( A^{1999} \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix} \)
The square of A is already the zero matrix:
\( \displaystyle A^{2}=\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix} 0\cdot 0+1\cdot 0 & 0\cdot 1+1\cdot 0\\ 0 & 0\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix} \)
Since \( A^{2}=O \), every higher power is also O:
\( A^{1999}=A^{2}\cdot A^{1997} \)
\( =O\cdot A^{1997}=O \)